Instagram for Chrome is a simple browser extension that allows users to view and interact with their Instagram powered by Google Chrome.
There was no official Instagram site, so it can see and interact with the users fotografijamaInstagram and feedsTheirmobile phones. Still. Instagram for Chrome allows users to view their feed Instagram from the browser. You can "how" and comment on the photos. Users can even access photos and popular with Instagram.
Udoka Instagram for Chrome is good, there are some problemsInterface. That is, in general, the application is nothing more than a copy of the Instagram interface on the phone, but users can not return to the site. You will need to select a page, were on top of the menijuje. While you can view photos in Instagram for Chrome, you can download them and view themIn higher resolution.
Another thing, it should be noted that there is no way to get out of the extension. Of course, vammozhna remove Instagram for Chrome, but this is not an ideal option if you want to switch accounts.
In general, Instagram for Chrome is a good way to see more photosInstagram score, but annoying interface whim quickly.


text:
Q: If $E(Xi'Xi)$ is invertible and $\Sigma>0$, is $E(Xi'\Sigma^{-1}Xi)$ invertible? $\Sigma>0$ means that it is a positive definite matrix, and $E$ is simply the expectations operator over observations. Dimensions are correct. Thanks!

A: Usually there is an additional assumption like $rk(X)=K$ which implies that both $E(X^TX)$ and $E(X^T \Omega^{-1}X)$ are full rank. The empirical averages $E_N(X_i^TX_i)$ and $E_N(X_i^T \Omega^{-1}X_i)$ are invertible provided that $N$ is sufficiently large (by the law of large numbers).

redpajama_set_name: RedPajamaStackExchange


Your music as you want it. In time.
Don't you just hate it when you're listening to music and have to stop in the middle of the song? With inTime, you don't have to! Just connect to Spotify and we'll take it from there.
Paul is taking his 9 year old son on a trip to Memphis. They want to listen to music on the six hour drive, and they want the satisfaction of having a song end exactly when they reach their destination. Paul uses inTime.
inTime was first created at the 2015 Tufts PolyHack. We took first place and are looking forward to continued development.
inTime can integrate with other APIs. Use with Google maps to coordinate behind-the-wheel listening. Combine with video editing software and you'll never have to worry about finding songs to fit your exact video length. In development.
Based on sound mathematical theory to provide optimized times for you. Click here to view full paper.


text:
package com.maxifier.mxcache.config;

import org.slf4j.Logger;
import org.slf4j.LoggerFactory;

import javax.xml.bind.JAXBContext;
import javax.xml.bind.JAXBException;
import java.io.File;
import java.io.StringReader;
import java.net.URL;
import java.util.*;

import static com.maxifier.mxcache.impl.CacheProviderImpl.registerMBean;

/**
 * @author Alexander Kochurov (alexander.kochurov@maxifier.com)
 */
public class MxCacheConfigProviderImpl implements MxCacheConfigProvider {
    private static final Logger logger = LoggerFactory.getLogger(MxCacheConfigProviderImpl.class);

    private final Map<ClassLoader, MxCacheConfig> configs = new WeakHashMap<ClassLoader, MxCacheConfig>();

    public MxCacheConfigProviderImpl(boolean needsMBean) {
        if (needsMBean) {
            registerMBean(new ConfigurationControl(this), "com.maxifier.mxcache:service=ConfigurationControl");
        }
    }

    @Override
    public Rule getRule(Class cls, String group, String[] tags) {
        JaxbRule result = new JaxbRule();
        for (RuleWithSelector rule : getConfig(cls.getClassLoader()).getRules()) {
            if (rule.matches(cls, group, tags)) {
                result.addRuleName(rule.getName());
                result.override(rule);
            }
        }
        return result;
    }

    @Override
    public List<ResourceConfig> getResources() {
        return getConfig(null).getResources();
    }

    public List<RuleWithSelector> getRules() {
        Set<RuleWithSelector> res = new LinkedHashSet<RuleWithSelector>();
        for (MxCacheConfig config : configs.values()) {
            res.addAll(config.getRules());
        }
        return new ArrayList<RuleWithSelector>(res);
    }

    // may be overriden in tests
    MxCacheConfig loadBootstrapConfig() {
        MxCacheConfig config = new MxCacheConfig();
        try {
            JAXBContext context = getJAXBContext();
            Enumeration<URL> configs = ClassLoader.getSystemResources("META-INF/mxcache.xml");
            while (configs.hasMoreElements()) {
                overrideFrom(config, context, configs.nextElement());
            }
            String path = System.getProperty("mxcache.xml.path");
            if (path != null) {
                overrideFrom(config, context, new File(path));
            }
            String url = System.getProperty("mxcache.xml.url");
            if (url != null) {
                overrideFrom(config, context, new URL(url));
            }
            File f = new File("mxcache.xml");
            if (f.exists()) {
                overrideFrom(config, context, f);
            }
        } catch (Exception e) {
            logLoadingError(e, "unknown");
        }
        return config;
    }

    private MxCacheConfig loadConfig(ClassLoader classLoader) {
        if (classLoader == null) {
            return loadBootstrapConfig();
        }
        MxCacheConfig parentConfig = getConfig(classLoader.getParent());
        MxCacheConfig config = new MxCacheConfig();
        config.merge(parentConfig);
        try {
            JAXBContext context = getJAXBContext();
            Enumeration<URL> configs = classLoader.getResources("META-INF/mxcache.xml");
            while (configs.hasMoreElements()) {
                URL url = configs.nextElement();
                if (!parentConfig.hasSource(url)) {
                    overrideFrom(config, context, url);
                }
            }
        } catch (Exception e) {
            logLoadingError(e, "unknown");
        }
        return config;
    }

    private synchronized MxCacheConfig getConfig(ClassLoader classLoader) {
        MxCacheConfig config = configs.get(classLoader);
        if (config == null) {
            config = loadConfig(classLoader);
            configs.put(classLoader, config);
        }
        return config;
    }

    private static void overrideFrom(MxCacheConfig config, JAXBContext context, File file) {
        try {
            MxCacheConfig newConfig = load(context, file);
            newConfig.setSource(file.toString());
            config.merge(newConfig);
        } catch (Exception e) {
            logLoadingError(e, file.toString());
        }
    }

    private static void overrideFrom(MxCacheConfig config, JAXBContext context, URL url) {
        try {
            MxCacheConfig newConfig = load(context, url);
            newConfig.setSource(url.toString());
            config.merge(newConfig);
        } catch (Exception e) {
            logLoadingError(e, url.toString());
        }
    }

    private static void logLoadingError(Exception e, String source) {
        // we don't throw exceptions cause it will cause fatal error and MxCache couldn't be loaded
        logger.error("MxCache configuration loading failed (" + source + ")", e);
    }

    static MxCacheConfig load(JAXBContext context, URL url) throws JAXBException {
        MxCacheConfig res = (MxCacheConfig) context.createUnmarshaller().unmarshal(url);
        res.addSource(url);
        return res;
    }

    static MxCacheConfig load(JAXBContext context, File path) throws JAXBException {
        return (MxCacheConfig) context.createUnmarshaller().unmarshal(path);
    }

    static JAXBContext getJAXBContext() throws JAXBException {
        return JAXBContext.newInstance(MxCacheConfig.class.getPackage().getName());
    }

    public static JaxbRule loadRule(String... xmlRules) throws JAXBException {
        JAXBContext context = getJAXBContext();
        JaxbRule r = new JaxbRule();
        for (String rule : xmlRules) {
            r.override((JaxbRule) context.createUnmarshaller().unmarshal(new StringReader(rule)));
        }
        return r;
    }
}

redpajama_set_name: RedPajamaGithub


text:
Silent Night at the Liberty
Swansea's Liberty Stadium will once again play host to the Silent Night Christmas Carol Service on Tuesday, December 1 (7pm).
Following last year's highly successful event, the evening promises to be a creative mixture of carols, music and Christmas readings.
Swansea City and The Ospreys will take part, alongside a number of special guests - including our matchday host Kev Johns - while music will be provided by the Salvation Army Band, The Ospreys Supporters Choir and Mal Pope.
Tickets are priced at £3, while there will be collection buckets in aid of SwansAid, The Ospreys wheelchair rugby team and Swansea Hope projects.
Tickets are available from the Liberty Stadium or online here
For further information, please contact event organiser Mandy Harvey on 07738238922 or via email at swanseahope@gmail.com.
The Boot Room's 'boot drive' to return for Birmingham match
Swansea City is once again joining up with The Boot Room to host a boot drive at the upcoming Championship match against Birmingham City at the Swansea.com Stadium on February 4.
Club statement | Pyrotechnics
Swansea City reminds supporters the club has a zero tolerance approach to the handling and use of pyrotechnics following an incident at last weekend's Championship fixture against Queens Park Rangers.
Swansea.com Stadium accessibility a lifeline for Sue and McKenzie
Swansea City season ticket holder Sue has found attending matches at the Swansea.com Stadium with grandson McKenzie has helped her manage her Frontotemporal dementia diagnosis.
Swansea City saddened to learn of passing of Peter Davies
Swansea City is saddened to learn of the passing of the club's former player Peter Davies at the age of 86.

redpajama_set_name: RedPajamaCommonCrawl


Rushwar is a web-based strategy game, with real time construction and military simulation. The game features various different buildings, heroes, and items to play with.
Roman Setting: Rushwar is empire-building MMORTS based on and set in the ancient Roman Empire. Players start with a small town and must gradually build it up into a thriving city-state.
Leagues: Players can form alliances and compete with one another in a larger world. Competition and cooperation both take place at the small and large scale, both economically and martially. Choose your allies well, as enemies will make themselves for you.
Story, Daily and Barter Quests: Players can follow along the main storyline at their own pace, learning more about the game's features and higher strategies as they go. Daily quests offer experienced players options for progression and resource-hoarding. Barter quests give options outside of combat and military challenges, breaking up the routine.
This video features gameplay footage from Rush War. It also shows off some cinematic content and music.


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text:
All Fields Collection Title Creator Identifier Subject Type Contributing Institution Transcription Place search for
API and Data Sets
Collections Guide
Institutions Guide
Essays and Exhibits
Introducing Women's Activism
Biographical Profiles
Mourning Lucretia Mott
By James Truitt, Archives Digitization Technician at the Friends Historical Library of Swarthmore College
This post was originally published on 11 November 2019; 139 years since the death of Quaker minister and activist Lucretia Mott (on 11 November 1880). Born in 1793, Mott was a founder of the women's rights movement and a leading advocate for abolition, peace, and other causes.
In a letter to Mott's children shortly after her death (November 14, 1880), her friend Elizabeth Cady Stanton described her legacy thus: "One who has lived eighty-eight years, reflecting ever the sober virtues of the true wife & mother, the earnest reformer, the religious teacher, both in the school room & 'Friends Meeting,' must have a strong influence for good on our young impressible nation."
Stanton added that Mott's spirit would live on, saying that her "great words & deeds, & this holy influence, are immortality. . . . As long as memory serves us, & we can recall your noble mother's nice opinions in all questions, she will live to us though her sweet presence be forever withdrawn."
Mott never gave up her activist spirit, pushing for what she felt was right even in the disposition of her body. A few months before she passed, her son-in-law recorded her feelings about her own death and her wish to be buried plainly in protest of extravagant interments. "My only dread of death," she told him, "is in having to leave so much enjoyment. Even that I do not dread. Indeed I dread nothing. I am ready to stay or to go, but I feel that my faculties are giving way & that it is quite time I should be quietly laid away. Edward, I charge thee that I be put in a simple box. Avoid all show, parade, cost, in box or clothing. It is not right & I want to protest against it. . . . My life has been a simple one, let simplicity mark the last done for me."
Mott's correspondence was digitized for the In Her Own Right project, and is available for browsing.
Belva Ann Lockwood: Vote Belva!
Caroline LeCount & the Ohio Street School
Chloe Cudsworth Littlefield: A Different Type of Activism
Deborah Norris Logan: Historian, Diarist, Republican
Dora Kelly Lewis: Philadelphia's Voice in the Suffrage Movement
Mary Kite: Mary and the Devil
Lucretia Mott: 19th Century Foodie
M. Carey Thomas and Mary Elizabeth Garrett


A change in 'cultural identity' can be quite rapid (think modern Americans who are a mix of almost every nation on the planet) or it can be slow as demonstrated in the Mitchell Case Study which can be downloaded and studied by clicking here. Today, upon commercial ancestral DNA testing males that carries the R-M222 paternal genetic maker will be of American (including African American), Canadian, Australian, and of course of Irish or Scottish origin (to name but a few). Yet they are all the descendants of 'Celts' who lived in Mainland Europe and who began arriving in waves into Britain and Ireland from about 800BC onwards. Their Celtic ancestors would have been indistinguishable from those described by Julius Caesar, and their spread into Britain and Ireland almost completely changed the genetic makeup of both islands, virtually replacing the pre-historic inhabitants (think European impact on the Native Americans). By 500AD the founding ancestor of R-M222 was living near the ancient structure of Aileach near the modern City of Derry/Londonderry in the far Northwest of Ireland. A relatively short time after the R-M222-Adam lived, some of his descendants had spread overland through the Ancient Irish Kingdom of Breifne (Leitrim/Cavan), and into Westmeath and as far south as the Galway and Tipperary borderlands, where they would later contribute to ancient Irish Clans like the O'Reillys, Whelehans, O'Mannions and McEgans.
However, the most notable spread of R-M222 males began with the arrival of the Vikings in Ireland, whose cutting edge seafaring technology allowed the Gaels of Northwest Ireland to spread into what would become 'Galloway' in Southwest Scotland in 1100AD. The Irish Gaels took with them to Southwest Scotland their surnames (Mac), language (Gaelic) and their genetic markers (like R-M222). But by 1600AD they had completely changed! English had become their primary language (Gaelic would be extinct in Southwest Scotland by 1760AD), and although they remained Christian they were now Protestant rather than Catholic, and they regarded themselves as Lowland Scots rather than 'Foreign Gaels' (Galloway means 'land of the foreign Gael,' Galloway was incorporated into the Kingdom of Scotland in 1240AD). However, their distinctive genetic markers (most notably R-M222) remained in the local population, as did their distinctly Gaelic surnames. But, their surnames were not immune from change; with essentially all Gaelic surnames being extensively anglicised! The Mitchell case study illustrates this quite beautifully. Mitchell is a Norman surname common throughout Britain and Ireland, yet the test subject carries the Gaelic R-M222 marker and his DNA places his paternal ancestry among the Irish colonisers of Galloway. By studying his genetic matches it demonstrates how his paternal ancestor moved from the Gaelic world of Galloway into a Norman area of Central Ayrshire, which also corresponded to his ancestor either willingly or unwillingly changing his surname from Gaelic-MacMichael to Norman-Mitchell. In fact his most recent (Scottish) paternal ancestral genetic homeland lies in an area where Mitchells and MacMichaels live (and farm) side by side. Scottish Lowlander Mitchells and MacMichaels would continue to have lived oblivious to their common ancestry if not for modern DNA science.
What will your DNA reveal? If you have not taken a commercial ancestral DNA test but wish to find out what your DNA will reveal then contact Scottish Origenes here (I'll recommend a suitable test). Or if you would like a FREE CONSULTATION on your DNA results then contact Scottish Origenes here.
Remember folks, your DNA test results are generated in a scientific lab, using complex scientific equipment and trained scientists. Interpretation of your scientific results should always be left to a professional scientist. ALWAYS CHECK THAT YOUR GENETIC GENEALOGIST HAS A SCIENTIFIC QUALIFICATION! IT WILL SAVE YOU BOTH TIME AND MONEY!


We understand that paying for a course can be expensive if you are funding it yourself. To register for a course we ask that you pay a deposit of 15% and the full balance to be paid no later than 6 weeks before the course is due to start.
If you require assistance with funding for the Certificate in Culinary Skills, we can offer a number of options for paying in installments. Please contact us on 01202 280050 to discuss your requirements.
If you need help with funding, you can apply for a Professional & Career Development Loan. It is a Government funded loan and you can borrow between £300 and £10,000. The Government will pay the interest whilst you are studying and for one month after you finish. The loan can be used to contribute to course fees, travel and living expenses. Find out if you are eligible here.
The National Careers Service can also help with funding options. They have different routes available depending on your individual situation, including bursaries, loans and grants. To find out more about the types of funding available to you contact the National Careers Service.


In this demo, the player spawns in the first level and tries to reach the exit. When he does, he respawns in the second level, reaches the exit again, spawns back in the first one and so on. The 2 tilemaps are edited manually using the Tiled editor and loaded in game.
It's been a couple of months since the last project update. For this fifth update, I decided to follow up on my previous post about the Tiled map editor. At the time, I implemented a custom tilemap exporter. I refined this text format and implemented a loader in the game engine.
At the moment, the information contained in the tileset (line 6 to 13 in the 2nd map) is not used. It essentially provides information that is directly linked to a specific tile. It will be useful to implement collision with the walls, player interaction with the food and more.
This entry was posted in Project update and tagged alpagos, gamedev on September 15, 2017 by Guillaume Riby.


Luxurious villa built to the highest of standards and located frontline in the prestigious Guadalmina Baja Golf Estate. Just 200 meters from the beach.Orientated to the South West and set in a prime position enjoying views over the private gardens, pool and golf course.
The villa offers a beautiful drive-in courtyard with access to two double garages and villa main entrance hall leading to an elegant modern lounge with fireplace, high wooden beamed ceilings with large windows and interconnecting dining area, magnificent fully fitted kitchen with central island, large open and covered terraces, a laundry room, 2 guest cloakroom, 5 bedrooms with bathrooms en-suite, master suite has His & Hers.
The plot is 3.022 m² and the total built area is 1.018 m² plus a 168 m² terrace and a 90 m² overflow saltwater swimming pool.


I really thought I had already told you guys that you should be reading things by Jeannine Hall Gailey. I met her at WonderCon, and read her book of poems, Becoming the Villainess, which I great enjoyed, and which is full of beautiful poems about life and fairy tales and gender and geeky things and just read them because it will so much more awesome than me trying to tell you about them. But now I can't find where I said those things - shame! So I say now that you really should be reading things by Jeannine Hall Gailey. Like Timekeeping, or "I Forgot to Tell You the Most Important Part..." Excellent, no? Her blog lives over here, in case you missed the link in my "library" over there.
Thanks for the very kind shout out, Elisa!


Obiageli Ezekwesili, the presidential candidate of the Allied Congress Party of Nigeria (ACPN), says the economy under the present administration of the All Progressives Congress (APC) is "disgusting".
According to the former Minister of Education under the Peoples Democratic Party (PDP) administration led by former President Olusegun Obasanjo, bad governance has led to increasing in the poverty index of the country, as many youths are now jobless.
She spoke at a town hall meeting in Akure, Ondo State, as part of activities for her presidential campaign.
Speaking on the "very disgusting" state of the Nigerian economy, as she referred to it, she said: "Nigerians are fed up with the failures of governance, which has become too costly for Nigerians. Just look at us, we are now the world capital of poverty.
We have overtaken India, which is even seven times of our population, as a country with the largest number of poor people. There are 87 million poor people in Nigeria, which is a country of fewer than 200 million people.
In India, there are about 1.3 billion people and there are 79 million people in extreme poverty and that is a shameful situation.
"We cannot allow it to continue. The more that bad governance is given to us by APC and PDP, the more that people drop into poverty. It is time for a change of direction in our country and it is time for us too to take ourselves out of mediocrity and poor performance.
It is time to become a country that is governed on the basis of knowledge and by someone who has character and competence and knows the kind of policies that can grow an economy.
According to Ezekwesili, ACPN is the only alternative that can take Nigerians out of poverty.
"We are not a party for politicians; our party is for the people and we shall focus more on the economy. Also, the government of ACPN will focus on education, instead of oil and will ensure that children of the poor compulsorily attend schools," she added.
She, however, explained that she is contesting in the presidential election to mobilize people for a new Nigeria. Ezekwesili advised Nigerians to be watchful on the antics of attempts by APC to rig the polls.
She continued: "I believe the Nigerian people must be on the alert. I do not trust this APC government of President Muhammadu Buhari. He (Buhari) has not shown himself to be trustworthy with the fact that he refused to sign into law, the electoral amendment bill of 2018.
"He should have taken that bill and signed it into law. If it was signed into law, it means we can believe in the electoral integrity of our elections and electoral umpires. Today, he (Buhari) has basically weakened the capacity of INEC.
So, I do not believe in the peace accord that he signed and if it was in good faith. I don't trust the fact that Amina Zakari, who is the daughter of a man that married the sister of the president is part of the conduct of 2019 election.
"I don't believe a lot of things around the way that the government is trying to execute the 2019 elections, but I believe in the Nigeria people and they would not allow anyone dismantles our hard-won democracy.
This democracy, people died and were also maimed fighting for this democracy. This democracy is greater than President Buhari and his administration. This democracy will survive Buhari.
"The Nigerian people are saying both the APC and PDP governments have failed and they must go, so that we can build a new nation. ACPN too would also do a proper restructuring of the country in order to allow every region to develop its own resources.


Lunar Lander Challenge with Janet's PlanetMake Your Own Lunar Lander Design and build a shock-absorbing system that will protect two "astronauts" when they land. Follow the Design Process to: (1) design and build a shock-absorbing system out of paper, straws, and mini-marshmallows; (2) attach their shock absorber to a paper plate platform; and (3) improve their design based on testing results.
Shock Absorbers aka (folded index card) • Fold an index card into a spring. When you jump off a high step, you bend your back and knees to absorb some of the energy and break your fall. That's what a shock absorber does—absorbs the energy of an impact. Soft things, like marshmallows, cotton balls, foam, and bubble wrap absorb shock well. You can also use paper, like this index card made into a spring by folding it like an accordion.
What kind of shock absorber can you make from these materials to help soften a landing? Marshmallows can serve as soft footpads. Cards can be folded into springs.
Straws can provide a flexible structure. Rubber bands can flex and hold things together. Now make sure the lander doesn't tip over as it falls through the air.
When you make the parts below the platform weigh more than the parts on the top the lander falls straight down. Also, it helps to evenly distribute the weight on top of the platform.
Build, test, evaluate, and redesign.
If the lander tips over when it drops—Move the cup slightly away from the side that's tipping. Or, reposition the parts of the shock-absorbing system to better balance the weight.
Check out the results from a Lunar Lander Challenge with Janet's Planet at the Hendersonville Relay for Life Event.
Potential and kinetic energy—When the lander hits the surface, its motion (kinetic) energy is changed into stored (potential) energy, which gets stored in the shock absorbers.
Acceleration due to gravity—The lander accelerates (speeds up) as it falls due to Earth's gravitational pull.
Air resistance—Air exerts a force on the lander as it falls, slowing it down.
Kids measure the various heights from which they drop the lander. Change the size, position, or the number of shock-absorbing parts.
You can also add mini-marshmallows for landing-pad feet. Or, you can use marshmallows at key junctions in the lander's frame to help absorb energy.
What forces affected your lander as it fell? (It accelerated [sped up] as it fell due to the pull of gravity. Air also pushed on it, and this air resistance slowed it down.) After testing, what changes did you make to your lander? Engineers' early ideas rarely work out perfectly. How do you think testing helps them improve their design?
Testing actually helps you see what works and what doesn't. Knowing this lets you improve a design by fixing the things that aren't working well or could work even better.


text:
\section{introduction}
In 1976, G.B. Seligman \cite{Se} showed that finite dimensional simple Lie algebras containing a nonzero  maximal toral subalgebra have a decomposition as $$(*)\hspace{1cm}\LL=(\gg\ot \aa)\op(\ss\ot \bb)\op(\v\ot \cc)\op\dd$$ in which $\gg$ is a finite dimensional split  simple Lie algebra and
$\v,$ $\ss$ are specific irreducible $\gg$-modules. The Lie algebra structure on $\LL$ induces an  algebra structure on   the vector space $\fb:=\aa\op\bb\op\cc$ and $\dd$ is a Lie subalgebra  of $\LL$ isomorphic to a specific subalgebra of inner derivations of $\fb.$ Following G.B. Seligman, S. Berman and R. Moody  \cite{BM} introduced the notion of a Lie algebra graded by an irreducible  reduced finite root system $R$ which  is  a Lie algebra containing a finite dimensional split  simple Lie subalgebra $\gg$ of type $R$  with a Cartan subalgebra $\hh$  such that $\LL$ has a weight space decomposition with respect to $\hh$ and as an algebra, $\LL$ is  generated by weight spaces corresponding to "nonzero" weights.
Next Allison, Benkart and Gao generalized \cite{BM}'s definition    to non reduced case by letting $\gg$ be a finite dimensional split simple Lie algebra of type $B,C$ or $D$ if $R$ is of type $BC.$
In  \cite{BM}, \cite{BZ} and \cite{ABG}, the authors stated recognition theorems for root graded Lie algebras to classify such  Lie algebras up to centrally  isogeny, indeed any root graded Lie algebra $\LL$ has a decomposition as $(*)$ with a prescribed  algebra structure on the so called coordinate algebra $\fb:=\aa\op\bb\op\cc.$  In 1999, N. Stumme  studied  locally finite  Lie algebras containing a splitting Cartan subalgebra and called such algebras, locally finite split  Lie algebras. She  showed that, a reduced locally finite root system,  the root system of  a locally finite split semisimple Lie algebra, is  the direct union of finite root subsystems  of semisimple types. She also proved that a locally finite split simple Lie algebra is the direct union of finite dimensional simple subalgebras.  Locally finite root systems  appear in the theory of Kac-Moody Lie algebras as well, more precisely, countable irreducible  reduced locally finite root systems are the root systems of infinite rank affine  Lie algebras
\cite[\S 7.11]{K}.  In 2003, O. Loos and E. Neher \cite{LN} introduced    locally finite root systems axiomatically and gave a complete description   of these root systems.  Next Neher \cite{N} generalized the definition of a root graded Lie algebra to Lie algebras graded by a  locally finite root system. In Neher's sense, if $R$ is a locally finite root system, $S$ is a subsystem of $R$ and $\Lam$ is an abelian group,  an $(R,S,\Lam)$-graded Lie algebra is a compatible  $\hbox{span}_\bbbz(R)$-graded and $\Lam$-graded Lie algebra whose support with respect to $\hbox{span}_\bbbz(R)$-grading is contained in $R,$ for  every $0\neq \a\in S,$ the homogeneous space $\LL_\a^0$
contains a so called invertible element and $\LL_0=\sum_{0\neq\a\in R}[\LL_\a,\LL_{-\a}].$ In this work, we first study  the direct union $\cup_{n\in\bbbn}\gg_n$ of finite dimensional simple Lie algebras $\gg_n,$ $n\in\bbbn,$ containing a nonzero maximal toral  subalgebra, these Lie algebras are  $(R,R_{sdiv},0)$-graded Lie algebras (see Definition \ref{def-root}) in Neher's sense for a locally finite root system $R.$ Here we study a class of Lie algebras, called  admissible Lie algebras. An admissible Lie algebra  $\LL$ has  a nonzero toral subalgebra $\hh$ contained in the subalgebra of $\LL$ generated by the weight spaces   such that for any nonzero weight vector $x,$ there is a weight vector $y$ such that $[x,y]\in\hh$ and  $(x,[x,y],y)$ is an $\mathfrak{sl}_2$-triple. We show that if the Lie algebra $\LL$ is locally finite simple Lie algebra whose weight spaces are finite dimensional, then $\hh$ is a maximal toral subalgebra, the root system of $\LL$ with respect to $\hh$ is a locally finite root system and $\LL$ is the direct union of finite dimensional simple subalgebras.
\section{admissible Lie algebras}
Throughout   this work $\bbbn$ denotes the set of all nonnegative
integers and $\bbbf$ is a field of characteristic zero. Unless
otherwise mentioned, all vector spaces are considered over
$\bbbf.$ In the present paper, we denote the dual space of a
vector space $V$ by $V^\star$ and by $GL(V),$ we mean the group of
automorphisms of $V.$
For a matrix $A,$ $tr(a)$ denotes the trace of $A.$ Also for a Lie algebra $\LL,$   we mean by $Z(\LL),$ the center of
$\LL$ and if $\LL$ is finite dimensional, we denote the Killing
form of $\LL$ by $\kappa.$ We also make a convention that for
elements  $x_1,\ldots,x_m$ of a Lie algebra, by an expression of
the form $[x_1,\ldots,x_m],$ we always mean $[x_1,\ldots[x_{m-1},
x_m]\ldots].$
to be zero.
\begin{deft}\label{def-root}
{\rm Let $\v$ be a nontrivial vector space and $R$ be a subset of
$\v,$ $R$ is said to be a {\it locally finite root system in $\v$}
if

(i) $0\not\in R,$ $R$ is locally finite and spans $\v,$

(ii) for every $\a\in R,$ there exists $\check\a\in\v^\star$ such
that $\check\a(\a)=2$ and $s_\a(\b)\in  R$ for $\a,\b\in  R$ where
$s_\a:\v\longrightarrow\v$ maps $v\in\v$ to $v-\check\a(v)\a.$

(iii) $\check\a(\b)\in\bbbz,$ for $\a,\b\in R.$

Set  $R_{sdiv}:=R\setminus\{\a\in R\mid 2\a\in R\}$ and call it  the semi-divisible  subset of $R,$ the root system $R$ is called {\it reduced} if $R=R_{sdiv}.$

}
\end{deft}

Suppose that $R$  is a locally finite root   system. A nonempty
subset $S$ of $R$  is said to be  a {\it subsystem} of $R$ if
$s_\a(\b)\in S$ for $\a,\b\in S.$ Following \cite[\S 2.6]{LN}, we
say two roots $\a,\b$ are {\it connected} if there exist finitely
many roots $\a_1=\a,\a_2,\ldots,\a_n=\b$ such that
$\check\a_{i+1}(\a_{i})\neq 0,$ $1\leq i\leq n-1.$ Connectedness
is an equivalence relation on $R.$  The root system $R$ is the
disjoint union of its connected components. A nonempty subset $X$
of $R$ is called {\it irreducible,} if each two elements $x,y\in
X$ are connected and it is called {\it closed} if for $x,y\in X$
with $x+y\in R,$ we would have $x+y\in X.$ It is easy to see that
each connected component of the locally finite root system $R$ is
a closed subsystem of $R.$ Also using \cite[Cor. 3.15]{LN}, $R$
is a direct limit of its finite subsystems, and if $R$ is
irreducible, it is a direct limit of its irreducible finite
subsystems.
\begin{deft}
{\em Let $\hh$ be a Lie algebra. We say an $\hh-$module $\m$ has a
{\it weight space decomposition with respect to $\hh,$} if
\begin{equation*}
\m=\op_{\a\in \hh^\star}\m_\a \;\hbox{where}\; \m_\a:=\{x\in\m\mid
h\cdot x=\a(h)x;\;\;\forall\; h\in \hh\};\;\a\in\hh^\star.
\end{equation*}
The set $R:=\{\a\in\hh^\star\setminus\{0\}\mid \m_\a\neq \{0\}\}$
is called the {\it set of weights} of $\m$ with respect to $\hh$
and $\m_\a,$ $\a\in R,$ is called a {\it weight space}, also any
element of $\m_\a$ is called a {\it weight vector} of {\it weight}
$\a.$ If a Lie algebra $\LL$ has a weight space decomposition with
respect to a subalgebra of $\LL$ via  the adjoint representation,
the set of weights is called the {\it root system } and weight
spaces are called {\it root spaces.}}
\end{deft}

An easy verification proves the following lemma.
\begin{lem}
\label{rest3} Let $\hh$ be a Lie algebra and $\m$ be  an
$\hh$-module admitting a weight space decomposition
$\m=\op_{\a\in\hh^\star}\m_\a$ with respect to $\hh.$ Let $T$ be a
subalgebra of $\hh$ and take $\pi:\hh^\star\longrightarrow T^\star$ to be
defined by $\pi(\a)=\a\mid_{_T},$ the restriction of $\a$ to $T.$
For $\b\in T^\star,$ define $\m'_\b:=\{v\in\m\mid t\cdot
v=\b(t)v;\;\forall t\in T\},$ then $\m=\op_{\b\in T^\star}\m'_{\b}$
and for $\b\in T^\star,$ $\m'_\b=\displaystyle{\bigoplus_{\{\a\in
\hh^\star,\;\pi(\a)=\b\}}\m_\a.}$
\end{lem}
\begin{deft}\label{split}
{\em  Let $\LL$ be a Lie algebra.  A nontrivial subalgebra $\hh$
of $\LL$ is called a {\it  split toral subalgebra}  if $\LL$ as an
$\hh-$module, via the adjoint representation, has a weight space
decomposition with respect to $\hh.$ One can see  that a split
toral subalgebra of a Lie algebra is abelian. Throughout the
present paper  by a toral subalgebra, we always mean a split toral
subalgebra.}
\end{deft}

Now suppose that $\LL$ is a Lie
algebra containing a nonzero toral subalgebra $\hh$ with
corresponding root system $R.$ The Lie algebra $(\LL,\hh)$ (or
$\LL$ for simplicity) is called {\it admissible } if $\LL$
satisfies the following property:

\begin{equation}\label{con}
\parbox{4.2in}{$\hh\sub\sum_{\a\in R}[\LL_\a,\LL_{-\a}]$ and for  $\a\in R,$  $0\neq x\in\LL_\a,$ there is $y\in\LL_{-\a}$
such that $h:=[x,y]\in\hh$ and $(x,h,y)$ is an
$\mathfrak{sl}_2$-triple. }
\end{equation}

An element $h\in\hh$ is called  a {\it splitting element
corresponding to $\a\in R$} if there are  $x\in\LL_\a$ and
$y\in\LL_{-\a}$ such that $(x, h:=[x,y],y)$ is an
$\mathfrak{sl}_2$-triple.  A subset $\D$ of $R$ is called {\it
connected} with respect to a fix set $\{h_\a\mid\a\in R\}$ of
splitting elements if for any $\b,\gamma\in\D,$ there is a finite
sequence $\a_1,\ldots,\a_n$ of elements of $\D$ such that
$\a_1=\b,$ $\a_n=\gamma$ and $\a_{i+1}(h_{\a_i})\neq 0,$ $1\leq
i\leq n-1.$  A root $\a\in R$ is called {\it integrable} if there
are $e_\a\in \LL_\a,$ $f_\a\in\LL_{-\a}$ such that
$h_\a:=[e_\a,f_\a]\in \hh,$ $(e_\a,h_\a,f_\a)$ is an
$\mathfrak{sl}_2$-triple and that $ad_{e_\a}$ and $ad_{f_\a}$ act
locally nilpotently on $\LL.$ We denote by $R_{int},$  the set of
integrable roots of $\LL$ and note that if $\LL$ is a locally
finite admissible Lie algebra, then $R=R_{int}.$ A subset $\D$ of
$R$ is called {\it symmetric } if $\D=-\D$ and it is called {\it
closed } if $(\D+\D)\cap R\sub \D.$

\begin{exa}\label{exa1}{\rm
Let $\LL$ be  a finite dimensional semisimple Lie algebra
containing a maximal toral subalgebra  $\T.$ Take $\Phi$ to be the
root system of $\LL$ with respect to $\T.$ Using \cite[Lem. I.3
and Lem. I.5]{Se}, we get that $(\LL,\T)$ is an admissible Lie
algebra. Moreover   Corollary to Lemma I.4 of \cite{Se} shows
that for $\b\in\Phi,$ there is a unique $k_\b\in\T$ with
$$
\b(k_\b)=2 \hbox{ and $k_\b=[u,v]$ for some
$u\in\LL_\b,v\in\LL_{-\b}.$}
$$
Using \cite[Lem. I.5 and Lem. I.6]{Se}, we get that
\begin{equation}\label{span}
\T=\hbox{span}_\bbbf\{k_\b\mid\b\in \Phi\}\andd
\T^\star=\hbox{span}_\bbbf\{\b\mid\b\in \Phi\}. \end{equation}
Following the proof of Lemma I.5 of \cite{Se}, we get that if
$\b\in\Phi,$  then  $\kappa(k_\b,k_\b)\neq 0$  and for $t\in\T,$
$\b(t)=2\kappa(t,k_\b)/\kappa(k_\b,k_\b),$ now (\ref{span})
implies that  the linear transformation form $\T$ to $\T^\star$
mapping $t\mapsto\kappa(t,\cdot),$ $t\in\T,$ is onto and so is one
to one, this in particular implies that the Killing form
restricted to $\T$ is non-degenerate and that for $\b\in\Phi,$
$2k_\b/\kappa(k_\b,k_\b)$ is the unique element of $\T$
representing $\b$ through the Killing form. Therefore one
concludes that
\begin{equation}\label{lin}
\parbox{4in}{\begin{center}if $\a,\b_1,\ldots,\b_n\in \Phi$ are such that $\a\in\hbox{span}_\bbbf\{\b_1,\ldots,\b_n\},$ then
$k_\a\in\hbox{span}_\bbbf\{k_{\b_1},\ldots,k_{\b_n}\}.$
\end{center} }
\end{equation}}
\end{exa}
\medskip

{\it From now on we assume $(\LL,\hh)$ is an admissible Lie
algebra with nonempty root system $R.$}

Using $\frak{sl}_2$-module  theory, one can easily prove the following proposition.
\begin{pro}
\label{loc5} Let $\a\in R.$ Take $e\in\LL_{\a}$ and
$f\in\LL_{-\a}$ to be such that $(e,h:=[e,f],f)$ is an
$\mathfrak{sl}_2$-triple and set
$\fg:=\hbox{span}_\bbbf\{e,h,f\}.$
Suppose    $ad_e$ and $ad_f$  act  locally nilpotently on
$\LL,$ then the followings are satisfied:

(i) For $\b\in R,$ $\b(h)\in\bbbz$  and $\b-\b(h)\a\in R.$

(ii) If $\b\in R$ is such that  $\a+\b\in R,$ we have
$[e,\LL_\b]\neq\{0\},$ in particular, $[\LL_\a,\LL_\b]\neq \{0\}.$

(iii) For $\b\in R,$ $\{k\in\bbbz\mid \b+k\a\in R\cup\{0\}\}$  is
an interval.

(iv) For $\b\in R,$ if $\b(h)>0,$ then $\b-\a\in R\cup\{0\}$ and if $\b(h)<0,$ then $\b+\a\in R\cup\{0\}.$

(v) If $k\in\bbbf$ and $k\a\in R,$ then $k\in \bbbz/2.$

(vi)  $\{k\a\mid k\in\bbbf\}\cap
R_{int}\sub\{\pm\a,\pm(1/2)\a,\pm2\a\}.$

\end{pro}

Now use  Proposition \ref{loc5}$(i),(iv)$ together with the same argument as in \cite[\S 10.2]{LN} to prove the following Corollary:

\begin{cor}\label{cor0}
If $R=R_{int,}$ $R$ has the partial sum property, in the sense that if $n\in\bbbn\setminus\{0\}$ and $\b,\a_1,\ldots,\a_n\in R\cup\{0\}$ are such that $\a_1+\cdots+\a_n=\b,$  there is a permutation $\pi$ of $\{1,\ldots,n\}$ such that $\a_{\pi(1)}+\cdots+\a_{\pi(i)}\in R\cup\{0\}$ for all $i\in\{1,\ldots,n\}.$
\end{cor}

\begin{pro}\label{semi-sim1} Suppose that $R=R_{int}.$

(a) For  an ideal $I$ of $\LL,$ set  $ R_{_I}:=\{\a\in R\mid
I\cap\LL_\a\neq \{0\}\},$ then  $ R_{_I}$ is a symmetric closed
subset of $ R$ and  $I= (I\cap\LL_0)\op \sum_{\a\in R}(I\cap
\LL_\a)=(I\cap\LL_0)\op \sum_{\a\in  R_{_I}}\LL_\a.$ Moreover if
$I$ is  simple  as   a Lie algebra, $I\cap\LL_0=\sum_{\a\in
R_{_I}}[\LL_\a,\LL_{-\a}].$
\smallskip

(b) If $\LL$ is semisimple (in the sense that it is a direct sum
of simple ideals), then $\LL_0=\sum_{\a\in R}[\LL_\a,\LL_{-\a}].$
\smallskip

(c) Let  $I$ be an ideal of $\LL.$ Define $\D:=R\setminus  R_{_I}$
and set $J:=\sum_{\a\in \D}[\LL_\a,\LL_{-\a}]\op\sum_{\a\in
\D}\LL_\a.$ If $\LL_0=\sum_{\a\in R}[\LL_\a,\LL_{-\a}],$ then $J$
is an ideal of $\LL,$ $\LL=I+J$ and $I\cap J\sub Z(\LL).$
\end{pro}

\pf $(a)$ Since $I$ is an ideal of $\LL,$ we have
$I=(I\cap\LL_0)\op\sum_{\a\in R}(I\cap\LL_\a)$ by \cite[Pro. 2.1.1]{MP}. Now
if $\a\in
 R_{_I}$ and $0\neq x\in I\cap\LL_\a,$ then by (\ref{con}), there
exists $y\in \LL_{-\a}$ such that $(x,h:=[x,y],y)$ is an
$\mathfrak{sl}_2$-triple. Now as $x\in I,$ one gets that $h,y\in
I$  and for all $z\in\LL_\a,$ $2z=\a(h)z=[h,z]\in I.$ So we have
\begin{equation}
\label{com}R_{_I}=-R_{_I}\andd \LL_\a\sub I;\;\;\a\in R_{_I}.
\end{equation} This  in particular implies that  $I=(I\cap\LL_0)\op\sum_{\a\in
R_{_I}}\LL_\a.$ Now this together with Proposition
\ref{loc5}$(ii)$ implies that $R_{_I}$ is closed. For the last
assertion, one can easily check that $I_c:=\sum_{\a\in
R_{_I}}[\LL_\a,\LL_{-\a}]\op\sum_{\a\in R_{_I}}\LL_\a$ is an ideal
of $I$ and so   we are done as $I$  is simple.
\smallskip

($b$) Since $\LL$ is semisimple, there are an index set $A$ and
simple ideals $\LL^i,$ $i\in A,$  of $\LL$ such that
$\LL=\op_{i\in A}\LL^i.$ For $i\in A,$ use the notation as in the
previous part and set $R_i:=R_{_{\LL^i}},$ since $\LL^i$ is
simple,  part $(a)$ implies  that $\LL^i=\sum_{\a\in
R_i}[\LL_{\a},\LL_{-\a}]\op\sum_{\a\in R_i}\LL_\a.$ Now we are
done as $\LL=\op_{i\in A}\LL^i.$

\smallskip

($c$) Suppose $\a,\b\in R,$ then thanks to   (\ref{com}),   we
have
\begin{equation}\label{closed}\begin{array}{l}
\; [\LL_\a,\LL_\b]\sub\LL_{\a+\b}\cap I\hbox{ if $\a\in  R_{_I}$
or $\b\in R_{_I}$}\\\\ \;[\LL_{\a+\b},\LL_{-\a}]\sub\LL_\b\cap
I\hbox{ if $\a\in R_{_I}$ or $\a+\b\in  R_{_I}$}.\end{array}
\end{equation}
So using  Proposition \ref{loc5}$(ii),$ we have
\begin{equation} \label{clos1}
[\LL_\a,\LL_{\b}]\sub\left\{\begin{array}{ll}
I& \hbox{if $\a,\b\in R_{_I},$}\\
J& \hbox{if $\a,\b\in\D,$}\\
\{0\}& \hbox{if $\a\in R_{_I},\;\b\in\D.$}
\end{array}\right.
\end{equation}
This in particular implies that
\begin{equation}
[\sum_{\a\in R_{_I}}[\LL_\a,\LL_{-\a}]+\sum_{\a\in
R_{_I}}\LL_\a,\sum_{\b\in\D}[\LL_\b,\LL_{-\b}]]=\{0\}.
\label{lc2}\end{equation} Now (\ref{clos1}),(\ref{lc2}) imply that
\begin{eqnarray}
[\LL,J]=[\sum_{\a\in R}\LL_\a+\sum_{\a\in R}[\LL_\a,\LL_{-\a}],J]
&=&[\sum_{\a\in \D}\LL_\a+\sum_{\a\in
\D}[\LL_\a,\LL_{-\a}],J]\nonumber\\
&\sub& [J,J]\sub J\label{ideal}
\end{eqnarray}
which means that $J$ is an ideal of $\LL.$ Also as $\sum_{\a\in
R_{_I}}[\LL_\a,\LL_{-\a}]\sub I,$ one gets that
\begin{equation}\label{dec}\LL=I+J.
\end{equation}

Now suppose that  $x\in I\cap J,$ then $x\in
I\cap\sum_{\a\in\D}[\LL_\a,\LL_{-\a}],$ so by (\ref{lc2}),
\begin{equation}\label{cen1}[x,\sum_{\a\in R_{_I}}\LL_\a\op\sum_{\a\in R_{_I}}[\LL_\a,\LL_{-\a}]]=\{0\}.
\end{equation}
On the other hand since $x\in I\cap\LL_0,$  we have
$$[x,\LL_\a]=\{0\};\;\;\a\in\D.$$
This in particular implies that
$[x,\sum_{\a\in\D}[\LL_\a,\LL_{-\a}]+\sum_{\a\in\D}\LL_\a]=\{0\}.$
Now this together with (\ref{dec}) and  (\ref{cen1})  implies
that $x\in Z(\LL).$  \qed

\begin{cor}
\label{semi-sim} Suppose that $R=R_{int}.$ Set
$\LL_c:=\LL_{0,0}\op\sum_{\a\in R}\LL_\a$ where
$\LL_{0,0}:=\sum_{\a\in R}[\LL_\a,\LL_{-\a}].$ Then
$\LL_{cc}:=\LL_c/Z(\LL_c) $ is a semisimple Lie algebra. In
particular, if $\LL$ is finite dimensional, $\LLc$ is semisimple.
\end{cor}
\pf It follows from  (\ref{con}) that   the canonical projection
map $\LL_c\longrightarrow \LL_{cc}$ restricted to $\sum_{\a\in
R}\LL_\a$ is injective, so we identify $\sum_{\a\in R}\LL_\a$ as a
subspace of $\LL_{cc},$ also it is easy to see that
$(\LL_{cc},\frac{\hh+Z(\LLc)}{Z(\LLc)})$  is an admissible  Lie
algebra whose root system can be identified with  $R.$ More
precisely, we have  the following  weight space decomposition for
$\LL_{cc}$ with respect to $(\hh+Z(\LLc))/{Z(\LLc)}:$
\begin{equation}
\label{llc}\begin{array}{c}\LL_{cc}=(\LL_{cc})_0\op\sum_{\a\in
R}(\LL_{cc})_\a \hbox{ with }\vspace{3mm}\\
 (\LL_{cc})_0=\LL_{0,0}/Z(\LL_c),
(\LL_{cc})_\a=\LL_\a,$ $\a\in R.\end{array}\end{equation}
Moreover we have  $R=R_{int},$ $(\LLcc)_0=\sum_{\a\in
R}[(\LLcc)_\a,(\LLcc)_{-\a}]$ and $Z(\LLcc)=\{0\}.$
 Using
Proposition \ref{semi-sim1}($c$), we get that for any ideal $I$ of
$\LLcc,$ there is an ideal $J$ of $\LLcc$ such that $\LL=I\op J,$
this results in that  $\LLcc$ is semisimple (see \cite[\S
3.5]{J}). Now suppose $\LL$ is finite dimensional and take
$\mathfrak{r}$ to be the solvable radical of $\LLc,$ then
$Z(\LLc)\sub \mathfrak{r}.$ Also as $\mathfrak{r}$ is solvable, we
get  that $\mathfrak{r}/Z(\LLc)$ is a solvable ideal of the
semisimple Lie algebra $\LLcc.$  So $\mathfrak{r}/Z(\LLc)=\{0\},$
i.e., $\mathfrak{r}=Z(\LLc).$ This means that $\LLc$ is a
reductive Lie algebra, now  as $\LLc$ is perfect, one concludes
that $\LLc=[\LLc,\LLc]$ is semisimple. This completes the
proof.\qed

\begin{lem}
\label{rest2} For a  subset  $\D$ of $ R,$ set
$\LL_{_\D}:=\sum_{\a\in\D}[\LL_\a,\LL_{-\a}]\op\sum_{\a\in
\D}\LL_\a$ and $\hh_{_\D}:=\hh\cap\sum_{\a\in
\D}[\LL_\a,\LL_{-\a}].$ Suppose that $\D\sub R_{int}$ is symmetric
and closed and take $\pi_{_\D}:\hh^\star\longrightarrow
\hh_{_\D}^\star$ to be defined by
$\a\mapsto\a_{\mid_{\hh_{_\D}}},$ $\a\in\hh^\star.$ Then
$\pi_{_\D}$ restricted to $\D\cup\{0\}$ is injective. Also
identifying $\D$ with  $\pi_{_\D}(\D),$ we have that
$(\LL_{_\D},\hh_{_\D})$ is an admissible Lie algebra with root
system $\D.$   If moreover   $\D$  is connected with respect to a
fixed  set  of splitting elements, then $\LL_{_\D}$ is  a simple
Lie subalgebra of $\LL.$
\end{lem}
\pf Set $\pi:=\pi_{_\D}.$ Suppose that $\a,\b\in \D\cup\{0\}$ and
$\pi(\a)=\pi(\b).$ We first suppose that $\gamma:=\a-\b\in R,$
then since $\D$ is symmetric and closed, we have  $\gamma\in \D,$
so (\ref{con}) guarantees the existence  of
$t\in\hh\cap[\LL_\gamma,\LL_{-\gamma}]\sub\hh\cap\LL_{_\D}=\hh_{_\D}$
with $\gamma(t)=2,$ so $(\a-\b)(t)=2$ which contradicts the fact
that $\a\mid_{\hh_{_\D}}=\b\mid_{\hh_{_\D}}.$ Next suppose that
$\a-\b\not \in R\cup\{0\},$ then $\a\neq 0$ and $\b\neq0 .$ Since
$\a$ is integrable, there are $e\in\LL_\a,$ $f\in\LL_{-\a}$ such
that $(e,h:=[e,f],f)$ is an $\mathfrak{sl}_2$-triple and $ad_e,$
$ad_f$ act locally nilpotently on $\LL.$ Now since $\a-\b\not\in
R\cup\{0\},$ Proposition \ref{loc5}$(iv)$ implies that
$\b(h)\leq 0.$ This contradicts the fact that $\pi(\a)=\pi(\b)$ as
$h\in\hh\cap[\LL_\a,\LL_{-\a}]\sub\hh_{_\D},$ $\a(h)=2$ and
$\b(h)\leq 0.$ These all together imply that
$\pi\mid_{_{\D\cup\{0\}}}$ is injective. Next we note that since
$\D$ is closed, $\LL_{_\D}$  is a subalgebra of $\LL.$ One also
sees that $\LL_{_\D}$ is an $\hh$-submodule of $\hh$-module $\LL$
admitting the weight space decomposition $\LL_{_\D}=\sum_{\a\in
\D\cup\{0\}}(\LL_{_\D})_\a$ with respect to $\hh$ where
$(\LL_{_\D})_\a=\LL_\a$ for $\a\in \D$ and
$(\LL_{_\D})_0=\sum_{\a\in\D}[\LL_\a,\LL_{-\a}].$ From Lemma
\ref{rest3}, we know that $\LL_{_\D}$ admits a weight space
decomposition
$\LL_{_\D}=\op_{\pi(\a)\in\pi(\D\cup\{0\})}(\LL_{_\D})_{\pi(\a)}$
with respect to $\hh_{_\D}$ where for $\a\in \D\cup\{0\},$
$$(\LL_{_\D})_{\pi(\a)}=\{x\in\LL_{_\D}\mid [h,x]=\a(h)x;\;\forall
h\in\hh_{_\D}\}=\displaystyle{\bigoplus_{\{\b\in
\D\cup\{0\},\pi(\b)=\pi(\a)\}}(\LL_{_\D})_\b}.$$ Now this together
with  the injectivity of $\pi_{\mid_{_{\D\cup\{0\}}}}$  and  the
fact that $(\LL,\hh)$ is an admissible Lie algebra implies that
$(\LL_{_\D},\hh_{_\D})$ is an admissible Lie algebra with root
system $\D.$ The last sentence   follows from Proposition
\ref{semi-sim1}.\qed

\begin{pro}
\label{max-tor} Suppose that $\LL$ is a semisimple Lie algebra and
$\hh$ is a nonzero maximal toral subalgebra of $\LL$ such that
$(\LL,\hh)$ is an admissible Lie algebra with root system
$R=R_{int}.$ Take $I$ to be an index set  such that $\LL=\op_{i\in
I}\LL^i$ where for $i\in I,$ $\LL^i$ is an ideal of $\LL$ which is
simple as a Lie algebra. Set  $R_i:=\{\a\in R\mid \LL_\a\sub
\LL^i\}$ and $\hh_i:=\LL^i\cap\hh,$ $i\in I,$ then

(a) $\hh=\op_{i\in I}\hh_i,$

(b) for $i\in I,$ $\hh_i$ is a maximal toral subalgebra of $\LL^i$
and $(\LL^i,\hh_i)$ is a admissible Lie algebra with root system
$R_i.$
\end{pro}
\pf ($a$) We first mention that for $i,j\in  I,$ we have
\begin{equation}\label{sem}[\LL^i,\LL^j]\sub\d_{i,j}\LL^i.
\end{equation} Using Proposition \ref{semi-sim1}($a$), we get that
for $i\in I,$ $\LL^i=\sum_{\a\in
R_i}[\LL_\a,\LL_{-\a}]\op\sum_{\a\in R_i}\LL_\a,$ and
$\LL_0=\sum_{i\in I}\sum_{\a\in R_i}[\LL_\a,\LL_{-\a}].$ For $i\in
I,$ take $\pi_i:\LL\longrightarrow \LL^i$ to be the natural
projection map and set $T_i$ to be the image of $\hh$ under
$\pi_i.$ Take $T'_i$ to be a subspace of $T_i$ such that
$T_i=(\hh\cap T_i)\op T'_i.$ Let $h\in\hh,$ then  there is a
unique expression $h=\sum_{i\in I}t_i$ with a finitely many
nonzero terms $t_i\in\LL^i,$ $i\in I.$ Now fix $i\in I,$ using
(\ref{sem}), we have for $x\in\LL_\a$ with $\a\in R_i$ that
\begin{equation}\label{ext}
\begin{array}{c}
[\pi_i(h),x]=[t_i,x]=[\sum_{j\in I}t_j,x]=[h,x]=\a(h)x.
\end{array}
\end{equation}
This in particular implies that
\begin{equation*}\label{par2}
\parbox{4in}{if $h,h'\in\hh$  and $\pi_i(h)=\pi_i(h'),$ then   for $\a\in R_i,$ $\a(h)=\a(h')$}
\end{equation*}
which allows us to define $f_{t,\a}$ for $t\in T'_i$ and $\a\in
R_i$ as follows:
\begin{equation}
\label{tor4}
\parbox{4in}{ $f_{t,\a}:=\a(h)$ where $h\in \hh$ is such that $\pi_i(h)=t,$}
\end{equation}
also define $f_{t,\a}:=0$ if $t\in T'_i$ and $\a\in
(R\cup\{0\})\setminus R_i.$
 Next suppose that $\a\in R\cup\{0\}$ and
extend $\a$ to the functional $\a^i\in(\hh\op T'_i)^\star$ defined
as follows:
\begin{equation}
\label{func} \a^i(h+t)=\a(h)+f_{t,\a};\;h\in\hh,\; t\in T'_i.
\end{equation}
Now one can see that $\hh\op T'_i$ is a toral subalgebra of $\LL$
with  corresponding root system $\{\a^i\mid \a\in R\}.$ But
$\hh\op T'_i$ contains the  maximal toral subalgebra $\hh,$
therefore $T'_i=\{0\}$ and so $T_i=T_i\cap\hh$ which implies that
$T_i\sub \hh.$ This gives  that $T_i\sub\hh\cap\LL^i=\hh_i,$ on
the other hand, if $h\in\hh_i,$ then $h=\pi_i(h)\in T_i,$ so we
get that $T_i=\hh_i.$ Now we have $\hh=\op_{i\in I} T_i,$ and so
$\hh=\op_{i\in I}\hh_i.$

$(b)$ Suppose that $i\in I.$ Using Preposition
\ref{semi-sim1}$(a)$ and Lemma \ref{rest2}, we get that
$(\LL^i,\hh_i)$ is an admissible Lie algebra with root system
$R_i.$ Next we note that any toral subalgebra $T_i$ of $\LL_i$
larger than $\hh_i$ would automatically be toral in $\LL$ and
centralizes $\hh_j$ for  $j\in I\setminus\{i\}.$ Now $\hh+T_i$ is
a toral subalgebra of $\LL$ larger than $\hh$ which is a
contradiction. This completes the proof.\qed

\begin{lem}
\label{interesting}  If $(\LL,\hh)$ is a finite dimensional
semi-simple admissible Lie algebra, then $\hh$  is a maximal toral
subalgebra.
\end{lem}
\pf
 Let $\T$ be a maximal toral subalgebra of $\LL$ containing
$\hh.$  We first suppose that $\LL$ is  simple. Since $\T$ is a
toral subalgebra of $\LL,$  $\LL=\sum_{\b\in \T^\star}\LL'_\b$
where for $\b\in \T^\star,$ $\LL'_\b:=\{x\in\LL\mid
[t,x]=\b(t)x;\;\forall t\in \T\}.$ Take $\Phi$ to be the root
system of $\LL$ with respect to $\T.$ Using Example \ref{exa1},
one knows that for $\b\in\Phi,$ there is a unique element
$k_\b\in\T$ satisfying
\begin{equation}\label{property}
\b(k_\b)=2 \hbox{ and $k_\b=[u,v]$ for some
$u\in\LL'_\b,v\in\LL'_{-\b}.$}
\end{equation}

Using Lemma \ref{rest3}, we get  that for $\a\in R\cup\{0\},$
$\LL_\a=\bigoplus_{\b\in A_\a}\LL'_\b$ where
$A_\a:=\{\b\in\T^\star,\;\b\mid_{_\hh}=\a\}.$ Now let $\a\in R$
and $\b\in\Phi$ be such that   $\b\in A_\a.$ Suppose that  $0\neq
x\in\LL'_\b\sub\LL_\a,$ then (\ref{con}) guarantees the existence
of $y\in\LL_{-\a}$ such that $h:=[x,y]\in\hh$ and $(x,h,y)$  is an
$\mathfrak{sl}_2$-triple. Since
$y\in\LL_{-\a}=\bigoplus_{\gamma\in A_\a}\LL'_{-\gamma},$ we have
that $y=\sum_{\gamma\in A_\a}y_{-\gamma}$ with
$y_{-\gamma}\in\LL'_{-\gamma}.$ Now
\begin{eqnarray*}
h=[x,y]=[x,\sum_{\gamma\in A_\a}y_{-\gamma}]= \sum_{\gamma\in
A_\a}[x,y_{-\gamma}]\end{eqnarray*} and so $ h=[x,y_{-\b}] $ as
$h\in\hh\sub\T\sub\LL'_0.$  Also as $h\in \hh,$ $\b(h)=\a(h)=2. $
Now considering (\ref{property}), we get from the uniqueness of
$k_\b$ that $k_\b=h,$ therefore we have proved
\begin{equation}\label{belonh}
k_\b\in\hh;\;\hbox{ for $\b\in\Phi$  with $\b\mid_{_\hh}\neq 0.$}
\end{equation}
Now set $A:=\{\b\in\Phi\mid \b\mid_{_ \hh}\neq 0\}$ and
$I:=\sum_{\b\in A}\LL'_\b\op\sum_{\b\in A}[\LL'_\b,\LL'_{-\b}].$
We claim that $I$ is an ideal of $\LL.$ For this, it is enough to
show that $(\Phi+A)\cap\Phi\sub A.$ Suppose that $\gamma\in\Phi$
and $\b\in A$ are such that
 $\gamma+\b\in\Phi.$ If $\gamma\mid_{_\hh}=0,$ then  $\gamma+\b\mid_{_\hh}\neq0$ and so $\gamma+\b\in A$
 and if
$\gamma\mid_{_\hh}\neq 0,$   then (\ref{lin}) together with
(\ref{belonh}) implies that $k_{\gamma+\b}\in\hh,$ now as
$(\gamma+\b)(k_{\gamma+\b})=2\neq 0,$ we get that
$\gamma+\b\mid_{_\hh}\neq 0,$ i.e., $\gamma+\b\in A.$ This shows
that $I$ is an ideal of $\LL,$ but $\LL$ is simple, so $I=\LL$
which  in particular implies that $A=\Phi.$ Now (\ref{span})
together with (\ref{belonh}) implies that $\T\sub\hh$ and so
$\hh=\T$ is a maximal toral subalgebra. Next suppose that
$\LL=\op_{i=1}^n\LL_i$ where for $1\leq i\leq n,$ $\LL_i$ is a
simple ideal of $\LL.$ So by  Proposition \ref{semi-sim1}($a$) and
Lemma \ref{rest2}, $(\LL_i,\hh\cap\LL_i)$ is a finite dimensional
admissible simple Lie algebra and so using the first part of the
proof, we get that $\hh\cap\LL_i$ is a maximal toral subalgebra of
$\LL_i.$ Now if $\T$ is a maximal toral subalgebra of $\LL$
containing $\hh,$ Example \ref{exa1} gets that $(\LL,\T)$ is a
semi-simple admissible Lie algebra. So Proposition \ref{max-tor}
implies that $\T_i:=\LL_i\cap\T$ is a maximal toral subalgebra of
$\LL_i$ and that $\T=\T_1\op\T_2\op\cdots\op\T_n.$ But
$\hh\cap\LL_i\sub \T_i,$ therefore $\hh\cap\LL_i=\T_i.$ Now  we
have  $\T=\T_1\op\T_2\op\cdots\op\T_n\sub \hh$ and so we are
done.\qed

\begin{pro}\label{simple} (a) Suppose that   $\D$ is a  symmetric closed finite subset of $R_{int}$ such that $\LL_\a$  is finite dimensional for any $\a\in \D,$ then for $\a\in\D,$
there is a unique $h_\a\in\hh$ such that $$\a(h_\a)=2\andd
[x_\a,x_{-\a}]=h_\a\hbox{ for some $x_{\pm\a}\in\LL_{\pm\a}$}.$$
Moreover if  $\a,\b_1,\ldots,\b_m\in\D$ are such that
$\a\in\hbox{span}_\bbbf \{\b_1,\ldots,\b_m\},$ then
$h_\a\in\hbox{span}_\bbbf\{h_{\b_i}\mid1\leq i\leq m\}.$

(b) If  $R=R_{int}$ and all the weight spaces are finite
dimensional, then there is a unique set $\{h_\a\mid \a\in R\}$ of
splitting elements which we refer to as the splitting subset of
$\LL.$
\end{pro}
\pf $(a)$ Set
$\LL_{_\D}:=\sum_{\b\in\D}[\LL_\b,\LL_{-\b}]\op\sum_{\b\in\D}\LL_\b$
and $\hh_{_\D}=\hh\cap\LL_{_\D}.$ Using Propositions \ref{rest2}
and Corollary \ref{semi-sim}, we get that $(\LL_{_\D},\hh_{_\D})$
is a finite dimensional  admissible semi-simple Lie subalgebra
with root system $\D.$ Using  Lemma \ref{interesting}, one  gets
that $\hh_{_\D}$ is a maximal toral subalgebra  of $\LL_{_\D}.$
Now the result follows from  Example \ref{exa1}.

($b$)
 Let $\a\in R$ and take $\D:=R\cap\bbbf\a.$ Using Proposition
\ref{loc5}$(vi)$ together with part ($a$), we are done. \qed

\begin{pro}
\label{bound} Suppose that $(\LL,\hh)$ is a  locally finite
admissible Lie algebra with corresponding root system $R$  such
that all weight spaces are finite dimensional. Take
$\{h_\a\mid\a\in R\}$ to be  the splitting  subset of $\LL$ and
$\v:=\hbox{span}_\bbbf R.$ For $\a\in R,$ define
$\check\a:\v\longrightarrow\bbbf$ by $v\mapsto v(h_\a),$ $v\in\v$
and $s_\a:\v\longrightarrow\v$ by $v\mapsto v-\check\a(v)\a,$
$v\in\v.$ For  $M\sub R,$  set $M_\pm:=M\cup-M$ and take
$\gg_{_M}$ to be the subalgebra of $\LL$ generated by $\cup_{\a\in
M_\pm}\LL_\a,$ finally set $\D_{_M}:=\{\a\in R\mid
\gg_{_M}\cap\LL_\a\neq\{0\}\},$ then we have the followings:

(a) For $M\sub R,$ $\D_{_M}$ is a symmetric closed subset of $R,$
in particular
$\LL_{_{\D_{_M}}}=\sum_{\a\in\D_{_M}}[\LL_\a,\LL_{-\a}]\op\sum_{\a\in\D_{_M}}\LL_\a$
is a Lie subalgebra of $\LL.$

(b) If  $M$ is a  subset of $R$ and $\a,\b\in \D_{_M},$ then
$s_\a(\b)\in\D_{_M}.$ Also if $M$ is a finite subset of $R,$
$\gg_{_M}$ is a finite dimensional Lie subalgebra of $\LL$ and
$\D_{_M}$ is a finite root system in its span where the reflection
based on $\a\in \D_{_M}$ is $s_\a$ restricted to
$\hbox{span}_\bbbf\D_{_M}.$

(c) If $\a,\b\in R,$ then $\b(h_\a)\in\bbbz\cap[-4,4].$

(d) $R$ is a locally finite root system in its span where the
reflection based on $\a\in R$ is $s_\a$ and if $\LL$ is
semi-simple,  the necessary and sufficient condition  for $R$ to
be  irreducible is that $\LL$ is simple. Also for $M\sub R,$
$\D_{_M}$
 is  a closed subsystem of $R$ which is irreducible if $M$ is irreducible.
\end{pro}

\pf $(a)$ We first recall  that as $\LL$ is locally finite, for
any weight vector $x,$ $ad_x$ acts locally nilpotently on $\LL.$
Now we  show that $\D_{_M}$ is symmetric.  One knows that each
element of $\gg_{_M}$ is a sum of elements of the form
$[x_n,\ldots,x_1]$ where $n\in\bbbn\setminus\{0\}$ and for $1\leq
i\leq n,$ $x_i\in\LL_{\a_i},$  for  some $\a_i\in M_{\pm}.$ Now if
$\a\in \D_{_M},$ $\gg_{_M}\cap \LL_\a\neq \{0\},$ so there are
$\a_1,\ldots,\a_n\in M_\pm$ such that $\a=\sum_{i=1}^n\a_i$ and
there are  $x_i\in\LL_{\a_i},$ $1\leq i\leq n,$
 such that $0\neq
x:=[x_n,\ldots,x_1]\in\gg_{_M}\cap \LL_\a.$ We use induction on
$n$ to show that $\D_{_M}$ is symmetric. If $n=1,$ then $\a\in
M_{\pm}$ and so $-\a\in M_{\pm}\sub\D_{_M}.$ Now let
$\a\in\D_{_M}$ be such that there are $n\in\bbbn^{\geq 2},$
$\a_1,\ldots,\a_n\in M_\pm$ and    $x_i\in\LL_{\a_i},$ $1\leq
i\leq n,$  such that $\a=\sum_{i=1}^n\a_i$ and $0\neq
x:=[x_n,\ldots,x_1]\in\gg_{_M}\cap\LL_\a.$ Therefore  $0\neq
[x_{n-1},\ldots,x_1]\in\gg_{_M}\cap \LL_\b$ where
$\b:=\sum_{i=1}^{n-1}\a_i.$ Now  we have
$\b,\a_n,\a=\b+\a_n\in\D_{_M}\sub R.$ Using  the induction
hypothesis, we get that $-\b\in\D_{_M}$ and so there is a nonzero
$y\in\LL_{-\b}\cap\gg_{_M}.$ Now contemplating   Proposition
\ref{loc5}$(ii),$ we have  that $\{0\}\neq
[y,\LL_{-\a_n}]\sub(\gg_{_M}\cap\LL_{-\a_n-\b})$ which shows that
$-\a\in\D_{_M}.$ Next we show that  $\D_{_M}$ is closed. We first
show that
\begin{equation}\label{new1}
\parbox{4.3in}{if $\a\in \D_{_M}$ and $\b\in M_\pm$ are
such that $\a+\b\in R,$ then $\a+\b\in\D_{_M}.$ }\end{equation}
Take $\a\in \D_{_M}$ and $\b\in M_\pm$ to be  such that $\a+\b\in
R$ and x to be a nonzero element of $\gg_{_M}\cap\LL_\a.$ Using
Lemma \ref{loc5}$(ii$), we get that $0\neq
[x,\LL_\b]\sub[\gg_{_M},\gg_{_M}]\sub\gg_{_M}$ and so we are done
as $[x,\LL_\b]\sub\LL_{\a+\b}.$   Now suppose $\a,\b\in \D_{_M}$
and $\a+\b\in R.$  Since $\a,\b\in \D_{_M},$ there is $\{\a_i\mid
1\leq i\leq n\}\cup\{\b_j\mid 1\leq j\leq m\}\sub M_\pm$ such that
$\a=\sum_{i=1}^n\a_i$ and $\b=\sum_{j=1}^m\b_j.$ For $1\leq i\leq
m+n,$ set
$$\gamma_i:=\left\{
\begin{array}{ll}
    \a_i, & \hbox{if } 1\leq i\leq n \\
    \b_{i-n}, & \hbox{if } n+1\leq  i\leq n+m, \\
\end{array}
\right.    $$ then $\gamma_i\in M_\pm,$ $1\leq i\leq m+n$ and
since $\a+\b\in R,$ $\sum_{i=1}^{n+m}\gamma_i\in R.$ Now by
Corollary  \ref{cor0}, there exists a permutation $\pi$ on
$\{1,\ldots,m+n\}$ such that  all partial sums
$\sum_{i=1}^t\gamma_{\pi(i)}\in R\cup\{0\},$ $1\leq t\leq m+n.$
Now using (\ref{new1}) together with an  induction process, we get
that $\a+\b=\sum_{i=1}^{n+m}\gamma_i\in\D_{_M}.$ This means that
$\D_{_M}$ is a closed. This completes the proof.

 $(b)$ Since $\a,\b\in \D_{_M}\sub R,$ by Proposition \ref{loc5}$(i),$
$\check\a(\b)\in\bbbz$ and $s_\a(\b)\in R.$ Now take
$e_\a\in\LL_\a$ and $f_\a\in\LL_{-\a}$ to be  such that
$h_\a=[e_\a,f_\a]$ and $(e_\a,h_\a,f_\a)$ is an
$\mathfrak{sl}_2-$triple, setting  $\theta_\a:=exp(ad_{e_\a
})exp(ad_{-f_\a })exp(ad_{e_\a }),$  we get that $\{0\}\neq
\theta_\a(\LL_\b)\sub(\LL_{s_\a(\b)}\cap\LL_{_{\D_{_M}}})$ and so
$s_\a(\b)\in \D_{_M}.$ Now if $M$ is a finite set, since $\LL$ is
locally finite, $\gg_{_M}$ is a finite dimensional subalgebra of
$\LL$ and so $\D_{_M}$ is finite.  This completes the proof.

$(c)$ Set $M:=\{\a,\b\},$ by part $($b$),$ $\D_{_M}$ is a finite
root system in $\hbox{span}_{\bbbf}\D_{_M}$ where the reflection
based on $\gamma\in\D_{_M}$ is defined by $v\mapsto
v-\check\gamma(v)\gamma,$ $v\in\hbox{span}_{\bbbf}\D_{_M}.$ Using
the finite root system theory, we get  that
 $\{\check\gamma(\eta)\mid
\gamma,\eta\in \D_{_M}\}\sub\bbbz\cap[-4,4],$ in particular
$\b(h_\a)=\check\a(\b)\in\bbbz\cap[-4,4].$

 $(d)$ Using Proposition \ref{loc5}$(i),$ it is  enough to prove that $R$ is locally finite. Suppose
that $M$ is a finite subset of $ R,$ then by part ($a$),
$\D:=\D_{_M}$ is a finite root system in $\hbox{span}_\bbbf\D$
where the reflection based on $\gamma\in\D_{_M}$ is defined by
$v\mapsto v-\check\gamma(v)\gamma$ for $v\in\hbox{span}_\bbbf\D.$
Take $\Pi:=\{\a_1,\ldots,\a_n\}$ to be a base of $\D.$ One knows
that the Cartan matrix  $(\check\a_i(\a_j))$ is an invertible
matrix. So for any choice of $\{k_1,\ldots,k_n\}\sub\bbbf,$ the
following system of equations
$$\sum_{i=1}^n\check\a_j(\a_i)x_i=k_j;\;\; 1\leq j\leq n$$ has a
unique solution. Next suppose that  $(r_1,\ldots,r_n)\in\bbbf^n$
is such that $\eta:=\sum_{i=1}^nr_i\a_i\in R,$ then for any $1\leq
j\leq n,$
$$\sum_{i=1}^nr_i\check\a_j(\a_i)=\sum_{i=1}^nr_i\a_i(h_{_{\a_j}})=\eta(h_{_{\a_j}}).$$
This means that $(r_1,\ldots,r_n)$  is a solution for the
following system of equations
\begin{equation}\label{system}\sum_{i=1}^n\check\a_j(\a_i)x_i=\eta(h_{_{\a_j}});\;\; 1\leq
j\leq n.\end{equation}  But we know that by part ($c$),
$\eta(h_{_{\a_j}})\in [-4,4]\cap\bbbz,$ so there are a finitely
many choice for $\eta(h_{_{\a_j}}),$ $ 1\leq j\leq n.$ This
together with the fact that (\ref{system}) has a unique solution
implies that  there are a finitely many choice for
$(r_1,\ldots,r_n)\in\bbbf^n$ such that $\sum_{i=1}^nr_i\a_i$ is a
root. So $(\hbox{span}_\bbbf M)\cap R=(\hbox{span}_\bbbf\D)\cap
R=(\hbox{span}_\bbbf\{\a_1,\ldots,\a_n\})\cap R$ is a finite set.
Now if $\w$ is a finite  dimensional subspace  of
$\v=\hbox{span}_\bbbf R,$ then there is a finite set $M$ of $R$
such that $\w\sub \hbox{span}_\bbbf M,$ so $\w\cap R\sub
(\hbox{span}_\bbbf M)\cap R$ and so  $\w\cap R$ is a finite set.
This shows that $R$ is a locally finite root system in $\v.$ It
follows easily from  Proposition \ref{semi-sim1}($a$) that if
$\LL$ is semi-simple, $\LL$  is simple if and only if $R$ is
irreducible. Now let $M$ be  a subset of $R,$ parts $(a)$ and
$(b)$ shows that $\D_{_M}$ is a subsystem of $R.$ Next suppose
that $M$ is irreducible but $\D_{_M}$ is not irreducible, then
there is $\a\in\D_{_M}$ such that  $\a$ is connected to none of
the elements of $M,$ in particular,
\begin{equation}\label{conect}
\a(h_\b)=0;\;\;\;\forall\b\in M.
\end{equation}
But   we know there are $\b_1,\ldots,\b_m\in M$ such that
$\a=\sum_{i=1}^m\pm\b_i.$ Take $M':=\{\a,\b_1,\ldots,\b_m\},$ we
know from parts $(a)$ and $(b)$ that  $\D_{_{M'}}$ is a finite
closed  subsystem of $R,$ so Proposition \ref{simple} implies that
$h_\a=\sum_{i=1}^m r_ih_{\b_i}$ for some $r_i\in\bbbf, $ $1\leq
i\leq m.$  This together with (\ref{conect}) implies that
$\a(h_\a)=0$ which is a contradiction. Therefore $\D_{_M}$ is
irreducible. \qed

\bigskip
\begin{pro}
\label{max-tor-ori} Suppose that  $\LL$ is simple and locally
finite such that  all the weight spaces are finite dimensional,
then $\hh$ is a maximal toral subalgebra of $\LL.$
\end{pro}
\pf Take $\{h_\a\mid \a\in R\}$ to be the splitting  subset of
$\LL.$ By Proposition  \ref{bound}$(d)$, $R$ is an irreducible
locally finite root system in its span where the  reflection based
on $\a\in R$ is defined by $v\mapsto v-v(h_\a)\a,$
$v\in\hbox{span}_\bbbf R.$ Now let $T$ be a toral subalgebra of
$\LL$ containing $\hh.$ Take $\Phi$ to be the root system of $\LL$
with respect to $T$ and for $\b\in\Phi\cup\{0\},$ denote by
$\LL'_\b,$ the "weight space" of $\LL$ corresponding to $\b.$ For
$\a\in R\cup\{0\},$ set $A_\a:=\{\b\in\Phi\cup\{0\}\mid
\b\mid_{_\hh}=\a\},$ then by Lemma \ref{rest3}, we have
$\LL_\a=\op_{\b\in A_\a}\LL'_\b.$ Now let $t\in T,$ since $\hh\sub
T$ and $T$ is abelian, $t\in\LL_0,$ but $\LL$ is simple, so
Proposition \ref{semi-sim1}($b$) implies that $t\in\sum_{\a\in
R}[\LL_\a,\LL_{-\a}].$ Therefore there is a finite subset $M$ of
$R$ such that $t\in\sum_{\a\in M}[\LL_\a,\LL_{-\a}].$ Now  take
$\D$ to be  a finite  irreducible closed subsystem of $R$
containing $M$ (see  \cite[Cor. 3.16]{LN}). Using Lemma
\ref{rest2}, we get that
\begin{equation}\label{decom1}\LL_{_\D}=\sum_{\a\in\D}[\LL_\a,\LL_{-\a}]\op\sum_{\a\in \D}\LL_\a=\sum_{\a\in\D}\sum_{\b,\gamma\in
A_\a}[\LL'_\b,\LL'_{-\gamma}]\op\sum_{\a\in \D}\sum_{\b\in
A_\a}\LL'_\b\end{equation} is  an finite
dimensional simple admissible Lie algebra.  Now take
$T_{_\D}:=T\cap\sum_{\a\in\D}[\LL_\a,\LL_{-\a}].$ It is read from
(\ref{decom1}) that $\LL_{_\D}$ has a weight space decomposition
with respect to $T_{_\D},$ in other words $T_{_\D}$ is a toral
subalgebra  of $\LL_{_\D}$ containing $\hh_{_\D}=\hh\cap\LL_{_\D}$
which results in $\hh_{_\D}=T_{_\D}$ contemplating Lemmas \ref{rest2}
and \ref{interesting}. Now we have
$$t\in T\cap \sum_{\a\in M}[\LL_\a,\LL_{-\a}]\sub T\cap
\sum_{\a\in\D }[\LL_\a,\LL_{-\a}]=T_{_\D}=\hh_{_\D}\sub\hh.$$ This
completes the proof.\qed

\medskip

Now one can use Propositions \ref{max-tor-ori}, \ref{bound}($d$)
and  Lemma \ref{rest2}  to get the following theorem:
\begin{thm}\label{direct}
Suppose that $(\LL,\hh)$ is a locally finite  simple admissible
Lie algebra whose weight spaces are finite dimensional. Take $R$
to be the root system of $\LL$ with respect to $\hh. $ Then $R$ is an irreducible  locally finite root system. Next take
$\{R_i\mid i\in I\}$  to be the class of finite irreducible closed
subsystems of $R.$  For $i\in I,$ set $\LL_i:=\sum_{\a\in
R_i}[\LL_\a,\LL_{-\a}]\op\sum_{\a\in R_i}\LL_\a$ and
$\hh_i:=\hh\cap\LL_i.$ Then  $\hh$ is a maximal toral subalgebra of $\LL$ and for $i\in I,$  $\hh_i$ is a maximal
toral subalgebra of $\LL_i,$ also  $(\LL_i,\hh_i)$ is  a finite
dimensional simple admissible Lie algebra with root system $R_i.$
Moreover $\{R_i\mid i\in I\}$ and  $\{\LL_i\mid i\in I\}$ are
directed systems with respect to inclusion,  $R$ is a direct limit
of $\{R_i\mid i\in I\}$ and $\LL$ is a direct limit  of
$\{\LL_i\mid i\in I\}.$
\end{thm}

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- Natural variations in leather adds to the artisan feel of the bag.
- 12" x 16" x 5"
- Multiple storage pockets inside and out for chargers, tablets, files, etc.


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Posted in: Cars. -1 people are following this item.


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Targeting Israeli Apartheid
Posted on March 6, 2010 Categories: BDS News, South African Context, Why Boycott?! Comments are off for this post
by Stephen Lendman / March 6th, 2010
Reports like the Cape Town, South Africa-based Human Sciences Research Council"s (HSRC) May 2009 one titled, "Occupation, Colonialism, Apartheid" highlight what many others understand, including former UN Special Human Rights Rapporteur for Occupied Palestine, John Dugard, stating in January 2007:
"Israel is clearly in military occupation of the OPT (Occupied Palestinian Territories). At the same time, elements of the occupation constitute forms of colonialism and of apartheid, which are contrary to international law."
Article 7(1)(j) of the Rome Statute of the International Criminal Court calls apartheid a crime, stating:
"For the purpose of this Statute, (a) "crime against humanity" means any of the following acts when committed as part of a widespread or systematic attack directed against any civilian population, with knowledge of the attack:
The crime of apartheid" includes murder, extermination, enslavement, torture, arbitrary arrest, illegal imprisonment, denial of the right to life and liberty, cruel, inhuman and degrading treatment, and other abusive acts imposed by one group on another.
In 2008, writing for the Campaign to End Israeli Apartheid, Karine MacAllister said in her article titled, "Applicability of the Crime of Apartheid to Israel" that exclusivism is key to understanding the essence of the Israeli-Palestinian conflict. It:
involves or necessitates the denial of the other; of their presence, rights and existence on the land and reconstruction of the past, namely that the land was empty before the advent of Zionist settlement, hence the movement"s slogan describing "a land without people for a people without land."
As implemented, Zionism"s essence is "a sophisticated legal, social, economic and political regime of racial discrimination that has led to colonialism and apartheid as well as the dispossession and displacement of the Palestinian people." Colonialism flourishes by separating indigenous people from their land and heritage.
Yet Fourth Geneva"s Article 49 states:
Individual or mass forcible transfers, as well as deportation of protected persons from occupied territory to the territory of the Occupying Power or to that of any other country, occupied or not, are prohibited, regardless of the motive." Neither shall "The Occupying Power… deport or transfer parts of its own civilian population into the territory it occupies.
The Convention on the Suppression and Punishment of the Crime of Apartheid (the Apartheid Convention) defines it as:
similar policies and practices of racial segregation and discrimination as practiced in southern Africa (for) the purpose of establishing and maintaining domination by one racial group of persons over any other racial group of persons and systematically oppressing them.
Apartheid is one of the worst forms of racism.
The Convention on the Elimination of Racial Discrimination defines it as:
any distinction, exclusion, restriction or preference based on race, color, descent, or national or ethnic origin which has the purpose or effect of nullifying or impairing the recognition, enjoyment or exercise, on an equal footing, of human rights and fundamental freedoms in the political, economic, social, cultural or any other field of public life.
The 1977 Additional Protocol to the Geneva Conventions (Protocol I) includes among other grave breaches:
"practices of apartheid and other inhuman and degrading practices involving outrages upon personal dignity, based on racial discrimination."
The October 2008 "UNITED AGAINST Apartheid, Colonialism and Occupation DIGNITY & JUSTICE for the Palestinian People" Palestinian Civil Society"s Strategic Position Paper for the April 20-24, 2009 Durban Review Conference called racism and foreign domination the root causes of Palestinian suffering under decades of "settler-colonialism, occupation and institutionalized racial discrimination."
It affirmed the Durban Declaration"s "principles of equal rights and self-determination of peoples and stress(ed) that states must protect such equality as a matter of highest priority."
It acknowledged that "no derogation from the prohibition of racial discrimination, genocide, the crime of apartheid and slavery is permitted (and recognized them as) crimes against humanity (and) major sources and manifestations of racism, racial discrimination, xenophobia and related intolerance (and) wherever and whenever they occurred, they must be condemned and their re-occurrence prevented."
Israeli Apartheid Week (IAW)
Launched in Toronto in 2005 by over 170 Palestinian civil society organizations, it"s an annual series of university lectures, rallies, multimedia events, cultural performances, films, and demonstrations held in cities worldwide to educate people about the nature and destructiveness of Israeli apartheid, and to strengthen the global Boycott, Divestment, and Sanctions (BDS) movement.
From March 1-14, 2010, they"ll be held in 40 cities:
Abu Dis in the West Bank, Amsterdam, Bard (NY), Berkeley (CA), Beirut, Bethlehem, Bogota, Bologna, Boston, Cape Town, Caracas, Chicago, Connecticut, Dundee (Scotland), Durban, Eastern Cape, Edinburgh, Edmonton, Gaza, Glasgow, Guelph (Canada), Hamilton, Houston, Ireland, Jenin, Johannesburg, Kingston, London (Canada), London (UK), Melbourne, Montreal, New York, Ottawa, Oxford, Peterborough (UK), Pisa, Pretoria, Providence, Puebla (Mexico), Rome, San Francisco, Seattle, Sudbury, Tilburg (The Netherlands), Toronto, Utrecht (The Netherlands), Vancouver, Waterloo (Canada), and Winnipeg.
Its supporters call it an expression of Palestinian solidarity, a call to boycott, divest and impose sanctions, and a demand that Israel be held accountable for decades of oppressive occupation, imperial wars, defiling international laws, expropriating Palestinian land, denying self-determination, the right of return, targeted killings, torture, illegal arrests and incarcerations, and the suppression of equal rights and social, political and economic justice.
They"ll also highlight apartheid"s environmental costs, the importance of ending a colonial occupation, and a vision for equality, justice and peace.
On March 2, Al Jazeera headlined, "Israeli Apartheid Week kicks off," explaining the annual event"s "condemnation of the Zionist regime"s suppression of the Palestinians" through protests and a host of related speeches and other activities.
Haaretz ran several articles, including Salman Masalha"s March 3 commentary headlined, "Israel"s apartheid doesn"t stop at the West Bank," saying:
Since Israel"s founding, it hasn"t "kept its promise "to preserve peace and participate in the upbuilding of the State on the basis of full and equal citizenship and due representation in all its provisional and permanent institutions."
Instead, it "continues to conduct itself like a Zionist occupation regime on every inch of the land. True," Israeli Arabs have some free movement "in their homeland and even send representatives to the Knesset (with no power) – but this is the sum of the equality that was formulated and promised," in contrast to the OPT where there"s none.
On the same day Haaretz"s Danna Harman headlined, "Universities across the globe mark Israeli Apartheid Week," highlighting Israeli participants and calling the events "some of the most important (ones) in the Palestine solidarity calendar, according to its organizers.
Harman also quoted Britain"s Jewish Board of Deputies" David Katz calling the participation of Jews in the events "atrocious…. They are free to do as they please, but it"s atrocious. I think they don"t understand the analogy they are making… which is insulting to those who suffered under apartheid."
Jewish South African immigrant Benjamin Pogrund agreed saying "Israelis (taking) part in this week should know better."
The Canadian Ontario legislature "unanimously condemned Israeli Apartheid Week, voting for a resolution that denounced the campus events." Will Ottawa, London and Washington be far behind?
Conservative legislator Peter Shurman told Shalom Life, a Toronto Jewish web site: "The use of the phrase "Israeli Apartheid Week" is about as close to hate speech as one can get without being arrested, and I"m not certain it doesn"t actually cross over the line."
The Canadian Parliamentary Coalition to Combat Anti-Semitism (CPCCA), a voluntary association of 22 MPs exploiting anti-Semitism for political purposes, calls "anti-Zionism… cover for anti-Semitism," and perpetrators should be held criminally liable.
In America, the New York Times was silent, but the Washington Post"s Richard Cohen, an unabashed Israeli flack, said Israeli Apartheid Week reflects anti-Semitism and "imaginary" not "legitimate" grievances "constructed out of lies about the Jewish state… denigrat(ing) the Palestinian cause… Israel has its faults, but it is not motivated by racism."
A Canada National Post commentary headlined, "A festival of bigotry (featuring) rabid expressions of hatred against Israel and its Jewish inhabitants… with extremist speakers whipping crowds into the sort of frenzy one more usually sees in newsreel footage from the streets of Cairo or Gaza City… IAW types don"t care about human rights. They care about smearing the Jewish state."
Pro-Israeli organizations denounced IAW, the Anti-Defamation League (ADL) for one highlighting the "extreme anti-Israel rhetoric… accusations of Israeli racism and apartheid… and allegation that Israel is committing war crimes and genocide against the Palestinian people," the ADL, of course, calling this hateful.
Organizers respond saying "Join us in making 2010 a year of struggle against apartheid and for justice, equality, and peace."
Stephen Lendman lives in Chicago. Contact him at: lendmanstephen[at]sbcglobal.net. Also visit his blog site and listen to The Global Research News Hour on Republic Broadcasting.org Mondays from 11AM-1PM US Central time for cutting-edge discussions with distinguished guests. All programs are archived for easy listening. Read other articles by Stephen, or visit Stephen's website.
Why Americans Should Support BDS
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text:
Q: Change GUC parameter before subquery PostgreSQL I'm currently doing a query like such:
SELECT * FROM (
    (SELECT * from A WHERE first_name % 'fakeFirstName')
    UNION
    (SELECT * from B WHERE last_name % 'fakeLastName')
) AS result;

Both A and B are views of the same underlying table C, with the exact same columns.
However the % operator uses the GUC parameter pg_trgm.similarity_threshold to compute both first_name % 'fakeFirstName' and last_name % 'fakeLastName', and my goal is to change this parameter before each sub query since the similarity thresholds are different for these two column.
To change the pg_trgm.similarity_threshold parameter, for example at the value 0.2, I have two options:

*

*SET pg_trgm.similarity_threshold = 0.2;

*SELECT set_limit(0.2);
I am very concerned with speed of execution, which means I'd prefer to use the % operation with a GIN index instead of the <-> operator with a GIST index.
I tried to do something like the following, but it didn't work since the function set_limit() wasn't called before the use of the % operator:
SELECT * FROM (
    (SELECT *, set_limit(0.2) from A WHERE first_name % 'fakeFirstName')
    UNION
    (SELECT *, set_limit(0.6) from B WHERE last_name % 'fakeLastName')
) AS result;

Any help is deeply appreciated.

A: I'm not sure, if I understand you correctly, but you can try joining A and B against CTE with set_limit(), eg:
t=# with a as (select set_limit(0.21))
, b as (select set_limit(0.22))
select show_limit(), 'a' from pg_database where datname = 't'
union
select show_limit(), 'b' from pg_database join a on true where datname = 't'
union
select show_limit(), 'c' from pg_database join b on true where datname = 't'
t-# order by 2;
 show_limit | ?column?
------------+----------
        0.2 | a
       0.21 | b
       0.22 | c
(3 rows)

also mind you don't change it back, so the value remains the last called set_limit()


Latvian politician and diplomat who served as the first Foreign Minister of Latvia from its independence in 1918 until 1924.
He grew up with his uncle's family in Sabile and then studied at the Riga Polytechnicum.
He was one of the founders of the Latvian Farmers' Union, one of Latvia's oldest political parties and he also served two terms as the Prime Minister of Latvia from June, 1921 to January, 1923 and from June 1923 to January, 1924.
He married Anna Fielhold and they had three children.
He and Huseynaga Sadigov were both diplomats.


text: Q: Gradle 'configuring projects' never ends Clean install of Android Studio 0.8. I create a new project with a blank activity and it will not let me run the app because there appears to be a gradle task still running. All it says is "Gradle: Configure projects" in the status bar of Android Studio and it just sits there. I'm not sure what's wrong but I've been trying to get an android dev environment setup on a couple of machines now and consistently hit this same problem.


text:
What can be done to reduce prosecutorial misconduct?
1 What can be done to reduce prosecutorial misconduct?
2 What does prosecutorial misconduct mean?
3 What are four types of prosecutorial misconduct?
4 Can you sue for prosecutorial misconduct?
5 How do you prove prosecutorial misconduct?
6 How do you prove malicious intent?
7 Can I sue for malicious intent?
8 What is a malicious act?
9 Can you sue police if found not guilty?
10 Can you sue without proof?
11 Can you sue if your case is dismissed?
12 Can you sue the police for emotional distress?
13 How do you deal with police misconduct?
14 What happens when you sue a police department?
15 How do I file a lawsuit against the police department?
Judge Kozinski on reforms that can help prevent prosecutorial misconductRequire open file discovery. Adopt standardized, rigorous procedures for dealing with the government's disclosure obligations. Adopt standardized, rigorous procedures for eyewitness identification. Video record all suspect interrogations.
What does prosecutorial misconduct mean?
Prosecutorial misconduct occurs when a prosecutor breaks a law or a code of professional ethics in the course of a prosecution. First and foremost, it is the prosecutor's job to seek justice and present the judge and jury with facts and legal arguments that result in the conviction of the guilty defendant.
What are four types of prosecutorial misconduct?
Four types of prosecutorial misconduct are offering inadmissible evidence in court, suppressing evidence from the defense, encouraging deceit from witnesses, and prosecutorial bluffing (threats or intimidation).
Can you sue for prosecutorial misconduct?
Prosecutors Can be Sued for Wrongful Convictions: Supreme Court Rules. On , the Supreme Court of Canada released a landmark ruling in Henry v. British Columbia, 2015 SCC 24 that damages in a civil claim may be awarded against the Crown for prosecutorial misconduct, absent proof of malice, under s.
How do you prove prosecutorial misconduct?
To show that prosecutorial misconduct requires dismissal of the indictment or a mistrial, the defendant usually has to show that the prosecutor willfully engaged in misconduct and that the misconduct "prejudiced" the defendant.
How do you prove malicious intent?
To win a suit for malicious prosecution, the plaintiff must prove four elements: (1) that the original case was terminated in favor of the plaintiff, (2) that the defendant played an active role in the original case, (3) that the defendant did not have probable cause or reasonable grounds to support the original case.
Can I sue for malicious intent?
Malicious Intent Unsurprisingly, malice towards the defendant needs to be present. A prosecution may have malice, even the utmost malice, but both malice and an absence of reasonable and probable cause must be satisfied before a Court finds that there has been a malicious prosecution.
What is a malicious act?
The term "Malicious acts" refers to risks of human origin, caused either deliberately or through voluntary lack of action, with the intent to harm a person, organization or property.
Can you sue police if found not guilty?
You may have a claim for the tort of malicious prosecution. If there was not probable cause then you can sue the police and if the complainant lied you can sue them too. You have only 90 days to file a notice of claim Against the police.
Can you sue without proof?
Without a written agreement, the "burden of proof" is on you. You must show that the amount you are claiming is owed to you. The person who owes you money can actually get away without saying a thing. That's because they don't have to prove they're innocent.
Can you sue if your case is dismissed?
If a prosecutor files such a case and the charges are dismissed, the defendant can sue for malicious prosecution and seek financial damages. The law that allows a malicious prosecution suit is aimed at preventing and addressing abuse of the legal process.
Can you sue the police for emotional distress?
Generally, citizens can (successfully) sue the police for infliction of emotional distress in one of two instances, when an officer: intentionally or recklessly acts in a way that causes emotional injury or. causes emotional distress through a negligent act.
How do you deal with police misconduct?
You must approach them in a calm and organized manner.Step 1: Write everything down. This step is extremely important and must be done as soon as possible after the incident. Step 2: Consult with an attorney. This step is essential if you were arrested following the incident. Step 3: File a Police Misconduct Report.
What happens when you sue a police department?
Suing a police officer is serious and takes a lot of work if you want to win your case. The better news is that if the police caused "damages" to you, the court may award you money from the defendant police officer or city to make up for the pain or other problems that the police officer caused you.
How do I file a lawsuit against the police department?
How to Sue the PoliceSpeak to a Civil Rights Lawyer. Preserve Evidence. File Complaints. Speak with a Personal Injury Attorney. File a Notice of Claim. Wait for a Response from the City. File a Lawsuit. Trial.
Can a passport be suspended?
Are school uniforms a good idea essay?
2 Do you need a lawyer to sue the government?
3 Can a state government be sued for negligence?
4 Can a federal employee sue the federal government?
5 Can you sue the king under the Federal Tort Claims Act?
6 What can a lawyer be sued for?
7 Can you sue your attorney?
8 Can the government Sue You?
9 Can states sue the federal government?
Judge Kozinski on reforms that can help prevent prosecutorial…
Require open file discovery.
Adopt standardized, rigorous procedures for dealing with the government's disclosure obligations.
Adopt standardized, rigorous procedures for eyewitness identification.
Video record all suspect interrogations.
Do you need a lawyer to sue the government?
Thus, mistakes and errors of government employees are, by law, the responsibility of their government employer. Because there are many restrictions, limitations, and exceptions to these rules, it is imperative that you consult a lawyer when you are considering bringing a lawsuit against any government entity.
Can a state government be sued for negligence?
Note on State Government Liability for Injury: State governments are entitled to the same sovereign immunity that is enjoyed by the federal government, but every state has also passed its own set of laws (often referred to as a "Tort Claims Act") in which the state has conditionally waived that immunity.
Can a federal employee sue the federal government?
Sovereign immunity has carried over to modern times in the form of a general rule that you cannot sue the government — unless the government says you can. Fortunately, the Federal Tort Claims Act ("FTCA") allows certain kinds of lawsuits against federal employees who are acting within the scope of their employment.
Can you sue the king under the Federal Tort Claims Act?
The Federal Tort Claims Act ("FTCA") Historically, under the doctrine of "sovereign immunity," you were not permitted to sue the king. Sovereign immunity has carried over to modern times in the form of a general rule that you cannot sue the government — unless the government says you can.
What can a lawyer be sued for?
Attorneys may also be sued for committing fraud on their clients. An attorney commits fraud if he makes a misrepresentation that he knows is false with the intent that the client act on it and the client eventually acts on it.
Can you sue your attorney?
A client can sue his or her attorney for negligence, breach of fiduciary duty and for breach of contract. The following are some common grounds for which you may sue your attorney. Your lawyer has abandoned your case. Your case was tossed out of court due to a lack of research and effort on the part of your attorney.
Can the government Sue You?
Can states sue the federal government?
Typically, you can't sue the federal government. However, the Federal Tort Claims Act (FTCA) provides a limited right for private citizens to file a lawsuit in federal court against a federal government agency for negligence or personal injury claims.
How do I make a claim against my housing association?
Who are some important people in the history of representation?


Using a comparative case study approach, the researcher followed 13 instructors for 2 years as they attempted to integrate the Visible Human Dissector, an educational software program, into their undergraduate anatomy courses. Instructors were motivated to use the software as a supplement for limited educational resources and because of its ability to provide students with novel educational experiences. Obstacles in technology access and services as well as organizational factors prevented integration. However, personal hesitancy and lack of confidence, posited to be a major obstacle to integration in the literature, played only a minimal role in slow integration for these instructors. The greatest obstacles to changes in instruction supported by the new technology were difficulties in finding computers to run the software in traditional anatomy laboratories.
Timothy Weston currently works as research associate for the Alliance for Technology, Learning and Society (ATLAS), Institute Evaluation and Research Group at the University of Colorado, Boulder. He received the Ph.D. from the University of Colorado in Research and Evaluation Methodology in Education and a B.A. in Psychology from Metropolitan State University (1993). His interests include the evaluation of educational technology and the use of web-based assessment for formative course evaluation in undergraduate science.
Center for Human Simulation. (2004). Overview. Retrieved March 1, 2004 from www.uchsc.edu/sm/chs/open.html.