This programme aims to offer learners with holistic information in baby developments, domesticate their professionalism and sensitivity in playgroup works and grasp the instructing and studying skills of infants aged 0-three. They help youngsters to develop their language, numeracy, motor and social skills by way of activities based on play, music and artwork that present the foundation for a kid's inventive improvement. The group additionally makes positive that the factors for certification to be eligible for funding from the Ministry of Schooling is met. This playgroup will fill a gap in services by supporting all children within the zero-5 age group and their parents/carers throughout the Borroloola Township and Town Camps," Minister Scullion stated.
Attending playgroup at a college permits parents and children to turn into familiar with the varsity prior to transition to formal schooling. Our playgroup sessions operate out of Moorleigh Community Village (next door to the kinder) on Monday, Tuesday, Wednesday and Friday mornings from 9:30am – eleven:30am. John Holt (1923-1985) was a trainer and controversial training writer who advocated college reform and in the end became one of many forces behind the homeschooling movement of the 70's and eighty's. We work closely in partnership with all major faculties in our native space to make sure we're guiding the children's education to movement into the systems they've in place as they transition into more formal schooling. Faculty or playgroup in your native area or one which your own youngsters have attended.
The framework covers the education and care of all kids in early years provision together with kids with particular academic wants and disabilities. If your little one was born between 2 July 2014 and 1 July 2015, you may apply for his or her pre-college schooling place beginning in September 2018. He wrote ten books about schooling and based the magazine 'Rising With out Schooling'. Playgroups are held at Clinton County RESA Education Center, 4179 S. US 27, St. Johns ( See map ). For extra information call 989-224-6831 Ext.
Every Wednesday for 5 weeks, mums and dads, grandparents and guardians of pre-schoolers with older siblings on the college are being invited to deliver their youngest kids along to participate in a playgroup facilitated by Deakin University's Faculty of Training.
The Steiner philosophy on kindergarten training goals to make sure each little one is bodily, emotionally, and spiritually nurtured. As a co-operative, we every share the accountability of holding the circle and contributing to the ongoing working of the playgroup in a variety of ways. To endeavour to create a vibrant and stimulating setting both indoors & outdoors, which promotes spontaneous and exploratory play utilizing a variety of actions and resources. It may not be potential to safe a spot at your first desire setting if more children apply to a setting than the locations out there. They're a spot where kids can play and develop while their dad and mom have an opportunity to fulfill other adults. Our education gives time for the youngsters's growth to take place in a wholesome method.


The most popular tourist attraction in Cairns is the Skyrail rainforest cableway, which travels nearly five miles (7.5km) from the Caravonica Terminal, the gondolas gliding above the canopy of the rainforest over magnificent waterfalls and lush gorges, to the village of Kuranda. En route passengers can alight temporarily at two stations to explore the boardwalks that have been constructed in the heart of the forest. Interpretative centres have been set up so visitors can learn about the ecology, fauna and flora of the fascinating ancient rainforest. The return journey takes about two and a half hours, and the gondolas seat up to six people. Booking in advance is advisable.


Who knows what lies down the road for hospital marketing? The convergence of factors such as the Affordable Care Act, the connected society, empowered patients, and many other dynamics could lead to radical changes in the ways that hospitals operate and market themselves. Increased competition for patients may lead to some very difficult choices.
Who knows what lies down the road for hospital marketing? The convergence of factors such as the Affordable Care Act, the connected society, empowered patients, and many other dynamics could lead to radical changes in the ways that hospitals operate and market themselves. Increased competition for patients may lead to some very difficult choices. The hospitals that come out ahead will be those which have laid a solid foundation for a strategic marketing plan.
Search Engine Optimization (SEO) strategy: Every segment of the population is now using the Internet to gather information on potential health care options. Hospitals that are not online are missing out on the opportunity to connect with, and engage, these potential patients. The fastest way to initiate patient interest is through high search engine rankings. These are achieved through consistent SEO strategies and dynamic website content.
E-mail communications plan: An increasing number of physicians are turning to digital tools to increase their productivity. Secure email messaging can be used to reduce patient visit time and answer basic medical questions. Targeted emails can be a remarkably effective way of sending health reminders, providing education on health issues, and maintaining patient contact.
A well-curated Facebook page: The Mayo Clinic has a Center for Social Media which exists to "improve health globally by accelerating effective application of social media tools throughout Mayo Clinic and spurring broader and deeper engagement in social media by hospitals, medical professionals and patients." The Mayo Clinic believes so strongly in social media that it has over 60,000 followers on Twitter and 20,000 connections on Facebook.
Health apps: Apple just generated a lot of publicity at its developers' conference with the introduction of a health-monitoring app for iOS 8, "Health." It logs user health statistics and then aggregates information into categories such as Diagnostics, Fitness, Lab Results, Medications, Nutrition, Sleep, and Vitals. Apple also introduced a HealthKit tool for developers, which can be incorporated into health and fitness apps to allow them to access the health data stored within the Health app. Users will be able to choose exactly what information is shared and can, potentially, choose to share data from a health-related app with a doctor.
Marketers might have to think "outside the box" to develop hospital marketing plans that can keep up with the rapid pace of change in the way patients expect to communicate with the medical community. They must research and participate in what works effectively now, but will also need to be aware of what is coming down the road in digital marketing options so they can be prepared to take advantage of them before their competition does.
Are You Ready for 2015 Readmission Penalties?


Westinghouse Electric Company has been selected by Eskom to provide three reloads of fuel for the Koeberg Nuclear Power Plant beginning in late 2011 and ending in 2015. The contract is worth $30 million.
Under terms of the contract executed with Westinghouse Electric Sweden AB, Westinghouse will produce fuel at its fabrication facility in Sweden.
The contract with Eskom is the most recent in a series of accomplishments for Westinghouse Sweden. Earlier this year, the company had received orders to provide fuel for five Swedish reactors beginning in 2011.


text:
Kim Clijsters
Numbers speak in Wozniacki's favor for the Australian Open quarterfinal against Clijsters
Dear readers, our now already famous stats expert Omair couldn't resist the urge and found time in his busy schedule to analyse the upcoming Australian Open quarterfinal between Kim Clijsters and Caroline Wozniacki. Numbers speak in Wozniacki's favor, but is everything in numbers? Let's see if we can conclude something.
There are so many things you plan on doing, but you you never get to complete them. That is what happened with me, when I had planned on making a probability analysis for the Australian Open 2012, it never got complete and then I got heck of busy. Being an audit trainee no doubt has its perks of excessive late sittings and no weekends, but I just could not resist my urge to analyse the Australian Open quarterfinals, using the performance of the quarterfinalists so far in the tournament. I will be using the stats to look into the quarterfinal showing, as to who has a better shot at making the semis. Let's first analyse Kim Clijsters and Caroline Wozniacki.
Kim got her ankle twisted in her fourth round epic battle against Li Na. Wozniacki won't get a better chance than this to avenge her previous losses to the former world No. 1, after all it was Clijsters who denied Wozniacki the taste of her first major when she beat the Dane in the US Open final in 2009.
Wozniacki has yet to lose a set in this tournament, while Clijsters has already lost one to Li, and was on the brink of defeat in straight sets but turned it around to emerge as a winner. Thousand dollar questions are: Will Clijsters' ankle hold to let her complete in the last Australian Open of her career? Will Clijsters have enough in tank to battle past Wozniacki who has no doubt the best defensive skills at the moment?
Clijsters is 2-0 against Wozniacki, but their last meeting at the year-end Championships in 2010 went the distance. Let us have a look at what the numbers say when we see each player's first four matches at the Aussie Open 2012:
Wozniacki has +8% advantage in the first serves, but what is important is that although Wozniacki has higher 1st serve percentage, Clijsters' 1st serve winning percentage is much higher than Wozniacki's, but so are her double faults. Clijsters has been serving 4+ double faults on average in each match so far, which means losing her serve at least once in each match by only double faults.
What is more worrying is Clijsters' ratio of winners and unforced errors. If you look at the number of winners, Wozniacki is not far behind, but if you look at the number of unforced errors Wozniacki comes out as a cleaner one by a huge margin. Clijsters has a differential of negative 24, whereas Wozniacki has a positive differential of 4. Break point conversion rate, receiving points won percentage again go in Wozniacki's favour. The department where Wozniacki lags is the points won at net. While Clijsters has lost only one point out of her 18 net trips (95%), Wozniacki managed to win only 68% of her net trips.
Numbers are in Wozniacki's favour and Clijsters' twisted ankle boosts Wozniacki's chances a lot, however, Clijsters does have a few things on her side as well:
Clijsters has the ability of lifting her game when really needed, and we have experienced that in the past as well.
Her head-to-head against Wozniacki is positive.
Of nine times Kim has played at the Australian Open, she has always won her quarterfinal match when she has made it to the quarterfinals. She made it to the semifinals or better six out of six times she reached the quarters.
My heart says Clijsters will win, as I am a Clijsters fan, but numbers say Wozniacki is the favorite. This makes me remeber Shakira's (if I am not wrong) song, "Hips Don't Lie", "Numbers Don't Lie".. I will say "sometimes, numbers do lie", that is why my heart says Clijsters will book a semifinal spot. What do you see? Who do u think will come out on top?
For a taste of Omair's previous analyses visit the following pages:
Part 1: Comparison of the five slamless WTA No.1s and the origin of the phenomenon
Part 2: Hingis and Davenport were just as slamless No.1s as Jankovic, Safina and Wozniacki
Comparative analysis of Top 10 players and their performance at big tournaments in 2011
Part 1: Top 10 players in 2011 – how they scored against fellow big opponents
Part 2: Top 10 players in 2011 – what do their losses suggest?
(photos: © Neal Trousdale)
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Aryna Sabalenka takes her Australian Open trophy on a cruise in Melbourne
Australian Open: Aryna Sabalenka beats Elena Rybakina, wins maiden Grand Slam title
Gregoire Gentil January 23, 2012 At 8:43 pm
Predicting a tennis match winner with past numbers is like trying to predict the next stock exchange hit with the past history! Tough. Some people think it can be done, others don't. Note that I didn't say that it's like predicting the next lotto numbers which would have been the real insult to this interesting work.
James January 23, 2012 At 11:21 pm
Although I partly agree with the comment above that it is difficult to predict a match based on numbers, it is a really well written article.
The thing that most surprised me was how Wozniacki has only hit five less winners than Clijsters throughout the championships! Wozniacki has definitely shown more intent to go after the ball at this Grand Slam, but in the second set against JJ she did go back into her defensive shell when things got tight, which she cannot afford to do against Clijsters.
I would have gone for Clijsters in straight sets, but with the uncertainty of her ankle injury i'm actually going for Wozniacki to win this in 3 sets. It should be an interesting match and i'm disappointed its in the middle of the night here in the UK 🙁
Omiar January 24, 2012 At 3:46 am
James and Gregoire Gentil
I agree with you in that it always is a new day and anything can happen on a given day. Why I did this was because coming into this match they are bringing a certain form in to this match, what does that form suggest??? I only intended to do this…

redpajama_set_name: RedPajamaCommonCrawl


1. Put Staff Development First…All The Time.
One reality that many business owners have to constantly remind themselves of is that their people are their greatest asset. If you're not consistently taking the time to help these individuals develop new skills, attain higher levels of self-esteem, or grow in some other significant way, you're stalling business growth as well as your personal development. Note that helping other people grow will empower you to fine-tune key capacities like emotional intelligence and listening skills. The more fine-tuned these skills become, the more effective you'll be when it comes to interfacing with people in an authentic yet professional way within the business setting. These are just a few of the reasons why you should put staff development first at all times if you want to become a better boss and see your company thrive.
One strategy you can implement to help your staff members grow is teaching them how to prioritize their personal and professional lives in ways that optimize productivity. There are many methodologies that can be used to optimize an individual's ability to prioritize, and one of them is the immersion/maintenance/minimal principle. This principle holds that there are certain activities that an individual should remain fully immersed in for the purpose of optimizing productivity. Other activities fall within the maintenance category and should be engaged in regularly to ensure things like mental health or work/life balance. Finally, there are some activities that should be kept to a minimum so that the employee can continue to grow personally and professionally.
2. Utilize Strategies That Help You Overcome Rejection.
Whether your spouse chooses to end the marriage, a long-standing client decides to end their business relationship with your company, or one of your employees opts not to take a promotion and leaves your organization entirely, you're going to face rejection throughout your career as a business leader. In recognizing that rejection is inevitable, it's important to begin implementing strategies that help you understand and overcome it. One thing to keep in mind is that your experience of rejection is simultaneously proof that you possess the bravery and self-knowledge necessary to try new things for the purpose of personal and professional growth. This understanding can help you avoid or overcome the fear that might prevent you from stepping outside of the box in the future.
Another strategy that you could use to become a better boss is continually updating your machines. Taking this step will empower you by ensuring that your employees can operate various equipment and devices easily and effectively. The end result will almost always be optimized productivity. In the event that your company makes use of the extrusion line , the professionals of Bergandi Machinery can empower you to locate the ideal equipment.
Three strategies that you can utilize to become a better boss are outlined above. Start using these strategies immediately so your organization can become increasingly innovative, exemplary, and excellent this year!


text:
\section{Introduction}

Lattices in rank 1 real (semi)simple Lie groups are still far from understood. A key notion is that of \emph{arithmetic lattice} which we will not define properly here but note that by a famous result of Margulis a lattice in such a Lie group is arithmetic if and only if it has infinite index in its commensurator.

Margulis' celebrated arithmeticity theorem states that every lattice of a simple real Lie group $G$ is arithmetic whenever the real rank of $G$ is at least two. Thus non-arithmetic lattices can only exist in real rank one, that is when the associated symmetric space is a hyperbolic space. In real hyperbolic space, where the Lie group is ${\rm PO}(n,1)$, Gromov and Piatetski-Shapiro  produced in \cite{GPS} a construction yielding non-arithmetic lattices in ${\rm PO}(n,1)$ for all $n \geqslant 2$ (see below for more details), in fact producing in each dimension infinitely many non-commensurable lattices, both cocompact and non-cocompact.
In quaternionic hyperbolic spaces (and the Cayley hyperbolic plane), work of Corlette and Gromov-Schoen implies as in the higher rank case that all lattices are arithmetic.

The case of complex hyperbolic spaces, where the associated Lie group is ${\rm PU}(n,1)$, is much less understood. Non-arithmetic lattices in ${\rm PU}(2,1)$ were first constructed by Mostow in 1980 in \cite{M1}, and subsequently by Deligne-Mostow and Mostow as monodromy groups of certain hypergeometric functions in \cite{DM} and \cite{M2}, following pioneering work of Picard. More recently, Deraux, Parker and the first author constructed new families of non-arithmetic lattices in ${\rm PU}(2,1)$ by considering groups generated by certain triples of complex reflections (see \cite{DPP1}, \cite{DPP2}).
Taken together, these constructions yield 22 commensurability classes of non-arithmetic lattices in ${\rm PU}(2,1)$, and only 2 commensurability classes in ${\rm PU}(3,1)$. The latter two are noncocompact; one is a Deligne-Mostow lattice and the other was constructed by Couwenberg-Heckman-Looijenga in 2005 and recently found to be non-arithmetic by Deraux, \cite{D}.
Major open questions in this area remain the existence of non-arithmetic lattices in ${\rm PU}(n,1)$ for $n \geqslant 4$, as well as the number (or finiteness thereof) of commensurability classes in each dimension.

The Gromov--Piatetski-Shapiro construction, which they call interbreeding of 2 arithmetic lattices (now often referred to as hybridization), produces a lattice $\Gamma < {\rm PO}(n,1)$ from 2 lattices $\Gamma_1$ and $\Gamma_2$ in  ${\rm PO}(n,1)$ which have a common sublattice $\Gamma_{12} < {\rm PO}(n-1,1)$. Geometrically, this provides two hyperbolic $n$-manifolds $V_1=\Gamma_1 \backslash H_\mathbb{R}^n$ and $V_2=\Gamma_2 \backslash H_\mathbb{R}^n$ with a hyperbolic $(n-1)$-manifold $V_{12}$ which is isometrically embedded in $V_1$ and $V_2$ as a totally geodesic hypersurface. This allows one to produce the hybrid manifold $V$ by gluing $V_1-V_{12}$ and $V_2-V_{12}$ along $V_{12}$ (more precisely, in case $V_{12}$ separates $V_1$ and $V_2$, by gluing $V_1^+ -V_{12}$ and $V_2^+ -V_{12}$ along $V_{12}$, with $V_i^+$ a connected component of $V_i - V_{12}$). The resulting manifold is also hyperbolic because the gluing took place along a totally geodesic hypersurface, and its fundamental group $\Gamma$ is therefore a lattice in ${\rm PO}(n,1)$. The main point is then that if $\Gamma_1$ and $\Gamma_2$ are both arithmetic but non-commensurable, their hybrid $\Gamma$ is non-arithmetic. Note that the resulting hybrid $\Gamma$ is algebraically an amalgamated free product of $\Gamma_1$ and $\Gamma_2$ over $\Gamma_{12}$ (say, in the case where $V_{12}$ separates both $V_1$ and $V_2$), and in all cases is generated by its sublattices $\Gamma_1$ and $\Gamma_2$.

It is not straightforward to adapt this construction to construct lattices in ${\rm PU}(n,1)$, the main difficulty being that there do not exist in complex hyperbolic space any totally geodesic real hypersurfaces. In fact, it has been a famous open question since the work of Gromov--Piatetski-Shapiro to find some analogous construction in ${\rm PU}(n,1)$. Hunt proposed the following construction (see \cite{Pau} and references therein). Start with 2 arithmetic lattices $\Gamma_1$ and $\Gamma_2$ in ${\rm PU}(n,1)$, and suppose that one can embed them in ${\rm PU}(n+1,1)$ in such a way that (a) each stabilizes a totally geodesic $H_\mathbb{C}^n \subset H_\mathbb{C}^{n+1}$ (b) these 2 complex hypersurfaces are orthogonal, and (c) the intersection of the embedded $\Gamma_i$ is a lattice in the corresponding ${\rm PU}(n-1,1)$. The resulting hybrid $\Gamma=H(\Gamma_1,\Gamma_2)$ is then defined as the subgroup of ${\rm PU}(n+1,1)$ generated by the images of $\Gamma_1$ and $\Gamma_2$. (See the end of Section~\ref{chg} for a more detailed and concrete description when $n=2$).

It is not clear when, if ever, such a group has any nice properties. One expects in general the hybrid group to be non-discrete, and in fact the first author showed in \cite{Pau} that this happens infinitely often among hybrids in ${\rm PU}(2,1)$ of pairs of Fuchsian triangle subgroups of ${\rm PU}(1,1)$. It was observed there that one can easily arrange for the hybrid to be discrete by arranging for the two subgroups $\Gamma_1,\Gamma_2$ to already belong to a known lattice. But even in the simplest case of arithmetic cusped lattices (where the matrix entries are all in $\mathcal{O}_d$, the ring of integers of $\mathbb{Q}[i\sqrt{d}]$ for some squarefree $d \geqslant 1$), it was not known whether the discrete hybrid $\Gamma$ could ever be a sublattice of the corresponding Picard modular group $\Gamma(d)={\rm PU}(2,1,\mathcal{O}_d)$, as opposed to an infinite-index (discrete) subgroup of $\Gamma(d)$. Following Sarnak (\cite{S}) we will call \emph{thin subgroup} of a lattice $\Gamma$ any infinite-index subgroup having the same Zariski-closure as $\Gamma$.

In this note we show that in fact both behaviors can occur, even among this simplest class of hybrids of sublattices of the Picard modular groups $\Gamma(d)$. More precisely, we consider for $d=3,1,7$ the hybrid subgroup $H(d)$ defined as the hybrid of two copies of ${\rm SU}(1,1,\mathcal{O}_d)$ inside the Picard modular group ${\rm PU}(2,1,\mathcal{O}_d)$.
These specific values of $d$ are those for which a presentation of ${\rm PU}(2,1,\mathcal{O}_d)$ is known (by \cite{FP}, \cite{FFP} and \cite{MP}). Our main results can be summarized as follows (combining Theorems~\ref{d=3main}, \ref{d=1main}, and \ref{d=7main} and Propositions~\ref{limitsets} and \ref{geominf}).

\begin{thm}\label{main}
\begin{enumerate}
\item The hybrid $H(3)$ is a thin subgroup of the Eisenstein-Picard lattice ${\rm PU}(2,1,\mathcal{O}_3)$. It has full limit set $\partial_\infty {\rm H^2_\mathbb{C}} \simeq S^3$ and is therefore geometrically infinite.
\item The hybrid $H(1)$ has index 2 in the Gauss-Picard lattice ${\rm PU}(2,1,\mathcal{O}_1)$.
\item The hybrid $H(7)$ is the full Picard lattice ${\rm PU}(2,1,\mathcal{O}_7)$.
\end{enumerate}
\end{thm}

{\bf Remarks:}

\smallskip

(a) We also give analogous results for two related hybrids $H'(3)$ and $H'(1)$ in Corollaries~\ref{d=3other} and \ref{d=1other}. In terms of Fuchsian triangle groups these groups are defined as the hybrids of two copies of the (orientation-preserving) triangle groups $(2,6,\infty)$ and $(2,4,\infty)$ respectively, as opposed to  $(3,\infty,\infty) \simeq {\rm PU}(1,1,\mathcal{O}_3)$ and $(2,\infty,\infty) \simeq {\rm PU}(1,1,\mathcal{O}_1)$ (so, replacing the elliptic generator by one of its square roots).
An interesting feature of $H'(3)$ is that it has infinite index in its normal closure in $\Gamma(3)$, whereas all other hybrids we consider are normal in $\Gamma(d)$.

(b) In all cases we also show that the hybrid $\Gamma$ is not an amalgamated free product of $\Gamma_1$ and $\Gamma_2$ over their intersection. In case $\Gamma$ is itself a lattice this follows from general considerations of cohomological dimension, and for $H(3)$ and $H'(3)$ we show this by finding sufficiently many relations among the generators for $\Gamma$, see Corollary~\ref{notamalgam}.

(c) One of the main geometric difficulties in analyzing these groups is understanding the parabolic subgroups. By construction the generators contain a pair of (opposite) parabolic isometries (as well as an elliptic isometry when $d=3$, two elliptic isometries when $d=1$, and two elliptic and two loxodromic isometries when $d=7$), however it seems hard in general to determine the rank of the parabolic subgroups of the hybrid. In the cases where the hybrid is a lattice we obtain indirectly that the parabolic subgroups must have full rank, but in the thin subgroup case we do not know what this rank is.

(d) The parabolic isometries appearing in the generators for our hybrids are by construction vertical Heisenberg translations, since they preserve a complex line (see Section~\ref{chg}). It turns out that Falbel (\cite{F}) and Falbel-Wang (\cite{FW}) studied a group formally similar to our hybrid $H(3)$, obtained by completely different methods, namely by finding all irreducible representations of the figure-eight knot group $\Gamma_8$ into ${\rm PU}(2,1)$ with unipotent boundary holonomy. Falbel showed in \cite{F} that there are exactly 3 such representations, one of which has image contained in  $\Gamma(3)={\rm PU}(2,1,\mathcal{O}_3)$ and the two others in  $\Gamma(7)={\rm PU}(2,1,\mathcal{O}_7)$. These are all generated by a pair of opposite horizontal Heisenberg translations. The image of the former representation is shown in \cite{F} and \cite{FW} to be, like our hybrids $H(3)$ and $H'(3)$, a thin subgroup of $\Gamma(3)$ with full limit set, whereas the images of the latter two representations are shown in \cite{DF} to have non-empty domain of discontinuity (and hence have infinite index in $\Gamma(7)$). We were inspired by some of the arguments of \cite{F} and \cite{FW}.

(e) Kapovich found in \cite{K1} the first examples of infinite-index normal subgroups of lattices in ${\rm PU}(2,1)$, among a family of four lattices first constructed by Livn\'e in his thesis (and predating the term \emph{thin subgroup}). Parker showed in \cite{Par0} (sections 6 and 7) that this description could be extended to the Eisenstein-Picard modular group ${\rm PU}(2,1,\mathcal{O}_3)$, and that  some of Kapovich's results extended to that case as well. An anonymous referee showed us by a sequence of clever computations that our hybrid $H(3)$ is in fact commensurable to the infinite-index normal subgroup of ${\rm PU}(2,1,\mathcal{O}_3)$ obtained in this way. It would be interesting to understand this relation geometrically.

(f) Discrete groups generated by opposite parabolic subgroups have been studied in higher rank by Oh, Benoist-Oh and others. A conjecture of Margulis states that if $G$ is a semisimple real algebraic group of rank at least 2 and $\Gamma$ a discrete Zariski-dense subgroup containing irreducible lattices in two opposite horospherical subgroups, then $\Gamma$ is an arithmetic lattice in $G$. Oh showed in \cite{O} that this holds when $G$ is a simple split real Lie group except ${\rm SL}(3,\mathbb{R})$,  Benoist-Oh extended this in \cite{BO} to the case of $G={\rm SL}(3,\mathbb{R})$, and very recently Benoist-Miquel treated the general case in \cite{BM}.

\smallskip

The paper is organized as follows. In section 2 we review basic facts about complex hyperbolic space, its isometries, subspaces and boundary at infinity. In Sections 3,4,5 we consider each of the hybrids $H(3)$, $H(1)$ and $H(7)$ respectively. In section 6 we review and apply basic facts about limit sets and geometrical finiteness to the non-lattice hybrid $H(3)$. We would like to thank Elisha Falbel for pointing out a simplification of the proof of Theorem~\ref{d=3main}, and an anonymous referee for several useful comments.

\section{Complex hyperbolic space, isometries and boundary at infinity}\label{chg}

We give a brief summary of basic definitions and facts about complex hyperbolic geometry, and refer the reader to \cite{G}, \cite{CG} or \cite{Par2} for more details.

\smallskip

{\bf Projective models of ${\rm H}_\mathbb{C}^n$:}

\smallskip

Denote $\mathbb{C}^{n,1}$ the vector space $\mathbb{C}^{n+1}$ endowed with a Hermitian form $\langle \cdot \, , \cdot \rangle$ of signature $(n,1)$. Define $V^-=\left\lbrace Z \in \mathbb{C}^{n,1} | \langle Z , Z \rangle <0 \right\rbrace$ and $V^0=\left\lbrace Z \in \mathbb{C}^{n,1} | \langle Z , Z \rangle =0 \right\rbrace$.
Let $\pi: \mathbb{C}^{n+1}-\{0\} \longrightarrow \mathbb{C}{\rm P}^n$ denote projectivization.
One may then define complex hyperbolic $n$-space ${\rm H}_\mathbb{C}^n$ as $\pi(V^-) \subset \mathbb{C}{\rm P}^n$, with the distance $d$ (corresponding to the Bergman metric) given by:

\begin{equation}\label{dist}
  \cosh ^2 \frac{1}{2}d(\pi(X),\pi(Y)) = \frac{|\langle X, Y \rangle|^2}{\langle X, X \rangle  \langle Y, Y \rangle}
\end{equation}

The boundary at infinity $\partial {\rm H}_\mathbb{C}^n$ is then naturally identified with $\pi(V_0)$.
 Different Hermitian forms of signature $(n,1)$ give rise to different models of ${\rm H}_\C^n$. Two of the most common choices are the Hermitian forms corresponding to the Hermitian matrices $H_1={\rm Diag}(1,...,1,-1)$ and:

\begin{equation}\label{H2matrix}
H_2=
\begin{bmatrix}
0 & 0 & 1 \\
0 & I_{n-1}& 0 \\
1 & 0 & 0
\end{bmatrix}
\end{equation}

 In the first case, $\pi(V^-) \subset \mathbb{C} P^n$ is the unit ball of $\mathbb{C}^n$, seen in the affine chart $\{ z_{n+1}=1\}$ of $\mathbb{C} P^n$, hence the model is called the \textit{ball model} of ${\rm H}_\C^n$.
In the second case,  we obtain the \emph{Siegel model} of ${\rm H}_\C^n$, which is analogous to the upper-half space model of ${\rm H}_\mathbb{R}^n$ and is likewise well-adapted to parabolic isometries fixing a specific boundary point. We will mostly use the Siegel model in this paper and will give a bit more details about it below. We will use the following \emph{Cayley transform} $J$ to pass from the ball model to the Siegel model (see \cite{Par2}); a key point for us is that $J \in {\rm GL}(3,\mathbb{Z})$, hence conjugating by $J$ preserves integrality of matrix entries.

\begin{equation}\label{cayley}
J = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & -1 \\ 1 & 1 & -1 \end{pmatrix}
\end{equation}

{\bf Isometries:}

\smallskip

It is clear from \eqref{dist} that ${\rm PU}(n,1)$ acts by isometries on ${\rm H}_\mathbb{C}^n$, denoting ${\rm U}(n,1)$ the subgroup of ${\rm GL}(n+1,\mathbb{C})$ preserving the Hermitian form, and ${\rm PU}(n,1)$ its image in ${\rm PGL}(n+1,\mathbb{C})$. It turns out that PU($n$,1) is the group of holomorphic isometries of ${\rm H}_\mathbb{C}^n$, and the full group of isometries is ${\rm PU}(n,1) \ltimes \mathbb{Z}/2$, where the $\mathbb{Z}/2$ factor corresponds to a real reflection (see below). A holomorphic isometry of ${\rm H}_\C^n$ is of one of the following three types:
\begin{itemize}
\item \emph{elliptic} if it has a fixed point in ${\rm H}_\mathbb{C}^n$
\item \emph{parabolic} if it has (no fixed point in ${\rm H}_\mathbb{C}^n$ and) exactly one fixed point in $\partial{\rm H}_\mathbb{C}^n$
\item \emph{loxodromic}: if it has (no fixed point in ${\rm H}_\mathbb{C}^n$ and) exactly two fixed points in $\partial{\rm H}_\mathbb{C}^n$
\end{itemize}

{\bf Totally geodesic subspaces:}

\smallskip

 A \emph{complex k-plane} is a projective $k$-dimensional subspace of $\mathbb{C} P^n$ intersecting $\pi(V^-)$  non-trivially (so, it is an isometrically embedded copy of ${\rm H}_\mathbb{C}^{k} \subset {\rm H}_\mathbb{C}^n$). Complex 1-planes are usually called \emph{complex lines}. If $L=\pi(\tilde{L})$ is a complex $(n-1)$-plane, any $v \in \mathbb{C}^{n+1}-\{ 0\}$ orthogonal to $\tilde{L}$ is called a \emph{polar vector} for $L$.

 A \emph{real k-plane} is the projective image of a totally real $(k+1)$-subspace $W$ of $\mathbb{C}^{n,1}$, i. e. a $(k+1)$-dimensional real linear subspace such that $\langle v,w \rangle \in \mathbb{R}$ for all $v,w \in W$.  We will usually call real 2-planes simply real planes, or $\mathbb{R}$-planes. Every real $n$-plane in ${\rm H}_\mathbb{C}^n$ is the fixed-point set of an antiholomorphic isometry of order 2 called a \emph{real reflection} or $\mathbb{R}$-reflection. The prototype of such an isometry is the map given in affine coordinates by $(z_1,...,z_n) \mapsto (\overline{z_1},...,\overline{z_n})$; this is an isometry provided that the Hermitian form has real coefficients.

We will need to distinguish between the following types of parabolic isometries. A parabolic isometry is called \emph{unipotent} if it has a unipotent lift to ${\rm U}(n,1)$. In dimensions $n>1$, unipotent isometries are either {\it 2-step} (also called \emph{vertical}) or {\it 3-step} (also called \emph{horizontal}), according to whether the minimal polynomial of their unipotent lift is $(X-1)^2$ or $(X-1)^3$ (see section 3.4 of \cite{CG}). Another way to distinguish these two types is that 2-step unipotent isometries preserve a complex line (in fact, any complex line through their fixed point) but no real plane, whereas 3-step unipotent isometries preserve a real plane (in fact, an entire \emph{fan} of these, see section 2.3 of \cite{PW}) but no complex line.

\smallskip

{\bf Boundary at infinity and Heisenberg group:}

\smallskip

In the Siegel model associated to the Hermitian form given by the matrix $H_2$ in (\ref{H2matrix}), ${\rm H}_\mathbb{C}^n$ can be parametrized by $\mathbb{C}^{n-1} \times \mathbb{R} \times \mathbb{R}^+$ as follows, denoting as before by $\pi$ the projectivization map: ${\rm H}_\mathbb{C}^n=\{\pi(\psi(z,t,u)) \, | \, z \in \mathbb{C}^{n-1}, t \in \mathbb{R}, u \in \mathbb{R}^+\}$, where:
\begin{eqnarray}\label{horocoord}
  \psi(z,t,u)= \left(\begin{array}{c}
    (-|z|^2-u+it)/2 \\
    z \\
    1
  \end{array}\right)
\end{eqnarray}
With this parametrization the boundary at infinity $\partial_\infty {\rm H}_\mathbb{C}^n$ corresponds to the one-point compactification:
$$\left\{\pi\!\left(\psi(z,t,0)\right) \, | \, z \in \mathbb{C}^{n-1}, t \in \mathbb{R} \right\} \cup \{\infty\}$$
where $\infty=\pi((1,0,...,0)^T)$. The coordinates $(z,t,u) \in \mathbb{C}^{n-1} \times \mathbb{R} \times \mathbb{R}^+$ are called the \emph{horospherical coordinates} of the point $\pi(\psi(z,t,u) \in {\rm H}_\mathbb{C}^n$.

The punctured boundary  $\partial_\infty {\rm H}_\mathbb{C}^n - \{ \infty \}$ is then naturally identified to the \emph{generalized Heisenberg group} ${\rm Heis}(\mathbb{C},n)$, defined as the set $\mathbb{C}^{n-1} \times \mathbb{R}$ equipped with the group law: $$(z_1,t_1)(z_2,t_2)=(z_1+z_2,t_1+t_2+2{\rm Im} \, (z_1 \cdot \overline{z_2}))$$ where $\cdot$ denotes the usual Euclidean dot-product on $\mathbb{C}^{n-1}$. This is the classical 3-dimensional Heisenberg group when $n=2$. The identification of  $\partial_\infty {\rm H}_\mathbb{C}^n - \{ \infty \}$ with ${\rm Heis}(\mathbb{C},n)$ is given by the simply-transitive action of ${\rm Heis}(\mathbb{C},n)$ on $\partial_\infty {\rm H}_\mathbb{C}^n - \{ \infty \}$, where the element $(z_1,t_1) \in {\rm Heis}(\mathbb{C},n)$ acts on the vector $\psi(z_2,t_2,0)$ by left-multiplication by the following \emph{Heisenberg translation} matrix in ${\rm U}(n,1)$:
\begin{equation}\label{heistrans}
  T_{(z_1,t_1)}=\left(\begin{array}{ccc}
    1 & -z_1^* & (-|z_1|^2+it_1)/2 \\
    0 & {\rm I}_{n-1} & z_1 \\
    0 & 0 & 1
  \end{array}\right)
\end{equation}
In other words: $T_{(z_1,t_1)}\psi(z_2,t_2,0)=\psi(z_1+z_2,t_1+t_2+2{\rm Im} \, (z_1 \cdot \overline{z_2}),0)$.

In the above terminology, the unipotent isometry (given by the projective action of) $T_{(z_1,t_1)}$ is 2-step (or vertical) if $z_1=0$ and 3-step (horizontal) otherwise.

\smallskip

{\bf The hybridization construction:}

\smallskip

We will first embed the pair of Fuchsian groups into  ${\rm SU}(2,1)$ in the ball model of $H_\mathbb{C}^2$; there, two preferred orthogonal complex lines $L_1$ and $L_2$ are given by (the coordinate axes in the standard affine chart) $L_1=\pi({\rm Span}(e_1,e_3))$ and $L_2=\pi({\rm Span}(e_2,e_3))$, where $(e_1,e_2,e_3)$ denotes the canonical basis of $\mathbb{C}^3$ and $\pi:\mathbb{C}^3 - \{ 0\}\longrightarrow \mathbb{C} P^2$ the projectivization map. These intersect at the origin $O=\pi(e_3)$.

We will embed ${\rm SU}(1,1)$ in the stabilizer of each of these complex lines in the obvious block matrix form, namely via the injective homomorphisms:

\begin{equation}
  \begin{array}{llll}
    \iota_1: & {\rm SU}(1,1) & \longrightarrow &{\rm SU}(2,1) \\
    & \left(\begin{array}{cc}
      a & b \\
      c & d
    \end{array}\right) &
    \longmapsto &
    \left(\begin{array}{ccc}
      a & 0 & b \\
      0 & 1 & 0 \\
      c & 0 & d
    \end{array}\right)
  \end{array}
\end{equation}

\begin{equation}\label{rho2}
  \begin{array}{llll}
    \iota_2: & {\rm SU}(1,1) & \longrightarrow &{\rm SU}(2,1) \\
    & \left(\begin{array}{cc}
      a & b \\
      c & d
    \end{array}\right) &
    \longmapsto &
    \left(\begin{array}{ccc}
      1 & 0 & 0 \\
      0 & a & b \\
      0 & c & d
    \end{array}\right)
  \end{array}
\end{equation}

In the notation from the introduction, given two lattices $\Gamma_1,\Gamma_2$ in ${\rm SU}(1,1)$, we consider the hybrid $H(\Gamma_1,\Gamma_2)$ to be the projective image in ${\rm PU}(2,1)$ of $\langle \iota_1(\Gamma_1), \iota_2(\Gamma_2) \rangle < {\rm SU}(2,1)$.

\section{A hybrid subgroup of the Eisenstein-Picard modular group $\rm{PU}(2,1,\mathcal{O}_3)$}

\begin{figure}[!hb]
    \centering
    \begin{tikzpicture}[x=1in,y=1in]
        \draw[thick] (0,0) circle (1);
        \coordinate (w) at ({1/2},{sqrt(3)/2});
        \coordinate (w-bar) at ({1/2},{-sqrt(3)/2});

        \draw[very thick] (w) arc (150:270:{1/sqrt(3)});
        \draw[very thick] (1,0) arc (90:210:{1/sqrt(3)});

        \draw[very thick] (0,0) -- (1,0);
        \draw[very thick] (0,0) -- (w);
        \draw[very thick] (0,0) -- (w-bar);

        \coordinate (R) at (0:0.25);
        \coordinate (U) at (0.5879,0.1823);
        \draw (R) node[anchor=north] {$R$};
        \draw (U) node[anchor=south west] {$U$};
        \draw[-latex] (R) arc (0:60:0.25);
        \draw[-latex] (U) arc (133.1736:226.8264:0.25);
    \end{tikzpicture}
    \caption{Fundamental domain for ${\rm PU}(1,1;\mathcal{O}_3)$}
\label{fig:d=3}
\end{figure}

Using the method for finding generators from \cite{MP} (using 4 as \emph{covering depth}), we obtain the following generators for ${\rm PU}(1,1;\mathcal{O}_3)$ in the disk model of ${\rm H}_{\mathbb{C}}^1$:
\begin{align*}
R = \begin{pmatrix} -\omega^2 & 0 \\ 0 & 1 \end{pmatrix}, \qquad U = \begin{pmatrix} 1 + i\sqrt{3} & -i\sqrt{3} \\ i\sqrt{3} & 1 - i\sqrt{3} \end{pmatrix},
\end{align*}
denoting $\omega=\frac{-1+i\sqrt{3}}{2}$. Thus $\operatorname{U}(1,1;\mathcal{O}_3) = \langle -{\rm Id}, R, U \rangle$. The index-2 subgroup $\langle R^2, U \rangle < {\rm PU}(1,1;\mathcal{O}_3)$ lifts to ${\rm SU}(1,1;\mathcal{O}_3) = \langle -{\rm Id},E, U \rangle$ where
\begin{align*}
E=\omega^2 R^{-2} = \begin{pmatrix} \omega & 0 \\ 0 & \omega^2 \end{pmatrix}.
\end{align*}
Note that ${\rm PSU}(1,1;\mathcal{O}_3)=\langle E, U \rangle$ is a $(3,\infty,\infty)$ triangle group while ${\rm PU}(1,1;\mathcal{O}_3)$ is a $(2,6,\infty)$ triangle group.

We consider the hybrid group $H(3)=H\left({\rm SU}(1,1;\mathcal{O}_3),{\rm SU}(1,1;\mathcal{O}_3)\right)$, which by definition is generated by $\iota_j(E),\iota_j(U)$ and $\iota_j(-\rm{Id})$ for $j=1,2$. By the following observation it will suffice for our purposes to study the subgroup $\tilde{H}(3)=\langle \iota_1(E), \iota_2(E), \iota_1(U), \iota_2(U) \rangle$, which is a hybrid of two copies of ${\rm PSU}(1,1;\mathcal{O}_3)$ (so, of two $(3,\infty,\infty)$ triangle groups, ignoring the central extension by $-{\rm Id}$).

\begin{lem}
  \label{HtildeinH} $\tilde{H}(3)$ is normal in $H(3)$, with index dividing 4.
\end{lem}

{\em Proof}.  Since the $\iota_j(E), \iota_j(-\rm{Id})$ are diagonal they all commute; moreover for each $j=1,2$ $\iota_j(-\rm{Id})$ commutes with $\iota_j(E), \iota_j(U)$. It suffices therefore to show that $\iota_j(-Id)$ conjugates $\iota_{3-j}(U)$ into  $\tilde{H}(3)$ for $j=1,2$; a straightforward computation gives: $\iota_j(-{\rm Id}) \iota_{3-j}(U) \iota_j(-{\rm Id}) =\iota_{3-j}((EUE)^{-1})$. Therefore $\tilde{H}(3)$ is normal in $H(3)$; the quotient group is a quotient of $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ as it is generated by the images of $\iota_1(-\rm{Id})$ and $\iota_2(-\rm{Id})$ which have order 2 and commute. \hfill$\Box$\vspace{.5cm}

It will be more convenient for us to work in the Siegel model, in other words to conjugate by the Cayley transform $J$ given in (\ref{cayley}); we will abuse notation slightly by denoting again $H(3), \tilde{H}(3)$ the conjugates $J^{-1}H(3)J, J^{-1}\tilde{H}(3)J$. Concretely, we consider $\tilde{H}(3)= \langle E_1,U_1,E_2,U_2 \rangle$, with:
\begin{align*}
  E_1 = J^{-1} \iota_1(E) J
  &=
  \begin{pmatrix}
    \omega^2 & \omega^2-1 & \omega+2\\
    i \sqrt{3} & 1+ i \sqrt{3} & \omega^2-1 \\
    i \sqrt{3} & i \sqrt{3} & \omega^2
  \end{pmatrix},
  &
  U_1 = J^{-1} \iota_1(U) J
  &=
  \begin{pmatrix}
    1 & 0 & i \sqrt{3} \\
    0 & 1 & 0 \\
    0 & 0 & 1
  \end{pmatrix},
  \\
  E_2 = J^{-1} \iota_2(E) J
  &=
  \begin{pmatrix}
    \omega^2 & -i \sqrt{3} & i \sqrt{3} \\
   \omega+2& 1+ i \sqrt{3} & -i \sqrt{3} \\
   \omega+2 &\omega+2 & \omega^2
  \end{pmatrix},
  &
  U_2 = J^{-1} \iota_2(U) J
  &=
  \begin{pmatrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    i \sqrt{3} & 0 & 1
  \end{pmatrix}.
\end{align*}
Likewise we will denote $I_1 = J^{-1} \iota_{1}(-{\rm Id}) J$ and $I_2 = J^{-1} \iota_2(-{\rm Id}) J$.

{\bf Remark:} Since $E_1, E_2$ are both regular elliptic of order 3 with the same eigenspaces, they are either equal or inverse of each other. It turns out that $E_2=E_1^{-1}$ in ${\rm PU}(2,1)$ (the matrices satisfy $E_2=\omega E_1^{-1}$). We will therefore omit the generator $E_2$ from now on.

In \cite{FP} the authors determine that the Eisenstein-Picard modular group ${\rm PU}(2,1;\mathcal{O}_3)$ has presentation:

\begin{align*}
  {\rm PU}(2,1;\mathcal{O}_3)=\left\langle P,Q,R \;|\;R^2,\, (QP^{-1})^{6},\,PQ^{-1}RQP^{-1}R,\,P^{3}Q^{-2},\,(RP)^{3}\right\rangle , \ \ {\rm where}
\end{align*}

\begin{align*}
  P=\begin{pmatrix}
  1 & 1 &\omega\\
  0 & \omega &-\omega \\
  0 & 0 & 1
  \end{pmatrix},
  \qquad
  Q=\begin{pmatrix}
  1 & 1 & \omega \\
  0 & -1 & 1 \\
  0 & 0 & 1
  \end{pmatrix},
  \qquad
  R=\begin{pmatrix}
  0 & 0 & 1 \\
  0 & -1 & 0 \\
  1 & 0 & 0
  \end{pmatrix}.
\end{align*}

The following result can be checked by straightforward matrix computation:

\begin{lem}\label{gens3}
  The generators for the hybrid $\tilde{H}(3)$ can be expressed in terms of the Falbel-Parker generators for ${\rm PU}(2,1;\mathcal{O}_3)$ as follows:
  \begin{align*}
    U_1 &= Q^2, \\
    U_2 &= R Q^2 R, \\
    E_1 &= P^2 (R Q^2)^2 P^{-2}.
  \end{align*}
\end{lem}

\begin{lem}\label{normal3}
  The hybrid $\tilde{H}(3)$ is a normal subgroup of ${\rm PU}(2,1;\mathcal{O}_3)$.
\end{lem}

{\em Proof}.  It suffices to check that generators of ${\rm PU}(2,1;\mathcal{O}_3)$ conjugate generators of $\tilde{H}(3)$ into $\tilde{H}(3)$. Straightforward computations give the following:

$$\begin{array}{lcr}
  \begin{array}{rcl}
    P^{-1}U_1 P &= &U_1 \\
    Q^{-1}U_1 Q &= & U_1 \\
    R^{-1}U_1 R &= &U_2
  \end{array}
  &
  \begin{array}{rcl}
    P^{-1}U_2 P &= &U_1^{-1}E_1 \\
    Q^{-1}U_2 Q &= &U_1^{-1}E_1\\
    R^{-1}U_2 R &= &U_1
  \end{array}
  &
  \begin{array}{rcl}
    P^{-1}E_1 P &= &U_2^{-1} E_1^{-1} U_1 \\
    Q^{-1}E_1 Q &= &U_2 U_1 \\
    R^{-1}E_1 R &= &E_1^{-1}
  \end{array}
\end{array}$$
\hfill$\Box$\vspace{.5cm}

We can then form the quotient group $G={\rm PU}(2,1;\mathcal{O}_3)/ \tilde{H}(3)$, which by Lemma~\ref{gens3} has presentation:

\begin{align*}
  G={\rm PU}(2,1;\mathcal{O}_3)/ \tilde{H}(3)
  = \left\langle P, Q, R \;\bigg|\;
  \begin{aligned}
    &R^2, (QP^{-1})^6, PQ^{-1}RQP^{-1}R, \\
    &P^3Q^{-2}, (RP)^3, Q^2
  \end{aligned}
  \right\rangle.
\end{align*}
(Note that the relation $Q^{2}$ makes the other three relators corresponding to the generators of $H(3)$ superfluous). The Tietze transformation $a=PQ^{-1}$, $b=Q$, $c=R$, yields the following presentation for $G$:

\begin{align*}
  G = \left\langle a,b,c \;|\; c^2,\, a^6,\, [a,c],\, (ab)^3,\,(cab)^3,\, b^2\right\rangle.
\end{align*}

\begin{thm}\label{d=3main}
  The hybrid $H(3)$ has infinite index in ${\rm PU}(2,1,\mathcal{O}_3)$.
\end{thm}

Note that $G / \langle\!\langle c \rangle\!\rangle$ is the $(2,3,6)$ triangle group, hence infinite. Therefore $G$ is also infinite, in other words $\tilde{H}(3)$ has infinite index in ${\rm PU}(2,1,\mathcal{O}_3)$. By Lemma~\ref{HtildeinH}, $H(3)$ also has infinite index in ${\rm PU}(2,1,\mathcal{O}_3)$. \hfill$\Box$\vspace{.5cm}

\begin{cor}\label{d=3Zdense}
  The hybrid $H(3)$ is a thin subgroup of ${\rm PU}(2,1,\mathcal{O}_3)$.
\end{cor}

{\em Proof}.  The only additional statement is that $H(3)$ is Zariski-dense in ${\rm PU}(2,1)$, which is simple to see in rank 1, as it reduces essentially to irreducibility. Indeed, by \cite{CG} if a discrete subgroup $\Gamma$ is not Zariski-dense then it preserves a strict subspace of ${\rm H}^2_\mathbb{C}$ or it fixes a point on $\partial_\infty {\rm H}^2_\mathbb{C}$. This is easily seen not to be the case, as $E_1$ does not preserve the unique complex line preserved by both $U_1$ and $U_2$. (This will also follow from the fact that $H(3)$ has full limit set, see Section 6).  \hfill$\Box$\vspace{.5cm}

We conclude this section with a few remarks about the algebraic structure of the hybrid $H(3)$. We do not know a complete presentation for $H(3)$, in fact it may be non-finitely presented as far as we know (see \cite{K2} and Proposition~4.2 of \cite{FW}). The following observations are obtained by direct computation using the generators in matrix form.

\begin{lem}\label{rels3}
  The following relations hold between the generators $E_1,U_1,U_2$ for $H(3)$:
  $$E_1^3=(U_1U_2)^3=(E_1U_1^{-1}U_2)^3=(E_1U_2U_1^{-1})^3=(E_1^{-1}U_1U_2^{-1})^3=(E_1^{-1}U_2^{-1}U_1)^3=1.$$
\end{lem}

\begin{cor}\label{notamalgam}
  The hybrid $H(3)$ has finite abelianization; in particular it is not isomorphic to the amalgamated product of $\iota_1({\rm SU}(1,1,\mathcal{O}_3))$ and $\iota_2({\rm SU}(1,1,\mathcal{O}_3))$ over their (projective) intersection.
\end{cor}

{\em Proof}.  Recall that $H(3)$ is generated by $E_1, U_1, U_2, I_1, I_2$ with $I_1, I_2$ both of order $2$. Observe that by Lemma~\ref{rels3}, the following relations hold in the abelianization $H(3)^{ab}$ (we slightly abuse notation by using the same symbol for elements of $H(3)$ and their image in $H(3)^{ab}$): $ E_1^3=U_1^6=1, \ U_1^3=U_2^3$. Therefore $H(3)^{ab}$ is a quotient of $(\mathbb{Z}/2\mathbb{Z})^2 \times \mathbb{Z}/3\mathbb{Z} \times (\mathbb{Z}/6\mathbb{Z})^2$.

For the second statement, recall from the beginning of Section 3 that ${\rm SU}(1,1,\mathcal{O}_3)$ is a central $\mathbb{Z}/2\mathbb{Z}$-extension of ${\rm PSU}(1,1,\mathcal{O}_3)$, a $(3,\infty,\infty)$ triangle group. Therefore ${\rm SU}(1,1,\mathcal{O}_3)$ has abelianization $\mathbb{Z} \times \mathbb{Z}/6\mathbb{Z}$. Now the projective images of $\iota_1({\rm SU}(1,1,\mathcal{O}_3))$ and $\iota_2({\rm SU}(1,1,\mathcal{O}_3))$ intersect along the cyclic group of order 3 generated by $E_1$ (indeed, this intersection is contained in the diagonal subgroup of ${\rm PU}(2,1,\mathcal{O}_3)$, and $E_1,E_1^2$ are the only non-trivial diagonal elements that can be scaled to $\{ 1,u,u^{-1}\}$ for some $u \in {\rm U}(1)$). Therefore the amalgamated product has abelianization $\mathbb{Z}^2 \times \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/6\mathbb{Z}$. \hfill$\Box$\vspace{.5cm}

It is interesting to note that this also tells us the behavior of a related hybrid group, namely the hybrid of two
$(2,6,\infty)$ triangle groups, rather than $(3,\infty,\infty)$ (which each $(2,6,\infty)$ group contains with index 2). A simple way to view this new hybrid $H'(3)$ as a subgroup of $\Gamma(3)={\rm PU}(2,1,\mathcal{O}_3)$ containing the previous hybrid $\tilde{H(3)}$ is to take the obvious square root of the previous generator $E_1$ in terms of the Falbel-Parker generators, in other words to take $H'(3)$ to be generated by $E_1'=P^2(RQ^2)P^{-2}$, and $U_1=Q^2,U_2=RQ^2R$ unchanged.

\begin{lem}\label{comm3}
  The hybrid $H'(3)$ is contained in $[\Gamma(3),\Gamma(3)]$.
\end{lem}

{\em Proof}.  From the Falbel-Parker presentation for $\Gamma_3$ we get (abusing notation slightly again by using the same symbol for elements of $\Gamma(3)$ and their image in $\Gamma(3)^{ab})$:
$$\Gamma(3)^{ab}=\Gamma(3)/[\Gamma(3),\Gamma(3)] = \langle P,Q,R \ | \ R=P^3=Q^2=[P,Q]=1  \rangle.$$
The result then follows by noting that the generators listed above for $H'(3)$ all become trivial in the abelianization. \hfill$\Box$\vspace{.5cm}

The following is Lemma~6 of \cite{FW}.
\begin{lem}\label{comm3ab}
  The commutator subgroup $[\Gamma(3),\Gamma(3)]$ has abelianization $\mathbb{Z} \oplus \mathbb{Z}$.
\end{lem}

\begin{lem}\label{H'fab}
  The hybrid $H'(3)$ has finite abelianization.
\end{lem}

{\em Proof}.  This follows from the relations given in Lemma~\ref{rels3} by noting that $H'(3)$ is generated by $E'_1,U_1,U_2$ with $(E'_1)^2=E_1$. \hfill$\Box$\vspace{.5cm}

The following is well known but we include it for completeness:

\begin{lem}\label{fiabs}
  If $K_1 < K_2$ are two groups with $[K_2:K_1]$ and $K_1^{ab}$ finite, then $K_2^{ab}$ is finite.
\end{lem}

{\em Proof}.  Denote $i$ the inclusion map from $K_1$ into $K_2$, and $\pi_i:K_i\longrightarrow K_i^{ab}$ the quotient maps for $i=1,2$. Then $\pi_2 \circ i$ is a homomorphism from $K_1$ to an abelian group, so by the universal property of abelianizations $\pi_2 \circ i$ factors through $K_1^{ab}$, i.e. there is a homomorphism $i_*:K_1^{ab} \longrightarrow K_2^{ab}$ such that $i_* \circ \pi_1=\pi_2 \circ i$. Since $K_1=i(K_1)$ has finite index in $K_2$ by assumption and $\pi_2$ is surjective, $\pi_2(K_1)=i_*(\pi_1(K_1))=i_*(K_1^{ab})$ has finite index in $K_2^{ab}$. The result follows since $K_1^{ab}$ is finite. \hfill$\Box$\vspace{.5cm}

Combining Lemmas~\ref{comm3}, \ref{comm3ab}, \ref{H'fab} and \ref{fiabs} gives the following:

\begin{cor}\label{d=3other}
  The hybrid $H'(3)$ has infinite index in $[\Gamma(3),\Gamma(3)]$, hence also in $\Gamma(3)$.
\end{cor}

It is interesting to note that, in contrast with the previous hybrid $H(3)$ which was normal in $\Gamma(3)$, $H'(3)$ now has infinite index in its normal closure $\langle \langle H'(3) \rangle \rangle =\Gamma(3)$ in $\Gamma(3)$ (the presentation of $\Gamma(3)/\langle \langle H'(3) \rangle \rangle$ obtained by adding the generators of $H'(3)$ to the presentation for $\Gamma(3)$ now gives the trivial group).

\section{A hybrid subgroup of the Gauss-Picard modular group $\rm{PU}(2,1,\mathcal{O}_1)$}

\begin{figure}[!hb]
    \centering
    \begin{tikzpicture}[x=1in,y=1in]
        \draw[thick] (0,0) circle (1);
        \coordinate (i) at (0,1);
        \coordinate (i-bar) at (0,-1);

        \draw[very thick] (i) arc (180:270:1);
        \draw[very thick] (1,0) arc (90:180:1);

        \draw[very thick] (0,0) -- (1,0);
        \draw[very thick] (0,0) -- (i);
        \draw[very thick] (0,0) -- (i-bar);

        \coordinate (R) at (0:0.20);
        \coordinate (U) at (0.5294,0.1176);
        \draw (R) node[anchor=north] {$R$};
        \draw (U) node[anchor=south west] {$U$};
        \draw[-latex] (R) arc (0:90:0.20);
        \draw[-latex] (U) arc (151.9275:208.0725:0.25);
    \end{tikzpicture}
\caption{Fundamental domain for ${\rm PU}(1,1;\mathcal{O}_1)$}
\label{fig:d=1}
\end{figure}

Using the method for finding generators from \cite{MP} (using again 4 as \emph{covering depth}), we obtain the following generators for ${\rm PU}(1,1;\mathcal{O}_1)$ in the disk model of $H_C^1$:
\begin{align*}
R = \begin{pmatrix} i & 0 \\ 0 & 1 \end{pmatrix}, \qquad U = \begin{pmatrix} 1 + i & -i \\ i & 1 - i \end{pmatrix}.
\end{align*}
Thus $\operatorname{U}(1,1;\mathcal{O}_1) = \langle -{\rm Id}, R, U \rangle$. The index-2 subgroup $\langle R^2, U \rangle < {\rm PU}(1,1;\mathcal{O}_1)$ lifts to ${\rm SU}(1,1;\mathcal{O}_1)$, which is therefore generated by $-{\rm Id},E,$ and $T$, with:
\begin{align*}
E=i R^{-2} = \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix}.
\end{align*}
Since $E^{2} = -{\rm Id}$, the group ${\rm SU}(1,1;\mathcal{O}_1)$ is generated by $E$ and $U$.

Note that ${\rm SU}(1,1;\mathcal{O}_1)$ is a central $\mathbb{Z}/2\mathbb{Z}$-extension of a $(2,\infty,\infty)$ triangle group while ${\rm PU}(1,1;\mathcal{O}_1)$ is a $(2,4,\infty)$ triangle group.

We now consider the hybrid group $H\left({\rm SU}(1,1;\mathcal{O}_1),{\rm SU}(1,1;\mathcal{O}_1)\right)$, which by definition is generated by  $\iota_1(E)$, $\iota_1(U)$, $\iota_2(E)$ and $\iota_2(U)$. It will be again more convenient for us to work in the Siegel model, in other words to conjugate by the Cayley transform $J$ given in (\ref{cayley}). We thus consider the group $H(1)= \langle E_1,U_1,E_2,U_2 \rangle$, where:

\begin{align*}
  E_1 = J^{-1}\iota_1(E)J
  &=
  \begin{pmatrix}
    i & -1+i & 1-i \\
    -2i & 1-2i & -1+i \\
    -2i & -2i & i
  \end{pmatrix},
  &
  U_1 =J^{-1}\iota_1(U)J
  &=
  \begin{pmatrix}
    1 & 0 & i \\
    0 & 1 & 0 \\
    0 & 0 & 1
  \end{pmatrix}, \\
  E_2 = J^{-1}\iota_2(E)J
  &=
  \begin{pmatrix}
    i & 2i & -2i \\
    1-i & 1-2i & 2i \\
    1-i & 1-i & i
  \end{pmatrix},
  &
  U_2 = J^{-1}\iota_2(U)J
  &=
  \begin{pmatrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    i & 0 & 1
  \end{pmatrix}.
\end{align*}

The following presentation for the Gauss-Picard lattice ${\rm PU}(2,1;\mathcal{O}_1)$ was first given in \cite{FFP} (Theorem 4):

\begin{align*}
  {\rm PU}(2,1;\mathcal{O}_1) = \left\langle I_0, Q, T \; \bigg| \;
  \begin{aligned}
    I_0^2,\, Q^2,\, (I_0 Q)^3,\, (I_0 T)^{12},\, (I_0 QT)^{8},\, [(I_0 T)^3,T],\, [Q,T]
  \end{aligned}
  \right\rangle
\end{align*}

where:

\begin{align*}
  I_0 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 0 \end{pmatrix},
  \qquad
  Q = \begin{pmatrix} 1 & 1-i & -1 \\ 0 & -1 & 1 + i \\ 0 & 0 & 1 \end{pmatrix},
  \qquad
  T = \begin{pmatrix} 1 & 0 & i \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}.
\end{align*}

The following result can be checked by straightforward matrix computation; the expression for $E_1$ was found by a computer search, using a different generating set.

\begin{lem}\label{d-1gens}
  The generators for the hybrid $H(1)$ can be expressed in terms of the above generators for ${\rm PU}(2,1;\mathcal{O}_1)$ as follows:
  \begin{align*}
    U_1 &= T, \\
    U_2 &= I_0 U_2 I_0, \\
    E_1 &= T^{-1} Q (I_0 T)^3 I_0 (T (I_0 T)^{-3} Q)^2 I_0, \\
    E_2 &= I_0 E_1 I_0.
  \end{align*}
\end{lem}

\begin{lem}\label{normal1}
  The hybrid $H(1)$ is a normal subgroup of ${\rm PU}(2,1;\mathcal{O}_1)$.
\end{lem}

{\em Proof}.  Since $U_1 = T$ and the generators come in pairs conjugate by $I_0$, it suffices to check conjugation by $Q$:
\begin{align*}
  Q^{-1} U_1 Q &= U_1 \\
  Q^{-1} U_2 Q &= (U_1 E_1) U_2 (U_1 E_1)^{-1} \\
  Q^{-1} E_1 Q &= (U_2 U_1) E_2 (U_2 U_1)^{-1} \\
  Q^{-1} E_2 Q &= (U_2 U_1) E_1 (U_2 U_1)^{-1}
\end{align*}
\hfill$\Box$\vspace{.5cm}

\begin{thm}\label{d=1main}
  The hybrid $H(1)$ has index 2 in the full Gauss-Picard lattice ${\rm PU}(2,1;\mathcal{O}_1)$.
\end{thm}

{\em Proof}.  Since $H(1)$ is normal in ${\rm PU}(2,1;\mathcal{O}_1)$, a presentation for the quotient ${\rm PU}(2,1;\mathcal{O}_1)/H(1)$ is obtained from the presentation for ${\rm PU}(2,1;\mathcal{O}_1)$, to which we add as relations the generators of the subgroup $H(1)$ written as the words given in Lemma~\ref{d-1gens}.
\begin{align*}
  {\rm PU}(2,1;\mathcal{O}_1)/H(1)
  &= \left\langle I_0, Q, T \; \bigg| \;
  \begin{aligned}
    I_0^2,\, Q^2,\, (I_0 Q)^3,\, (I_0 T)^{12},\, (I_0 QT)^{8},\, [(I_0 T)^3,T],\, [Q,T],\, T,\, I_0 T I_0, \\
    T^{-1} Q (I_0 T)^3 I_0 (T (I_0 T)^{-3} Q)^2 I_0,\, I_0 T^{-1} Q (I_0 T)^3 I_0 (T (I_0 T)^{-3} Q)^2
  \end{aligned}
  \right\rangle
\end{align*}
Since $T = 1$ in the quotient, the relations $(I_0 Q)^3 = 1$ and $(I_0 Q T)^8 = 1$ imply that $I_0 = Q$. The remaining relations are then satisfied, thus the presentation for the quotient simplifies to
\begin{align*}
  {\rm PU}(2,1;\mathcal{O}_1)/H(1) = \left\langle I_0, Q, T \;|\; I_0 = Q, I_{0}^2 = T = 1 \right\rangle = \mathbb{Z}/2\mathbb{Z} .
\end{align*}
\hfill$\Box$\vspace{.5cm}

We now consider the related hybrid $H'(1)$ as in the case of $d=3$, namely taking $H'(1)$ to be the hybrid of two copies of ($\mathbb{Z}/2\mathbb{Z}$-extensions of) the Fuchsian triangle group $(2,4,\infty)$, rather than $(2,\infty,\infty)$. We immediately get the following result by noting that $H'(1)$ contains $H(1)$, which has index 2 in the full lattice $\Gamma(1)$, as well as a new element of order 4 not belonging to $H(1)$.

\begin{cor}\label{d=1other}
  The hybrid $H'(1)$ is equal to the full lattice $\Gamma(1)={\rm PU}(2,1;\mathcal{O}_1)$.
\end{cor}

\section{A hybrid subgroup of the Picard modular group $\rm{PU}(2,1,\mathcal{O}_7)$}

\begin{figure}[!hb]
    \centering
    \begin{tikzpicture}[x=1in,y=1in]
        \draw[thick] (0,0) circle (1);
        \begin{scope}
        \coordinate (P1) at (0,0.4569);
        \coordinate (P2) at (0.625,0.3307);
        \coordinate (P3) at (1,0);
        \coordinate (P4) at (0.625,-0.3307);
        \coordinate (P5) at (0,-0.4569);

        \draw[very thick] (P1) arc (240 : 277.1808 : 1);
        \draw[very thick] (P2) arc (187.1807 : 270 : 0.37796);
        \draw[very thick] (0,0) -- (P3);
        \draw[very thick] (P3) arc (90: 172.8193 : 0.37796);
        \draw[very thick] (P4) arc (82.8192 : 120 : 1);

        \fill (0,0) circle (2pt);

        \coordinate (B) at (0:0.15);
        \coordinate (U) at (0.75,0.09448);
        \coordinate (A) at (0.4251,0.3257);
        \draw (B) node[anchor=north] {$B$};
        \draw (U) node[anchor=south west] {$U$};
        \draw (A) node[anchor=south] {$A$};
        \draw[-latex] (B) arc (0:180:0.15);
        \draw[-latex] (U) arc (138.592 : 221.408 : 0.14285);
        \draw[-latex] (A) arc (175.697 : 184.303 : 4.3411);
        \end{scope}
        \begin{scope}[xscale=-1,yscale=1
        \coordinate (P1) at (0,0.4569);
        \coordinate (P2) at (0.625,0.3307);
        \coordinate (P3) at (1,0);
        \coordinate (P4) at (0.625,-0.3307);
        \coordinate (P5) at (0,-0.4569);

        \draw[very thick] (P1) arc (240 : 277.1808 : 1);
        \draw[very thick] (P2) arc (187.1807 : 270 : 0.37796);
        \draw[very thick] (0,0) -- (P3);
        \draw[very thick] (P3) arc (90: 172.8193 : 0.37796);
        \draw[very thick] (P4) arc (82.8192 : 120 : 1);
        \end{scope}
    \end{tikzpicture}
\caption{Fundamental domain for ${\rm PU}(1,1;\mathcal{O}_7)$}
\label{fig:d=7}
\end{figure}

Using the method for finding generators from \cite{MP} (using 9 as \emph{covering depth}), we obtain the following generators for ${\rm PU}(1,1;\mathcal{O}_7)$ in the disk model of ${\rm H}_{\mathbb{C}}^1$:
\begin{align*}
  U = \begin{pmatrix} 1 + i \sqrt{7}& -i\sqrt{7} \\ i\sqrt{7} & 1-i\sqrt{7} \end{pmatrix}, \qquad
  A = \begin{pmatrix} -\frac{1}{2} + i\frac{\sqrt{7}}{2} & 1 \\ -1 & \frac{1}{2} + i\frac{\sqrt{7}}{2} \end{pmatrix},  \qquad
  B = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}.
\end{align*}
Thus $\operatorname{U}(1,1;\mathcal{O}_7) = \langle -\operatorname{Id},U,A,B \rangle$. As previously, the index-2 subgroup $\langle U, BA, AB, BUB \rangle < {\rm PU}(1,1;\mathcal{O}_7)$ lifts to ${\rm SU}(1,1;\mathcal{O}_7)$, which is therefore generated by $ -\operatorname{Id}, U, BA, AB,$ and $BUB$.

When embedding ${\rm SU}(1,1;\mathcal{O}_7)$ into ${\rm PU}(2,1)$ via $\iota_1$ and $\iota_2$, it turns out that $\iota_1(-{\rm Id}) = \iota_2(B)$ and $\iota_2(-\rm{Id}) = \iota_1(B)$, and so the hybrid group $H(7) = H({\rm SU}(1,1;\mathcal{O}_7),{\rm SU}(1,1;\mathcal{O}_7))$ admits the following simplified set of generators:

\begin{align*}
  U_1 = J^{-1}\iota_1(U)J
   &=
  \begin{pmatrix}
    1 & 0 & i \sqrt{7} \\
    0 & 1 & 0 \\
    0 & 0 & 1
  \end{pmatrix},
  &
  U_2 = J^{-1}\iota_2(U)J
  &=
  \begin{pmatrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    i \sqrt{7} & 0 & 1
  \end{pmatrix},
  \\
  A_1 = J^{-1}\iota_1(A)J
  &=
  \begin{pmatrix}
    -\frac{1}{2}+i\frac{\sqrt{7}}{2} & -\frac{3}{2}+i\frac{\sqrt{7}}{2} & \frac{1}{2}-i\frac{\sqrt{7}}{2} \\
    1 & 2 & -\frac{3}{2}+i\frac{\sqrt{7}}{2} \\
    1 & 1 & -\frac{1}{2}+i\frac{\sqrt{7}}{2} \\
  \end{pmatrix},
  &
  B_1 = J^{-1}\iota_1(B)J
  &=
  \begin{pmatrix}
    1 & 0 & 0 \\
    -2 & -1 & 0 \\
    -2 & -2 & 1
  \end{pmatrix},
  \\
  A_2 = J^{-1}\iota_2(A)J
  &=
  \begin{pmatrix}
    -\frac{1}{2} + i\frac{\sqrt{7}}{2} & -1 & 1 \\
    \frac{3}{2} - i\frac{\sqrt{7}}{2} & 2 & -1 \\
    \frac{1}{2} - i\frac{\sqrt{7}}{2} & \frac{3}{2}-\frac{i \sqrt{7}}{2} & -\frac{1}{2}+\frac{i \sqrt{7}}{2} \\
  \end{pmatrix},
  &
  B_2 = J^{-1}\iota_2(U)J
  &=
  \begin{pmatrix}
    1 & 2 & -2 \\
    0 & -1 & 2 \\
    0 & 0 & 1
  \end{pmatrix}.
\end{align*}

In ~\cite{MP} the authors determine that ${\rm PU}(2,1;\mathcal{O}_7)$ has presentation
\begin{align}\label{MPpres}
  {\rm PU}(2,1;\mathcal{O}_7)= \left\langle T_1, R, I\; \bigg| \;
  \begin{aligned}
    R^2 =  I^2 = (RI)^2 = R  T_1^{-1}  R  T_1 RT_1RT_1^{-1} =
    (T_1  I  T_1^{-1}  R)^4 = (T_1^{-1}  I  T_1  R)^4 =   \\
    T_1^{-1}  I  T_1^{-1}  I  T_1  I  T_1  I  T_1^{-3}  I  T_1
    I  T_1  I  T_1^{-1}  I  T_1^{-1} =
    (T_1^{-1}  I  T_1  I  T_1  I  T_1^{-1}  I  T_1^{-1}  I)^2 = \\

     (I  T_1^{-1}  R)^7 =
    T_1^{-1}  I  T_1  I  T_1  I  T_1^{-2}  I  T_1^{-1}  I  T_1
    I  T_1^2  I  T_1^{-1}  I  T_1^{-1}  I  T_1  I =  \\
    T_1^{-1}  I  T_1  I  T_1  I  R  T_1  I  R  T_1  I
    T_1  I  T_1^{-1}  I  T_1^{-1}  I  T_1  R  T_1^{-1}  I  R
    T_1^{-1}  I = \\
    R  T_1  I  R  T_1  I  T_1  I  T_1^{-1}  I  T_1^{-1}  I
     R  T_1^{-1}  I  R  T_1^{-1}  I  T_1^{-1}  I  T_1  I  T_1
    I  T_1^{-1} = \\
    R  T_1  I  R  T_1  R  T_1^{-1}  I  T_1  I  T_1  I
    R  T_1  I  T_1  I  T_1^{-1}  R  T_1  R  I  T_1  R
    T_1^{-1}  I  T_1  I  T_1  I  T_1  I  T_1^{-1} = 1
 \end{aligned}
  \right\rangle
\end{align}
where

\begin{align*}
  T_1 &= \begin{pmatrix}
    1 & -1 & -\frac{1}{2} + i\frac{\sqrt{7}}{2} \\
    0 & 1 & 1 \\
    0 & 0 & 1 \end{pmatrix}, &
  R &= \begin{pmatrix}
    1 & 0 & 0 \\
    0 & -1 & 0 \\
    0 & 0 & 1 \end{pmatrix}, &
  I &= \begin{pmatrix}
    0 & 0 & 1 \\
    0 & -1 & 0 \\
    1 & 0 & 0 \end{pmatrix}.
\end{align*}

The following result can be checked by straightforward matrix computation:
\begin{lem}\label{d=7gens}
In terms of the above generators, the generators for $H(7)$ can be written as follows:
\begin{align*}
  U_1 &= (R T_1)^2, \\
  U_2 &= I U_1 I, \\
  A_1 &= T_1 I T_1 R, \\
  A_2 &= I A_1 I,\\
  B_1 &= (IT_1) R (I T_1)^{-1},\\
  B_2 &= I B_1 I.
\end{align*}
\end{lem}

\begin{lem}\label{d=7normal}
  The hybrid $H(7)$ is a normal subgroup of ${\rm PU}(2,1;\mathcal{O}_7)$.
\end{lem}

{\em Proof}.
We have that $R = (A_1 A_2 B_1 A_1 B_2)^{-1} B_1 (A_1 A_2 B_1 A_1 B_2) \in H(7)$ and since the generators of $H(7)$ come in pairs conjugate by $I$, it suffices to check conjugation by $T_1$:
\begin{align*}
  T_1^{-1}A_1 T_1 &= (A_1 A_2^{-1} B_2 A_2^{-1} A_1)^{-1} A_2 (A_1 A_2^{-1} B_2 A_2^{-1} A_1) \\

  T_1^{-1}A_2 T_1 &= (B_2 A_2 A_1^{-1} A_2^{-1} B_1)^{-1} A_2 (B_2 A_2 A_1^{-1} A_2^{-1} B_1) \\

  T_1^{-1}B_1 T_1 &= (A_1^{-1} A_2^{-1} B_1)^{-1} B_1 (A_1^{-1} A_2^{-1} B_1) \\

  T_1^{-1}B_2 T_1 &= R \\

  T_1^{-1}U_1 T_1 &= U_1 \\

  T_1^{-1}U_2 T_1 &= (A_1^{2} A_2^{-1} B_2 A_2^{-1} A_1)^{-1} U_2 (A_1^{2} A_2^{-1} B_2 A_2^{-1} A_1) \\

\end{align*}
\hfill$\Box$\vspace{.5cm}

\begin{thm}\label{d=7main}
  The hybrid ${\rm H}(7)$ is the full lattice ${\rm PU}(2,1;\mathcal{O}_7)$.
\end{thm}

{\em Proof}.  As before, since $H(7)$ is normal in ${\rm PU}(2,1;\mathcal{O}_7)$, we obtain a presentation for the quotient ${\rm PU}(2,1;\mathcal{O}_7)/H(7)$ by adding the words from Lemma~\ref{d=7gens} as relations to the presentation (\ref{MPpres}). As noted in the proof of Lemma~\ref{d=7normal}, $R \in H(7)$ so $R=1$ in the quotient. As well, the relation corresponding to $A_1$ implies that $I = T_1^2$ in the quotient. The following relations in the presentation (\ref{MPpres})
\begin{align*}
I^2 = T_1^{-1} I T_1^{-1} I T_1 I T_1 I T_1^{-3} I T_1 I T_1 I T_1^{-1} I T_1^{-1} = 1
\end{align*}
imply that $T_1^4 = T_1^{13} = 1$ in the quotient, hence $T_1 = 1$. Thus
${\rm PU}(2,1;\mathcal{O}_7)/H(7)$ is trivial. \hfill$\Box$\vspace{.5cm}

\section{Limit sets and geometrical finiteness}

\subsection{Limit sets}

We first briefly recall the definition and two classical facts about limit sets of discrete groups of isometries, see e.g. \cite{B2} or \cite{K3}. The space we consider in this paper is the complex hyperbolic plane ${\rm H}^2_\mathbb{C}$, but these definitions and facts hold more generally in any negatively curved symmetric space (so, hyperbolic space of any dimension over the real or complex numbers or quaternions, or hyperbolic plane over the octonions).

{\bf Definiton:} Let $X$ be a negatively curved symmetric space, $\partial_\infty X$ its boundary at infinity, and $\Gamma$ a discrete subgroup of ${\rm Isom}(X)$. The \emph{limit set} $\Lambda(\Gamma)$ of $\Gamma$ is defined as the set of accumulation points in $\partial_\infty X$ of the orbit $\Gamma x_0$ for any choice of $x_0 \in X$; this does not depend on the choice of $x_0$.

A discrete subgroup $\Gamma$ of ${\rm Isom}(X)$ is called \emph{elementary} if $\Gamma$ has a global fixed point in $X \cup \partial_{\infty} X$ or preserves a geodesic. When $\Gamma$ is non-elementary, the limit set $\Lambda(\Gamma)$ is the minimal (nonempty) closed $\Gamma$-invariant subset of $\partial_\infty X$, in fact the orbit $\Gamma p_\infty$ is dense in $\Lambda(\Gamma)$ for any $p_\infty \in \Lambda(\Gamma)$. We will use the following two classical properties of limit sets:

\begin{prop} Let $X$ be a negatively curved symmetric space and $\Gamma$ a discrete subgroup of ${\rm Isom}(X)$.
  \begin{itemize}
  \item[(a)] If $\Gamma$ is a lattice in ${\rm Isom}(X)$ then $\Lambda(\Gamma)=\partial_\infty X$.
  \item[(b)] If $\Gamma'$ is a nonelementary normal subgroup of $\Gamma$ then $\Lambda(\Gamma')=\Lambda(\Gamma)$.
  \end{itemize}
\end{prop}

The following result is an immediate consequence of this and Lemmas~\ref{normal3}, \ref{normal1} (or Theorem~\ref{d=1main}).

\begin{prop}\label{limitsets} For $d=1,3,7$ the hybrid $H(d)$ has full limit set: $\Lambda(H(d))=\partial_\infty {\rm H^2_\mathbb{C}} \simeq S^3$.
\end{prop}

\subsection{Geometrical finiteness}

The original notion of geometrical finiteness for a Kleinian group $\Gamma < {\rm Isom \, (H}^3_\mathbb{R})$ was to admit a finite-sided polyhedral fundamental domain. This was later shown to admit several equivalent formulations, then systematically studied by Bowditch in higher-dimensional real hyperbolic spaces in \cite{B1}, and more generally in pinched Hadamard manifolds in \cite{B2}. In \cite{B1}, Bowditch labelled the five equivalent formulations of the definition of geometrical finiteness (GF1)-(GF5), with (GF3) corresponding to the original notion. He then showed in \cite{B2} that the four other formulations, now labelled F1,F2,F4, and F5, remain equivalent in the more general setting (but not the original one). The most convenient for our purposes will be condition F5, which we now recall.

Let as above $X$ be a negatively curved symmetric space and $\Gamma$ a discrete subgroup of ${\rm Isom}(X)$. The \emph{convex hull} ${\rm Hull}(\Gamma)$ of $\Gamma$ in $X$ is the convex hull of the limit set $\Lambda(\Gamma)$, more precisely the smallest convex subset of $X$ whose closure in $\overline{X}=X \cup \partial_\infty X$ contains $\Lambda(\Gamma)$. This is invariant under the action of $\Gamma$, and the \emph{convex core} ${\rm Core}(\Gamma)$ of $\Gamma$ in $X$ is defined as the quotient of ${\rm Hull}(\Gamma)$ under the action of $\Gamma$.

{\bf Definition:} We say that $\Gamma$ satisfies condition F5 if (a) for some $\varepsilon >0$, the tubular neighborhood $N_\varepsilon({\rm Core}(\Gamma))$ in $X/\Gamma$ has finite volume, and (b)  there is a bound on the orders of the finite subgroups of $\Gamma$.

\begin{prop}\label{geominf}
  The hybrid $H(3) < {\rm Isom}({\rm H}^2_\mathbb{C})$ is geometrically infinite.
\end{prop}

{\em Proof}.  We show that $H(3)$ does not satisfy condition F5. By Proposition~\ref{limitsets}, $\Lambda(H(3))=\partial_\infty {\rm H^2_\mathbb{C}}$, hence ${\rm Hull}(H(3))={\rm H^2_\mathbb{C}}$. Now by Theorem~\ref{d=3main}, $H(3)$ has infinite index in a lattice, therefore it acts on  ${\rm H^2_\mathbb{C}}$ with infinite covolume, in other words ${\rm Core}(H(3))$ has infinite volume hence so does any of its tubular neighborhoods. \hfill$\Box$\vspace{.5cm}

\raggedright
\frenchspacing

redpajama_set_name: RedPajamaArXiv


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In a quiet corner of the Ziibiwing Center, between permanent exhibits on the environmental changes that resulted from the effects of colonization and the sacred language and strength of the Anishinabe (first man), is a half-moon bench under a sign that explains the phenomenon of blood memory - the ancestral bond every person has to the language, music, spirituality, and other cultural components of their people.
On the wall across from the bench are three strikingly decorated traditional wood and fabric baby carriers. This room represents the Anishinabe peoples' seventh prophecy: the Anishinabek nation will be reborn and the newly born will rekindle the knowledge of their elders. The exhibits that come next highlight the ways the native people of the great lakes region are doing that, and contributing to their communities at the same time.
Though Ziibiwing Center didn't open until 2004, it has long been a dream of the Saginaw Chippewa Indian Tribe elders. "There are tribal council minutes from the 1950's where they talked about this," says Johnson. And the vision of those elders is taken seriously.
Unlike traditionally curated museums, the Ziibiwing Center is a community-led cultural center. This means that while Johnson's official title is "curator" he does more community engagement than simple curation. Everything from signage to programming to the very form of the mannequins on exhibit throughout the center are crowd-sourced.
"If you knew these people, you'd recognize them," Johnson explained, gesturing to a terra-cotta mannequin just outside the center's teaching lodge. That's because each mannequin in the center's Diba Jimooyung — the permanent exhibits which are collectively titled "Telling Our Story" — was custom-made in the likeness of a community member turned model.
Johnson says the decision to feature custom mannequins was out of necessity. When the center was being built mannequins that looked like Native Americans weren't available. Instead, they turned to the community.
"We invited anyone who wanted to model to come out and try," says Johnson. From there they chose the models that best fit the exhibits they had planned, and a company out of the U.K. built the forms in the community members' likeness. The decision has since proven beneficial in ways never imagined at the outset. Mannequins of those who have since passed away are a life-size monument for their families, and by using real people from the community as models the diversity of tribal members' features and builds can be appreciated throughout.
Showcasing that diversity and dispelling the stereotypes and myths about American Indian people is part of the center's main mission. Johnson says challenging notions that visitors have about Native Americans helps them think critically about historical depictions and recognize that tribal citizens are just like everyone else.
It's work that Johnson and the Ziibiwing Center team do in conjunction with the Michigan Anishinaabek Cultural Preservation and Repatriation Alliance (MACPRA), a group that represents all of the federally-recognized tribes in Michigan, plus two state-designated historic tribes. And it's made possible by the Native American Graves Protection and Repatriation Act (NAGPRA) of 1990.
NAGPRA addresses the ownership and possession of Native Indian and Native Hawaiian cultural items. Executed by the U.S. Department of the Interior, NAGPRA requires federal agencies and institutions that receive federal funding to return important cultural items and remains to their rightful owners - American Indian tribes and Native Hawaiian organizations. This means everything from artifacts to human remains that have historically been held in places such as universities can be brought home. For human remains, it's the chance to finally be laid respectfully to rest.
To honor the ancestors Ziibiwing helps repatriate, a Recommitment to the Earth Ceremony is held each time remains are brought back for burial. The ceremony begins before the remains are ever moved, with careful preparation for the journey. It continues with a sacred re-burial ceremony at the Chippewa Township Nibokaan Cemetery, and concludes with a journey feast.
The next Earth Ceremony will take place on May 10, 2019. Interested and respectful public are welcome to attend.
The Ziibiwing Center is located at 6650 E. Broadway in Mt. Pleasant, Michigan and is open Monday through Saturday for groups and individuals. The center's award-winning permanent exhibit is on display year round, with an additional temporary exhibit that changes periodically. The center features over 500 artifacts, two theaters displaying educational programming, and a gift shop.
For additional information visit the Ziibiwing Center website.


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Mizzou Professor Melissa Click Resigns Her Courtesy Position, Remains On Faculty As Communications Professor…
Symbolic resignation. Update to this story.
Via Independent Sentinel:
Irony of ironies! Mizzou has a journalism professor who is anti-free speech!
The anti-Free Speech journalism/communications Professor Click at Mizzou, who does research on Lady Gaga, has resigned her courtesy position at the school of journalism but don't be fooled, she's still at teaching at the school of communications. They are merely presenting a facade of repercussions for having an anti-free speech professor in the school of journalism.
She can't teach at the school of journalism but she can at the school of communications? She's going to communicate her fascist, anti-free speech blather. She can also review papers!
This is how far we've come under the so-called liberals. It's actually fascism. That's funny or ironical or insane. Choose one.
Dapandico | November 11, 2015 1:20 pm


Exquisitely hand painted Set of 5 Nordic Blue Wall PlatesFor all the wall plates, organic paint has been used to colour them. These colours have a matt texture, and may bleed if they come in contact with liquid. Hence, buyers are strongly advised not to wipe plates with a wet cloth. All plates should be lightly dusted.These should not be used to serve food, they are to be hung on walls for purely decorative purpose only. All plates are wall hanging and come with an attached fixture at the back.


What is Mars Kuja Dosha? Generally Vedic Astrologers (Jyotishis) will say that you should not marry someone if they have what is called a "Mars Dosha" (also called a "Mangal Dosh" or "Kuja Dosha"). This means they have the planet Mars in the 1st, 2nd (only according to Southern Indian Astrologers), 4th, 7th, 8th or 12th house of their primary chart. Marrying someone with this in their chart is said to cause disharmony and sometimes to lead to even more serious issues.
However, if the person they are considering marrying also has a Mars Dosha, then it is said to cancel out all the fiery, argumentative effects of Mars and so it is fine to marry them.
It is very common to have a Mars Dosha.
Some believe that if you do marry someone with a Mars Dosha, the effects can be reduced and maybe even cancelled if you have some Vedic priests perform a Yagya ceremony called a "Kumbh Vivah".
The bad effects are also said to be reduced through other remedies which can be assigned depending on your specific charts.
If we find that you a good match, we let you know in the Match report if Mars Kuja Dosha exists and, if it does exist, if it is cancelled.


The heart of the production is our aluminium foundry. It was the foundation of our business and looks back over decades of experience and tradition. Combined with the latest equipment in foundry technology and determined employees we challenge any requirements our customers demand.
We put a lot of effort into training our moulders internally, because of the large variety and complexity of our castings the best way to transfer the knowledge of our moulding techniques is through daily practice and training. On several moulding stations our employees produce sand moulds and sand cores with chemically bonded sand. After the moulds have been poured and the casting solidified, the sand moulds are broken down and the castings removed. The sand from the broken moulds is fully recycled.
On a regular basis, our staff in the foundry receives training internally through staff with decades of experienced. The employees in this department are responsible for each melt. This begins with melting the alloy, analysing its chemical composition, refining and modifying it according to specification and finally pouring the designated moulds.
We have developed specialised GDC SPM with tilting technology to cater to demand of highest quality castings for our customers. Our staff has been specially trained to operate on the GDC machines and produce the best quality castings. These castings are then visually inspected by the operator for any problems in castings.
The fettling shop performs the most individual job in the whole process and is to 100% a manual operation. Polishing or grinding shapes according to templates is routine procedure in our fettling shop. Each employee gives each casting the desired aesthetics, surface finish and also dimensional accuracy. Our employees can easily focus on the individual requirements of our customers.
Heat treatment of aluminium castings has seen strong further development in the past years and it is now possible to increase the mechanical properties of aluminium alloy up to 150%. To define the suitable parameters for the heat treatment process, we have to take into account the individual requirements of each customer and their aluminium castings.
We are setting up the state-of-art machining operation facilty in our company. We have taken up all highly accurate and high-end machines for the best possible solutions for the prodcuts. The facility houses all related testing and measuring equipments to check the machined surfaces and accuracy.
We are currently providing buffing (Steel Finish) type of surface finish for our customers. Further, State-of-art powder coating plant with imported booths and gun is proposed to be set-up by the end of FY 17-18. The capacity of the plant would be around 6 lacs pieces per month. This would cater to our future expansions.


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BioGTS's biogas reactors for the biorefinery plant were delivered by road transport from Central Finland to Säkylä on Wednesday, 15….
BioGTS Ltd., which operates in Jyväskylä and Laukaa, has signed a contract to supply a second biogas plant to China….
The inauguration of the biogas plant of Haminan Energia was held on Monday, 25.4. in Virolahti in cool spring weather….
Our company has been selected as one of the top five candidates for the Central Finland Marketing Deed Award 2016….
BioGTS has won the Kasvu Open Bioeconomy Metsä Group Growth Track, which was held in Espoo on Thursday, September 10….
Is BioGTS Ltd. energy sector's most promising start-up of the year?
An article recently published on the Finnish bioeconomy strategy news forum biotalous.fi is about the BioGTS's production of biogas plants….
The fourth annual Viljo agricultural and forestry exhibition will take place in Kiuruvesi, Finland, next week, on Thursday July 23rd….
BioGTS is pleased to announce some new appointments. This is an evidence of our continued commitment to excellence and growth….
Mr. Kauko Isomöttönen (M.Sc.Tech) has been appointed as BioGTS's Project Manager since January 12th 2015. Kauko has previously worked e.g….
The Mayor of the City of Jyvaskyla, Mr. Markku Andersson, and his official delegation was travelling in China last week….
BioGTS has been selected to deliver a biogas plant to the Sotkamo research station of the MTT Agrifood Research Finland….
BioGTS was featured at Taloussanomat, the Finnish economy journal last week. Please read the article (in Finnish) here….
The Former Ambassador of Finland to China, Mr. Pasi Rutanen, has joined our company as Advisor of the Board. Mr….


text:
A living myth
Seven year cycles
A living faith
The winter's tale
A living cathedral
The visionary messenger
The chapels of the seven tongues
The chapel of Saint Ansgar
The chapel of Saint Boniface
The chapel of Saint Columba
The chapel of Saint Savior
The chapel of Saint Martin
The chapel of Saint Ambrose
The chapel of Saint James
You are not only responsible for what you say, but also for what you do not say.
Martin Luther (1483-1546) monk, priest, member of the Protestant Reformation
Henry Vaughan, noted designer of the Washington Cathedral, designed three chapels at the cathedral: Saint Ansgar (1918), Saint Boniface (1916), and Saint James (1916). His architectural style and detailing is a fusion of fourteenth century English and thirteenth century French, with the resulting effect being twentieth century American.
This chapel contains a section of the columbarium, a room with niches for the storage of funeral urns. Immediately adjacent to the chapel and the columbarium is the baptistry, a room used for baptism. This is a profoundly symbolic use of sacred space with rooms devoted to first and last rites situated side by side.
The chapel is dedicated to people of Scandinavian descent and much of the iconography represents individuals important in Scandinavian history. This includes a series of intricately carved figures of St. Ansgar, Guastavus Adolphus, St. Olaf and Martin Luther found on the wall at the back of the altar.
Saint Ansgar was known as "The Apostle of Scandinavia" and "The Apostle of the North" and he is remembered as a true visionary. Throughout his life, Ansgar experienced waking visions and used those experiences to guide his life choices. He was a very little boy when he reported his first vision, shortly after his mother's death. Ansgar reported seeing his mother in the company of Mary, the mother of Christ. The feast day of Saint Ansgar is February 3.
The chapel of Saint Ansgar, The chapels of the seven tongues, New York, New York. 2013;
View from the chapel of Saint Ansgar into the adjacent baptistry.


text:
Петропавловський район — район, Алтайський край
 Петропавловський район — район, Воронезька область

ru:Петропавловский район

redpajama_set_name: RedPajamaWikipedia


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Senator Derryn Hinch has fallen off the wagon, according to his ex-girlfriend.
36-year-old Natasha Chadwick outed her former flame on Twitter for his drinking habits – five years after being given a lifesaving liver transplant.
'Derryn Hinch has been back on booze 12+ months. So much for respecting organ donors gift of life. Fave wine, cheap Gossips. No loyalty,' she posted on Sunday.
Derryn required a transplant after his alcohol use caused him to develop advanced cirrhosis and liver cancer.
According to The Herald Sun, the former shock jock admits he has enjoyed the odd drink since receiving his transplant, but insists that they were watered down and he sought the permission from his surgeon.
'I've had the odd glass of wine, perhaps one-third wine and two-thirds water, with Natasha at birthdays and at events like Christmas,' the 72-year-old admitted.
However, Natasha claims they only recently shared a glass of wine and she had seen him drink a light beer on the night of his maiden speech, where he humbly acknowledged the sacrifice his donor's had family made to save his life.
'Since then he's been back on it regularly, starting the day with non-alcoholic beer for breakfast, then moving to his Edenvale non-alcoholic wine and then moving on to harder stuff,' a source told the Herald Sun.
'He's even spent an afternoon at a pub in Redfern in Sydney and come home stinking of alcohol,' they added.


Donald Trump is facing widespread social media backlash after he was pictured greeting supporters with a triumphant double fist pump as he arrived to a 9/11 memorial service on Tuesday, the 17th anniversary of the terror attacks.
The president, 72, along with his wife, First Lady Melania Trump, 48, were in Pennsylvania, on their way to pay their respects to the fallen heroes of United Airlines Flight 93.
So the clueless, soulless phony-in-chief arrives on hallowed ground, attempting to portray a decent human being, and fails again. He defiles every space he inhabits.
"I can't believe he'd screw up memorializing 9/11," said nobody.
There is no bottom to @realdonaldtrump's lack of respect and absence of empathy.
Trump has also come under fire in the past for spreading the debunked claim that Muslims in New Jersey were cheering from their rooftops as they watched the Twin Towers collapse. He also said he helped "a little bit" to clear rubble on 9/11, a claim that has been challenged by others, according to The Washington Post.


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Q: WPF RichTextBox ScrollToEnd gets stuck after many lines I have a Rich text box that gets updated very fast, and I am adding a lot of paragraphs per second. Each update I use ScrollToEnd in order to see the latest addition.
After a few seconds it gets stuck. If i remove the  ScrollToEnd it doesn't get stuck. But, I have to manually scroll down, which is not what I want.
What can be the problem ? how can I solve it.
Thanx.

redpajama_set_name: RedPajamaStackExchange


text:
William Wallace Barron (December 8, 1911 – November 12, 2002) was an American Democratic politician in West Virginia. He was the state's 26th Governor from 1961 to 1965.

Life and career
He was born in Elkins, West Virginia.  He attended Washington and Lee University and the West Virginia University Law School. During World War II, he served in the United States Army. In 1949, he was elected mayor of Elkins. He became a member of the West Virginia House of Delegates in 1950 and was re-elected in 1952. He resigned his seat when appointed as Liquor Control Commissioner by Governor William C. Marland subsequent to the 1952 election. He was nominated to Attorney General in 1956.

In 1960, he was elected governor of West Virginia and continued the clean government and civil rights reforms that had been instituted by his predecessor, Cecil H. Underwood.

He died on November 12, 2002, in Charlotte, North Carolina.

Corruption trial and prison
On August 30, 1968, Barron was acquitted of federal charges concerning alleged money kickbacks and rigged state contract schemes in which he and several of his associates were involved. It was later realized that Barron and his wife, Opal Barron, had bribed the jury foreman. Barron was indicted, pleaded guilty and was sentenced to 25 years in prison. He served four years of his sentence.

See also
 Virginia Mae Brown

References

External links
 Biography of William W. Barron
 Inaugural Address of William W. Barron
 William W. Barron's Influence on the West Virginia State Centennial Celebration

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1911 births
2002 deaths
20th-century American lawyers
20th-century American politicians
American politicians convicted of bribery
Democratic Party governors of West Virginia
Disbarred American lawyers
Mayors of places in West Virginia
Democratic Party members of the West Virginia House of Delegates
Military personnel from West Virginia
People from Elkins, West Virginia
Presbyterians from West Virginia
United States Army personnel of World War II
United States Army soldiers
Washington and Lee University alumni
West Virginia Attorneys General
West Virginia University College of Law alumni