Our Vacation Bible school theme this year was Weird Animals from Group. The resounding lesson for Weird Animals VBS is that God loves you no matter if you're weird. In fact a synonym for weird is fantastic! The Lord made us all fantastic in his image.
This is my third summer decorating for vacation Bible school at Grace Lutheran. My girlfriend, Ann, and I accomplished this task together. We brainstormed and decided to do a rain forest in the narthex.
To create this look we used bulletin board paper along the bottom of the wall. We then cut the top of the paper about 3 inches to make the fringe.
Then we adhered previously cut fern leaves to the wall behind the fringe grass. We cut spirals out of pool noodles to create a 3-D effect with the grass. The flowers, butterflies, geckos, and such on were cut from scrapbook paper my cricuit machine.
We attached green table cloths to the ceiling with push pins. The waterfall and tree were created with bulletin board paper and Royal blue door fringe.
The character Fern from Weird Animals VBS lives in the ocean, so we thought it would tie in perfect to create an under water themed area. We selected the entrance to the family life center to decorate this way.
We covered the trophy case to look similar to an aquarium.We strung some blue streamers across the ceiling and added some balloon fish.
Another member of the church, Julie Ann and her mother Carmen, decorated the sanctuary. They did a great job incorporating the decor you can purchase from Group and creating their own. Overall our vacation bible school was decorated beautifully and the children enjoyed themselves immensely. They learned that no matter who they are or the situation, Jesus loves them. That my friends is the most important lesson of them all!
Of course we didn't come up with these decorating ideas all on our own! Pinterest was a huge help to our planning of how best to decorate. Click on over to see our inspiration!
I love the balloon fish! Very cute! Its also a great topic to cover, especially for the kiddos! Have fun!


text:
University students and youths from the Morgan Tsvangirai-led MDC-T attended the rally that also had People's Democratic Party (PDP) leader Tendai Biti as one of the speakers.
Addressing the rally, Mujuru said the economic crisis was a product of President Robert Mugabe's politics of hate.
"Every time he speaks, he is fighting everyone. He is always shouting at western countries," she said.
"Those people have pulled out their money, folded their hands and this is what has taken us where we are today."
Zimbabwe is experiencing a serious economic meltdown due to capital flight and poor foreign direct investment inflows.
Opposition political parties and economists blame Mugabe for the decline, citing inconsistent policies and inability to stamp out corruption.
But the 92-year-old leader has relentlessly accused the west of sabotaging him in order to effect a regime change. He blames western countries' restrictive measures for the economic collapse.
Mujuru said a country does not need enemies in order to develop economically. She said a Zimbabwe People First (ZimPF) government would embrace every country for the purpose of economic development.
"I believe every country can contribute to our economy in its own way for the good of our people," she said.
She said there was need for the country to change its agricultural policy in order to grow the economy as well as empower rural communities whose lives have been depending on the sector.
The former VP also claimed that during her years in government as Mugabe's deputy, she initiated several developmental projects which were sabotaged by fellow government ministers who viewed them as political projects to earn her political mileage for succession purposes.
She also warned her party supporters to be vigilant to avoid being infiltrated by Zanu PF.
Mujuru said she delayed coming to address in her home province because she did not want to have her supporters victimised by Zanu PF members as Mashonaland Central has become notorious for harbouring political thugs.
Several ZimPF supporters who fell victim to political violence after they were beaten or had their homes torched by suspected Zanu PF supporters, were paraded during the rally.
Suspected Zanu PF supporters were yesterday reportedly mounting roadblocks and forcing back some vehicles carrying ZimPF supporters to Mujuru's rally.
Meanwhile, Biti said Mujuru's entry into opposition politics was a blessing because it now gives the opposing forces the liberation war flavour that they lacked in previous elections.
He said opposition parties had failed in their quest to unseat Mugabe because they lacked the liberation war history.
"Most of us are born frees. We did not have the liberation pillar to lean on," Biti said.
"With Mujuru the past and the present equals the future. With the show of unity among opposition parties victory is now certain."
The former Finance minister brought along several PDP officials, including Solomon Madzore and Evelyn Masaiti.
Other ZimPF officials that attended the rally included, Rugare Gumbo, Dzikamai Mavhaire and Didymus Mutasa.-AMH
Latest Zimbabwe land Invasion: Mugabe Ally Physically Pushed 70 year Old Woman In Violent Takeover of 2 000-hectare Ranch
Mugabe fears ouster, attacks judges for allowing demonstrations against his government
Zanu PF Blasts Kasukuwere
WATCH: Zimbabweans hunted like dogs in South Africa

redpajama_set_name: RedPajamaCommonCrawl


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What are the closest stations to Cajero Automatico Coop Jep?
San Francisco is 195 meters away, 3 min walk.
Estévez De Toral is 249 meters away, 4 min walk.
Cenáculo is 249 meters away, 4 min walk.
Which Bus lines stop near Cajero Automatico Coop Jep?
These Bus lines stop near Cajero Automatico Coop Jep: 100, 12, 19, 27.
data-title="Cajero Automatico Coop Jep, Cuenca"
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You can get to Cajero Automatico Coop Jep by Bus. These are the lines and routes that have stops nearby - Bus: 100, 12, 19, 27.
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text:
Private Essential Sistine Chapel Tour, Vatican Museum & St. Peter's Basilica
Roma, Lazio
Tour Guide, Entrance Fees, Headsets
Transportation from your hotel or lodging to the activity check in.
Through E
several hours
Raphael Rooms
Preview Description
There are certain experiences that are guaranteed to change your life. Discover the Sistine Chapel before it opens its doors to the public where you can admire the detail and the figures that appear 3D in how much they are lifelike. Learn about the ideas and passion of Michelangelo, explore the characters he created on the iconic ceiling and in The Last Judgement without battling the crowds on our private Sistine Chapel Tour.
Treat yourself to the luxury of a private guide who will completely customize your visit according to your interests. Get an in-depth understanding of the beauty of Raffaello's Rooms, commissioned by Pope Julius II whilst Michelangelo laboured over the Sistine Chapel ceiling in an intense competition of artistic mastery. Explore the masterpieces of the Vatican Museum's collection, loved by patrons and artists of the Renaissance. Skip the queues and arrive directly at St. Peter's Basilica (except on Wednesday's) where you can appreciate the most important works of the Christian faith. Wonder at the great dome, the touching Pietà by Michelangelo and Bernini's majestic canopy above St. Peter's altar.
Walking Tour.
The works were collected by the church throughout the centuries and the collections include some of the most famous Renaissance masterpieces and sculptures from ancient Rome and Greece.
The Vatican Museums exhibit artworks from the Roman Catholic Church's Impressive collections. The works were collected by the church throughout the centuries and the collections include ... moresome of the most famous Renaissance masterpieces and sculptures from ancient Rome and Greece. The popes were great patrons of art and among the first sovereigns who opened their private art collections to the public.
Several sculptures collected by Pope Julius the second were the first items in what was to become the Vatican Museums collection. The Vatican Museums, as seen today are comprised of several pontifical galleries and museums that were assembled by various popes. Several masterpieces by Caravaggio, Raphael, Fra Angelico, Giotto, Nicolas Poussin the renown Sistine Chapel ceiling painted by Michelangelo, the Raphael rooms, and the impressive collection of maps are all part of the vast collection. They are the main attractions of the Vatican Museums and are considered treasures of immense value. There are also galleries displaying modern Christian religious art. Watch our 3-D virtual tours on www.vatican.com to see with your own eyes some of the Vatican's most astounding treasures including the Sistine Chapel.
The Vatican museums also have changing exhibitions, special programs with lectures on Thursdays and many different kinds of guided tours, some specially suited for children. In the educational tours for children, the children get to have a close look at some artworks and even create small "masterpieces" of their own. In addition The Vatican Museums have tours for the deaf and the blind upon request and pre- booking. If you are in need of these tours you will also get two free tickets for entry.
In order to skip the long entrance queues if you're visiting the Vatican Museums without a guide, it's best to pre-book your Vatican Museum tickets online. When booking online you will need to choose the time of entry. If you haven't booked online it's best to try and come in the afternoon, after 12:00. Take into account that you'll need at least three hours just to see the main attractions. The Vatican museums are the biggest museum complex in the world. Upon entry take into account that you may need to wait about 20 minutes to get through the security check as all visitors need to pass through metal detectors at the security check point. The entrance to the Vatican Museums is on Viale Vaticano.
Opening hours for the Vatican Museums and Sistine Chapel ticket offices are Monday to Saturday: 9 am to 4 pm. The Museums close at 6 pm. You must exit the rooms half an hour before closing time. The Vatican museums are closed on Sundays apart for the last Sunday of every month when there is free entrance to the museums from 9 am to 12.30 pm. On these Sundays the Vatican museums close at 2 pm.
Ticket Tarrifs: The price for a ticket to the Vatican Museums is 16 Euros and reduced price is 8 Euros. The added price for booking your Vatican museum tickets online is 4 Euros.
Public traansportaion to the Vatican Museums
By Metro: line A to Ottaviano-San Pietro Musei Vaticani station
by bus: number 49, stops right in front of the Museums Entrance.492, 990, Via Leone IV / Via degli Scipioni. 32, 81, 982, Piazza del Risorgimento, end of the line.
by tram: 19 to Piazza del Risorgimento.
Tours:
Private Rome in a Day Tour with Colosseum & Sistine Chapel: Essential Experience
Private VIP Vatican In a Day Tour: Experience Art and History As Never Before
All 4 Tours
Attraction, Cultural, Historic, Museum
Bathroom, Drinking water
The Raphael rooms (Stanze di Raphael) are four rooms which were used as the residence of the popes between 1507 and 1585.
The Raphael rooms (Vatican museums) The Raphael rooms (Stanze di Raphael) are four rooms which were used as the residence of the popes between 1507 and 1585. The first pope to use ... morethe Raphael rooms as his residence was Pope Julius II. Pope Gregory XIII was the last Pope to use these rooms as his residence. Today, the Raphael rooms are part of the Vatican museum.
The Raphael rooms form part of the apartment located on the second floor of the Pontifical Palace that was first chosen by Pope Julius II della Rovere as his own residence and was later used by his successors for the same purpose. The decoration of the rooms was done by Raphael and his pupils between 1508 and 1524. Although Raphael died in 1520, his pupils continued with the project until its completion in 1524.
The Raphael rooms are: Room of Constantine, Room of Heliodorus, Room of the Segnatura and Room of the Fire in the Borgo.
Room of Constantine: this is the largest room of the four Raphael rooms. It is the room which is dedicated to the victory of Christianity over paganism. The frescoes contained in this room represent the struggles from the life of the Roman Emperor Constantine. Some of the features in the Room of Constantine include: the Vision of the Cross, the Baptism of Constantine, the Battle of Milvian Bridge and the Donation of Constantine.
Room of Heliodorus: this is the private chamber that symbolizes the heavenly protection provided by Christ to the Church. The room contains four paintings, namely: the Expulsion of Heliodorus from the Temple, the Mass at Bolsena, the Meeting of Pope Leo I and Attila, and the Deliverance of Saint Peter from Prison.
Room of the Segnatura: it was the study room of Pope Julius II. The room symbolizes bringing into harmony the spirits of Antiquity and Christianity. Here is where most of the papal documents were signed and sealed among these were: The Disputation of the Holy Sacrament, the School of Athens, the Parnassus, the Cardinal Virtues, among others.
Room of the Fire in the Borgo: it shows Pope Leo IV making the sign of the cross to put out a raging fire in the Borgo. During the reign of Pope Leo X, this room was used as a music room. Most of the frescos in the room portray events that took place from the lives of Pope Leo III and Pope Leo IV. Some of the paintings in this room include: the Oath of Leo III, the Coronation of Charlemagne, Fire in the Borgo and the Battle of Ostia.
The Raphael rooms are decorated with paintings that make them look magnificent. The rooms are located a few steps away from the Sistine Chapel and St. Peter's Cathedral. The next time you visit Rome, pay a visit to these rooms and you will be astonished by the paintings within them. The Raphael rooms are one of the most visited parts of the Vatican museums and they have a grand history which is significant for the Roman Catholic Church.
Vatican Museum Night Tour
Named after Pope Sixtus IV, the Sistine Chapel is perhaps the most famous chapel in the world. It is renowned both for its function as the Papal Conclave and for its incredible artwork.... more
The ceiling and western wall of the Sistine Chapel are adorned with some of the most astonishing paintings in the world created by the famous artist – Michelangelo Buonarroti.
Michelangelo was commissioned to paint the ceiling of the Sistine Chapel in 1508 and he completed his work in 1512. The middle section depicts nine stories from the Book of Genesis, including the renowned 'Creation of Adam' where the hand of God is seen reaching to Adam to instill him with life.
In 1535 Michelangelo was commissioned once again to paint the western wall over the altar. He completed this spectacular painting of the'Last Judgment' in 1541. The painting depicts the end-of-days and second coming of Christ when according to Christian faith all souls will be judged by Jesus,the Saints and God and sent to heaven or hell.
Despite being one of the most celebrated works of art in history, Michelangelo was originally hesitant to paint the ceiling of the chapel for he considered himself as more of a sculptor than a painter.
The original 'Last Judgment' painting included fully nude human figures, but these were later covered by another artist due to the demands of some Papal members who claimed that the nudity was obscene and improper.
The ceiling of the Sistine Chapel was painted by Michelangelo who depicted nine stories from the book of Genesis including the famous 'Creation of Adam'.
The first three stories are:
1. The Separation of Light from Darkness: Genesis 1:1-5;
2. The Creation of the sun, moon and planets: Genesis 1:11-19;
3. The Separation of Land from Sea: Genesis 1,9-10)
Following these paintings of the Sistine Chapel ceiling in the Vatican are the mid-section patintings:
1. The Creation of Adam & Eve (Genesis 1:26- 2:25), with the figures of man and woman in their nakedness.
2. The Banishment from the Garden of Eden (Genesis 3:22-24).
The last three frescoes of Michelangelo include:
1. The Sacrifice of Noah: (Genesis 8:15-20)
2. The Flood (Genesis 6:5-8,20)
3. The Drunkenness of Noah (Genesis 9:20-27)
* All these show the fall and rebirth of humanity
The ceiling and western wall of the Sistine Chapel are adorned with some of the most celebrated paintings in the world created by the famous artist – Michelangelo Buonarroti. Michelangelo was commissioned to paint the ceiling in 1508 and he completed his work in 1512.
Despite being one of the most adored works of art in history, Michelangelo was originally hesitant to paint the ceiling of the chapel for he saw himself as more of a sculptor than a painter. The original 'Last Judgment' painting included fully nude human figures, but these were later covered by another artist due to the demands of some Papal members who claimed that the nudity was obscene and improper.
The middle section of the ceiling depicts 9 stories fromthe Book of Genesis, including the iconic 'Creation of Adam' where the hand ofGod is seen reaching to Adam to give him life. In 1535 Michelangelo was commissioned once again to paint thewestern wall over the altar.
Michelangelo completed this spectacular painting of the'Last Judgment' in 1541. The painting depicts the end-of-days and second comingof Christ when according to Christian faith all souls will be judged by Jesus,the Saints and God and sent to heaven or hell.
Rome in Two Days Tour: VIP Immersive Experience
St. Peter's Basilica - Entrance Gate
Erected in honor of the Prince of Apostles
The Latin inscription above the entrance states: "Paul V Borghese, Roman, Pontiff, in the year 1612, the seventh of his pontificate, erected in honour of the Prince of Apostles".... more
The entrance to Saint Peter's Basilica has spitirual significance as it is written:
"I am the gate. Whoever enters through me, will be safe. He will go in and out, and find pasture".
(John 10:9)
Above the entrance to St. Peter's Basilica is the central balcony where the election of the new pope is announced - the famous Latin saying: "Habemus Papum" which means "We have a Pope". It is also where the Pope gives the Urbi et Orbi blessing, and for this reason the balcony is also called the "Loggia of the Blessings",
Attraction, Church, Historic, Museum
2 People Private Tour
First 2 guests
Additional Guest
Car Transfer
Weekly on Monday, Tuesday, Thursday, Friday, Saturday 1:30 PM
Custom policy
Cancellations made 30 days before will be fully refunded except for a service fee of 3%.
Cancellations made 8 days before will be refunded 50% of the amount paid.
Cancellations made at a later date will not be refunded.
Was the tour accurately described on the webiste? In other words did it meet your expectations?
100 Viale Vaticano, Roma, Città Metropolitana Di Roma, Lazio, Italy 00192


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text:
Q: Did the initiation of the Higgs Field and the inflationary period occur simultaneously? My question is: Both of these concepts involve a phase transition and they both started, as far as I know,  early in time, so are the above two ideas linked in any way?
My first source for this query is Inflationary Period

The inflationary epoch lasted from 10$^{−36}$ seconds after the Big Bang to sometime between 10$^{−33}$ and 10$^{−32}$seconds. Following the inflationary period, the Universe continues to expand, but at a less rapid rate.

My second source is from Lisa Randall, who states, and I paraphrase her a bit here:

Early on, [After the Big Bang] particles had no mass, but later a phase  transition occurred that gave some of the particles mass (i.e. the Higgs field turned on).

My assumptions about these effects are that the inflation idea solves the horizon problem and the flatness problem and that the  Higgs Field provides mass, although  I am aware that, in a proton for example, most of the mass does not come from the Higgs  field, but from the "quark sea", as Randall puts it.
My initial guess is that they are connected in some way, but the probability of this diminishes, imo, if they did not occur simultaneously.
I self study, but I think I am at a level that, if I am lucky enough to receive any answers, I probably won't be able to completely follow them immediately. However,  I  may be able to understand them after a period of related reading, so being pushed a bit on implicit assumed knowledge is ok by me.

A: I expect you are familiar with the Big Bang model, seen here . It is a mathematical model using mathematical solutions from General Relativity and  the Standard Model of particle physics .
The BB developed to describe astronomical observations and the SM developed to describe particle physics observations. The SM describes how particles/nature behaves as the energy available gets larger, and the Big Bang describes the available energy as the time from the origin of the BB increases.
The electroweak symmetry is unbroken until a certain energy in  the interactions, as the energy for particle interaction gets lower  the Higgs field appears,  the particles get masses and nucleation can later start. This happened

Between 10^−12 second and 10^−6 second after the Big Bang

Before the Cosmic Microwave Background  data came out, and the horizon problem appeared,  the electroweak symmetry breaking was the phase transition appearing in the Big Bang chronology.
The quantization of gravity#  hypothesis for the very early universe could model the uniformity of the observed CMB spectrum. The inflationary period was introduced to model the CMB , in order to be consistent with the observation of a uniformity in the universe that could not be explained by thermodynamic arguments at other periods, except immediately after the BB.
The two phase transitions, (end of inflation period, electroweak breaking) are connected by the diminishing of the available energy per particle due to the expansion, but are independent of each other, the electroweak with the Higgs field appearing much later in the chronology of expansion.

# Please note that the quantization of gravity at the moment is an effective theory. There is research going on , no definitive answers.

redpajama_set_name: RedPajamaStackExchange


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the folded like his mother saying that. and jump anymore this mother came back. that's good said the mayor I was. length to keep me entertained and. study at university either despite doing. basically get money any way possible and.


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Household chlorine bleach and medicine dropper; when diluted, nine parts water to one part bleach, bleach can be used as a disinfectant. Or in an emergency, you can use it to treat water by using 16 drops of regular household liquid bleach per gallon of water. Do not use scented, color safe, or bleaches with added cleaners.


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text:
Q: How to adjust a table generation in a shiny app I can't generate the table correctly when I click in Table. The following message appears:
Warning: Error in f1: argument "var1" is missing, with no default.
I can't solve it, could you help me?
Executable code below:
library(shiny)
library(shinythemes)
library(dplyr)
library(tidyverse)
library(lubridate)
library(shinyWidgets)

function.test<-function(){

  df1 <- structure(
    list(date1= as.Date(c("2021-06-26","2021-06-26","2021-06-26","2021-06-26")),
         date2 = as.Date(c("2021-06-27","2021-07-01","2021-07-02","2021-07-03")),
         Category = c("ABC","ABC","ABC","ABC"),
         Week= c("Saturday","Wednesday","Thurday","Saturday"),
         DR1 = c(5,4,1,1),
         DR01 = c(8,4,1,0), DR02= c(7,4,2,0),DR03= c(6,9,5,0),
         DR04 = c(5,5,4,0),DR05 = c(5,5,4,0),DR06 = c(7,5,4,0),DR07 = c(2,5,4,0),DR08 = c(2,5,4,0)),
    class = "data.frame", row.names = c(NA, -4L))

  return(df1)

}

f1 <- function(df1, dmda, CategoryChosse, plot = FALSE, var1, var2, gnum=0) {

  x<-df1 %>% select(starts_with("DR0"))

  x<-cbind(df1, setNames(df1$DR1 - x, paste0(names(x), "_PV")))
  PV<-select(x, date2,Week, Category, DR1, ends_with("PV"))

  med<-PV %>%
    group_by(Category,Week) %>%
    summarize(across(ends_with("PV"), median))

  SPV<-df1%>%
    inner_join(med, by = c('Category', 'Week')) %>%
    mutate(across(matches("^DR0\\d+$"), ~.x +
                    get(paste0(cur_column(), '_PV')),
                  .names = '{col}_{col}_PV')) %>%
    select(date1:Category, DR01_DR01_PV:last_col())

  SPV<-data.frame(SPV)

  mat1 <- df1 %>%
    filter(date2 == dmda, Category == CategoryChosse) %>%
    select(starts_with("DR0")) %>%
    pivot_longer(cols = everything()) %>%
    arrange(desc(row_number())) %>%
    mutate(cs = cumsum(value)) %>%
    filter(cs == 0) %>%
    pull(name)

  (dropnames <- paste0(mat1,"_",mat1, "_PV"))

  SPV <- SPV %>%
    filter(date2 == dmda, Category == CategoryChosse) %>%
    select(-any_of(dropnames))

  if(length(grep("DR0", names(SPV))) == 0) {
    SPV[head(mat1,10)] <- NA_real_
  }

  datas <-SPV %>%
    filter(date2 == ymd(dmda)) %>%
    group_by(Category) %>%
    summarize(across(starts_with("DR0"), sum)) %>%
    pivot_longer(cols= -Category, names_pattern = "DR0(.+)", values_to = "val") %>%
    mutate(name = readr::parse_number(name))
  colnames(datas)[-1]<-c(var1,var2)
  datas$days <- datas[[as.name(var1)]]
  datas$numbers <- datas[[as.name(var2)]]

  datas <- datas %>%
    group_by(Category) %>%
    slice((as.Date(dmda) - min(as.Date(df1$date1) [
      df1$Category == first(Category)])):max(days)+1) %>%
    ungroup

  m<-df1 %>%
    group_by(Category,Week) %>%
    summarize(across(starts_with("DR1"), mean))

  m<-subset(m, Week == df1$Week[match(ymd(dmda), ymd(df1$date2))] & Category == CategoryChosse)$DR1

  if (nrow(datas)<=2){
  val<-as.numeric(m)
  }

  else{
    mod <- nls(numbers ~ b1*days^2+b2,start = list(b1 = 0,b2 = 0),data = datas, algorithm = "port")
    coef<-coef(mod)[2]
    val<-as.numeric(coef(mod)[2])
  }

  if(plot){

  maxrange <-  range(0, datas$numbers, na.rm = TRUE)
  maxrange[2] <- maxrange[2] + 10

  if (gnum) maxrange[2] <- maxrange[2] + 40
  max<-max(0, datas$days, na.rm = TRUE)+1
  limx = c(0,max)
  limy = c(0,maxrange[2])

  plot(numbers ~ days,  xlim= limx, ylim= limy, xlab = var1, ylab=var2,
  xaxs='i',data = datas,main = paste0(dmda, "-", CategoryChosse))

    if (nrow(datas)<=2){
      abline(h=m,lwd=2)
      points(0, m, col = "red", pch = 19, cex = 2, xpd = TRUE)
      text(.1,m+ .5, round(m,1), cex=1.1,pos=4,offset =1,col="black")

    }

    else{
      new.data <- data.frame(days = with(datas, seq(min(days),max(days),len = 45)))
      new.data <- rbind(0, new.data)
      lines(new.data$days,predict(mod,newdata = new.data),lwd=2)
      points(0, coef, col="red",pch=19,cex = 2,xpd=TRUE)
      text(.99,coef + 1,max(0, round(coef,1)), cex=1.1,pos=4,offset =1,col="black")

    }

  }

  return(val)

}

ui <- fluidPage(

  shiny::navbarPage(theme = shinytheme("flatly"), collapsible = TRUE,
                    br(),
                    tabPanel("",
                             sidebarLayout(
                               sidebarPanel(
                                 uiOutput("date"),
                                 uiOutput("mycode"),
                                 uiOutput('daterange')),

                               mainPanel(
                                 tabsetPanel(
                                 tabPanel("Graph1", plotOutput("graph",width = "100%", height = "600")),
                                 tabPanel("Table", dataTableOutput('table'))
                               )
                             ))
  )))

server <- function(input, output,session) {

  data <- reactive(function.test())

  output$date <- renderUI({
    req(data())
    all_dates <- seq(as.Date('2021-01-01'), as.Date('2021-01-15'), by = "day")
    disabled <- as.Date(setdiff(all_dates, as.Date(data()$date2)), origin = "1970-01-01")
    dateInput(input = "date2",
              label = h4("Choose"),
              min = min(data()$date2),
              max = max(data()$date2),
              value = NA,
              datesdisabled = disabled)

  })

  output$mycode <- renderUI({
    req(input$date2)
    df1 <- data()
    df2 <- df1[as.Date(df1$date2) %in% input$date2,]
    selectInput("code", label = h4("Category"),choices=unique(df2$Category))
  })

  output$graph <- renderPlot({
    req(input$date2,input$code)
    var1 = "Days"
    var2 = "Numbers"
    f1(data(),as.character(input$date2),as.character(input$code),var1,var2,plot=TRUE)
  })

  output$daterange <- renderUI({
    dateRangeInput("daterange1", "Period you want to see:",
                   start = min(data()$date2),
                   end   = max(data()$date2))
  })

  data_subset <- reactive({
    req(input$daterange1)
    days <- seq(input$daterange1[1], input$daterange1[2], by = 'day')
    df1 <- subset(data(), as.Date(date2) %in% days)
    df2 <- df1 %>% select(date2,Category)
    Test <- cbind(df2, coef = apply(df2, 1, function(x) {f1(data(),x[1],x[2])}))
    Test
  })

  output$table <- renderDataTable({
    data_subset()
  })

}

shinyApp(ui = ui, server = server)

A: You have many parameters in your function f1.  However, you are calling f1(data(),x[1],x[2]) and that seems to be the issue.  If you call with the other required paramters, it works fine. See below.
  data_subset <- reactive({
    req(input$daterange1)
    var1 = "Days"
    var2 = "Numbers"
    days <- seq(input$daterange1[1], input$daterange1[2], by = 'day')
    df1 <- subset(data(), as.Date(date2) %in% days)
    df2 <- df1 %>% select(date2,Category)
    Test <- cbind(df2, coef = apply(df2, 1, function(x) {f1(data(),x[1],x[2],plot=FALSE,var1,var2)}))
    Test
  })


text:
Game of Thrones Links
Who would win in a fight between Tolkien characters and Game of Thrones characters? GRRM weighs in
There's no سوال that GRRM's A Song of Ice and آگ کے, آگ کتابیں owe a huge debt to the works of J.R.R. Tolkien, but what would happen if Westeros declared war on Middle Earth? Martin himself weighs in on the ultimate high-fantasy smackdown.
کیا پیش سے طرف کی Saejima پہلے زیادہ سے سال ایک
Who'll Be the Three Dragon Riders on 'Game of Thrones'?
کیا پیش سے طرف کی Blacklillium پہلے زیادہ سے سال ایک
Who's Kinvara On 'Game Of Thrones'? The سیکنڈ Red Woman Is New On The Scene
Who's on whose side now in Game Of Thrones?
We break down all the allegiances at the end of Season 6.
Who's The Prince Who Was Promised On 'Game Of Thrones'? Jon Snow May Fulfill A Major Prophecy
Who's Dead and Who's Alive on Game of Thrones? A Quiz.
The fifth season of HBO's Game of Thrones is upon us. Winter is still coming. Millions will tune in for مزید of Tyrion's witty ripostes and for some مزید smoldering romance between Jon Snow and his Wildling paramour Ygritte.
کیا پیش سے طرف کی nermai پہلے زیادہ سے سال ایک
Who's Got the Best Beard on Game of Thrones?
We asked the cast to find out.
Why 'Game of Thrones' & 'The Walking Dead' Can't Fool Us
Why Arya Can Never Die On 'Game Of Thrones'
Why Arya Is The 'Game Of Thrones' Character Everyone Should Be Paying The Most Attention To
Game of Thrones کلب متعلقہ
A Song of Ice and آگ کے, آگ


text:
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Premier League round up, February 15th 2017
A chink of light in the title race?
Chelsea falter at fortress Burnley
The weekend brought another interesting round of results from the Premier League, not least, Chelsea being held to a 1-1 draw at Turf Moor on Sunday as Burnley's incredible home run continued.
The result may suggest that The Blues had a bit of an off-day, and there is an argument that the inclusion of a more attack-minded midfielder in Cesc Fabregas would have had an impact, but full credit must go to the locals who fought hard and were, at the end, fully deserving of a share of the spoils with the league leaders.
Victory would have moved the Londoners a dozen points clear at the top of the division, and although they dominated the play, and got themselves ahead after only seven-minutes through a Pedro opener, they found The Clarets a proud, and determined, opponent. Home fans may have feared the worst when watching their side fall behind so early in the afternoon, but record signing Robbie Brady proved his worth when scoring a spectacular free-kick to beat Courtois. Stats fans will be interested to know that is the first time the Chelsea keeper has been beaten by a direct free-kick in the Premier League.
Burnley find the formula for Premier League stability
Burnley manager Sean Dyche spoke to the press at full-time and thought it best to start by praising Chelsea and the way they play football. He called them a fine side, but his boys worked hard to limit their threat, while creating four or five really good opportunities of their own.
Burnley are now 12th in the table, with a recent record of two wins, two defeats and a draw from five, and have ensured their place in the Premier League for another season at least. That would have surely been their number one pre-season aim.
Title heading to the Bridge despite 2 dropped points
Chelsea hold their pole position, and are expected to stay there for the remainder of the season, and have an eight point advantage over nearest rivals Manchester City. The boys from Stamford Bridge are currently no better than 2/11 with bookmakers 888Sport to win the league. That is value however, when placed alongside William Hill's 1/7.
Top six in tight Champions League qualification battle
Liverpool remain out of the top-four Champions League places, but they caught everyone off-guard on the weekend when beating high-flying Tottenham 2-0 at Anfield on Saturday. Going in on the back of a five-match winless run in the division, Klopp's side were expected to struggle. The Reds never looked in danger of losing however and claimed the points courtesy of a stunning quick-fire double from Sadio Mane, who rifled home on 16 and 18-minutes to leave Spurs shell-shocked.
Manager Jurgen Klopp told fans after the match that he got the reaction he was desperate for from his players. He labelled it an outstanding offensive performance in the first-half, and an even better defensive one in the second-half. Liverpool are now 8/11 with bet365 to finish in the top-four, if you fancy them strongly enough to invest some of your cash.
Man City second after easy win
On Monday, Manchester City came through a tricky test away to Bournemouth with flying colours. The Citizens got the win needed, courtesy of goals from Sterling and a Mings OG, one either side of half-time, to move up to second in the table. The win did come at a price however as attacker Gabriel Jesus broke down with injury and his time on the park was cut-short at 14-minutes, replaced by Aguero. A metatarsal injury was diagnosed and he looks likely to miss the majority of the remaining season.
Previous Post is ‹ Premier League round up, February 8th 2017
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text:
Q: About Russell's paradox and Russell's anti-sets It is written on Wikipedia:

According to naive set theory, any definable collection is a set. Let R be the set of all sets that are not members of themselves. If R is not a member of itself, then its definition dictates that it must contain itself, and if it contains itself, then it contradicts its own definition as the set of all sets that are not members of themselves. This contradiction is Russell's paradox. Symbolically:
${{\text{Let }}R=\{x\mid x\not \in x\}{\text{, then }}R\in R\iff R\not \in R}$

I would like to define Russell's anti-set as:

A set $R$ is a Russell's anti-set if and only if it contains some sets that are members of themselves.

If we allow, in some set-theory, an existence of the (or a?) set of all sets and suppose that set of all sets is also a set, then, since set of all sets contains all sets and since set of all sets is a set, then it contains itself, so is a Russell's anti-set.
My question is:

Is set of all sets the only Russell's anti-set?

Also, is this a legal question in at least one reasonable set-theory (except in naive set-theory?)?

A: In any reasonable sense, no.
Either your theory does not permit any set to contain itself (e.g. $\sf ZF$), in which case the question is trivially answered.
Or there is a set $x$ that contains itself, but then $\{x\}$ contains a set which contains itself. So $\{x\}$ is an anti-Russell set. As is $\{x,\varnothing\}$ and $\{x\}\cup a$ for any set $a$.
Indeed, the existence of a universal set is irrelevant here. This holds in such theories, e.g. $\sf NF$, as well as in theories like $\sf ZF-Reg+\exists x(x\in x)$ which have no universal set.
The only thing you need is that every set is a member of a different set. For example singletons. If a theory cannot prove that, it's not much of a set theory.


text:
\section{Introduction}
$n$-Lie algebra, where $n\geq 2$, is a generalization of the notion of Lie algebra. An integer $n$ indicates the number of elements of algebra necessary to form a Lie bracket. Thus, a Lie bracket of $n$-Lie algebra $\frak g$ is an $n$-ary multilinear mapping $\frak g\times\frak g\times\ldots\times\frak g\;(n\; \mbox{times})\to\frak g$, which is skew-symmetric and satisfies the Filippov-Jacobi identity (also called fundamental identity).

\vskip.3cm
\noindent
A notion of $n$-Lie algebra was proposed by V.T. Filippov in \cite{Filippov}. Earlier Y. Nambu in \cite{Nambu} proposed a generalization of Hamiltonian mechanics by means of a ternary (or, more generally, $n$-ary) bracket of functions determined on a phase space. This $n$-ary bracket can be regarded as an analog of the Poisson bracket in Hamiltonian mechanics and later this $n$-ary bracket was called the Nambu bracket. The dynamics of this generalization of Hamiltonian mechanics is based on the Nambu-Hamilton equation of motion that contains $n-1$ Hamilton functions. It is worth to mention that Y. Nambu proposed and developed his generalization of Hamiltonian mechanics based on a notion of triple bracket with the goal to apply this approach to quarks model, where baryons are combinations of three quarks. Later it was shown that the Nambu bracket satisfied the Filippov-Jacobi identity and thus, the space of functions endowed with the $n$-ary Nambu bracket can be regarded as the example of $n$-Lie algebra. Further development this direction has received in the paper \cite{Takhtajan}, where the author introduced a notion of Nambu-Poisson manifold, which can be regarded as an analog of the notion of Poisson manifold in Hamiltonian mechanics. In the same paper the author also treated the problem of quantization of Nambu bracket and constructed the explicit representation of Nambu-Heisenberg commutation relations by means of a primitive cubic root of unity.

\vskip.3cm
\noindent
Another area of theoretical physics, where the 3-Lie algebras are applied, is a field theory. Particularly the authors of the paper \cite{Basu-Harvey} proposes a generalization of the Nahm's equation by means of quantum Nambu 4-bracket and show that their generalization of the Nahm's equation describes M2 branes ending on M5 branes.

\vskip.3cm
\noindent
The notion of 3-Lie algebra can be extended to a $\mathbb Z_2$-graded case by means of graded Lie bracket and the graded Filippov-Jacobi identity. The classification of simple linearly compact $n$-Lie superalgebras is given in \cite{Cantarini-Kac}.

\vskip.3cm
\noindent
The question of quantization of the Nambu bracket was considered in a number of papers and the first step in this direction was taken already by Y. Nambu, but so far this is the outstanding problem. In the paper \cite{Awata-Li-Minic-Yaneya} the authors propose the realization of quantum Nambu bracket given by the formula
\begin{equation}
[X,Y,Z]=\mbox{Tr}\,X\;[Y,Z]+\mbox{Tr}\,Y\;[Z,X]+\mbox{Tr}\,Z\;[X,Y],
\label{formula 1 in introduction}
\end{equation}
where $X,Y,Z$ are either square matrices or cubic matrices, and prove that this quantum Nambu bracket satisfies the Filippov-Jacobi identity. Inspired by (\ref{formula 1 in introduction}) in \cite{Abramov2017} we extend this approach to $\mathbb Z_2$-graded case by assuming that $X,Y,Z$ are supermatrices and taking supertrace of a matrix instead of the trace and $\mathbb Z_2$-graded commutator instead of ordinary commutator, i.e. we define the graded triple commutator of supermatrices by
\begin{equation}
[X,Y,Z]=\mbox{Str}\,X\;[Y,Z]+(-1)^{x\,\overline{yz}}\mbox{Str}\,Y\;[Z,X]+(-1)^{z\,\overline{xy}}\mbox{Str}\,Z\;[X,Y],
\label{formula 2 in introduction}
\end{equation}
where $X,Y,Z$ are square supermatrices, $x,y,z$ are the degrees of supermatrices, $\overline{xy}=x+y$ and $[\;,\;]$ is the graded commutator of two supermatrices. In analogy with (\ref{formula 1 in introduction}) the graded triple commutator (\ref{formula 2 in introduction}) is referred to as the \emph{quantum super Nambu bracket of square matrices}. We prove that this graded triple commutator satisfies the graded Filippov-Jacobi identity. Hence the super vector space of supermatrices equipped with the graded triple commutator (\ref{formula 2 in introduction}) becomes the 3-Lie superalgebra.

\vskip.3cm
\noindent
The aim of this paper is to extend (\ref{formula 2 in introduction}) to 3-dimensional supermatrices. For this purpose we define a $\mathbb Z_2$-graded structure of a cubic matrix and call a matrix endowed with this structure a \emph{cubic supermatrix}. We define the \emph{super trace of a cubic supermatrix} relative to one of three directions of a cubic matrix. Then we consider the binary product of two cubic matrices, which is formulated by means of products of sections of certain orientation of cubic matrices. Making use of this product we construct with the help of super trace of a cubic supermatrix the triple product of cubic supermatrices and find the identities for this triple product. Then we apply this triple product to quantum super Nambu bracket and obtain the graded triple commutator of cubic supermatrices, which we call \emph{quantum super Nambu bracket of cubic supermatrices}. We prove that quantum super Nambu bracket of cubic supermatrices satisfies the graded Filippov-Jacobi identity and thus induces the structure of 3-Lie superalgebra on the super vector space of cubic supermatrices. In our approach we use the notions and structures of the calculus of 3-dimensional matrices developed in \cite{Sokolov}. We also generalize (\ref{formula 1 in introduction}) by means of cochains of Lie algebra. Given a Lie algebra we construct the $n$-ary Lie bracket by means of a $(n-2)$-cochain of given Lie algebra and find the conditions under which this $n$-ary bracket satisfies the Filippov-Jacobi identity, thereby inducing the structure of $n$-Lie algebra. Particular case of this approach applied to matrix Lie algebra of $n$th order matrices $\frak gl(n)$ is the matrix 3-Lie algebra with quantum Nambu bracket, which is defined with the help of the trace of a matrix (\ref{formula 1 in introduction}). Similarly we propose a more general approach than given by (\ref{formula 2 in introduction}) by making use of cochains of Lie superalgebra and find the conditions under which a cochain induces the structure of $n$-Lie superalgebra.
\section{Quantum Nambu bracket and matrix 3-Lie algebra}
In \cite{Awata-Li-Minic-Yaneya} the authors show that given a matrix Lie algebra $\frak{gl}(n)$ of $n$th order square matrices one can construct the 3-Lie algebra by means of the trace of a matrix. In this section we generalize the approach proposed in \cite{Awata-Li-Minic-Yaneya} by means of cochains of Chevalley-Eilenberg complex of a Lie algebra.

\vskip.3cm
\noindent
A concept of \emph{$n$-Lie algebra}, where $n\geq 2$ is an integer, was independently introduced by V.T. Filippov \cite{Filippov} and Y. Nambu \cite{Nambu}. We remind that an $n$-Lie algebra is a vector space endowed with an $n$-ary Lie bracket
$$
(x_1,x_2,\ldots,x_n)\in \frak g\times\frak g\times\ldots\times\frak g\;(n\;\mbox{times})\to [x_1,x_2,\ldots,x_n]\in \frak g,
$$
where an $n$-ary Lie bracket is skew-symmetric and satisfies the Filippov-Jacobi or the fundamental identity (FI)
\begin{eqnarray}
[x_1,x_2,\!\!\!&\ldots&\!\!\!,x_{n-1},[y_1,y_2,\ldots,y_n]] =\nonumber\\
    &&\sum_{k=1}^n[y_1,y_2,\ldots,y_{k-1},[x_1,x_2,\ldots,x_{n-1},y_k],y_{k+1},\ldots,y_n].
   \label{n-ary Filippov identity}
\end{eqnarray}
Particularly a \emph{3-Lie algebra} is a vector space $\frak g$ equipped with a \emph{ternary Lie bracket} $[\;,\;,\;]:\frak g\times\frak g\times\frak g\to \frak g$, which satisfies
\begin{eqnarray}
[a,b,c]=[b,c,a]=[c,a,b],\;\;\;[a,b,c]=-[b,a,c]=-[a,c,b]=-[c,b,a],
\label{symmetries of ternary commutator 2}
\end{eqnarray}
and the \emph{ternary Filippov-Jacobi identity}
\begin{eqnarray}
 [a,b[c,d,e]] = [[a,b,c],d,e]+[c,[a,b,d],e]+[c,d,[a,b,e]].
\label{ternary Filippov identity}
\end{eqnarray}
Given the matrix Lie algebra $\frak{gl}(n)$ of $n$th order matrices one can introduce the ternary Lie bracket of $n$th order matrices as follows
\begin{equation}
[A,B,C]=\mbox{Tr}\,A\;[B,C]+\mbox{Tr}\,B\;[C,A]+\mbox{Tr}\,C\;[A,B].
\label{triple commutator of square matrices}
\end{equation}
Evidently this ternary Lie bracket obeys the symmetries (\ref{symmetries of ternary commutator 2}). It is proved in \cite{Awata-Li-Minic-Yaneya} that this ternary Lie bracket satisfies the ternary Filippov-Jacobi identity and the authors propose to call this ternary Lie bracket \emph{quantum Nambu bracket}. Hence the vector space of $n$th order matrices equipped with the quantum Nambu bracket (\ref{triple commutator of square matrices}) becomes the matrix 3-Lie algebra.

\vskip.3cm
\noindent
Now our aim is to extend this approach to a more general case than the trace of a matrix. Let $\frak g$ be a finite dimensional Lie algebra and $\frak g^\ast$ be its dual. Fix an element of the dual space $\omega\in \frak g^\ast$ and by analogy with (\ref{triple commutator of square matrices}) define the triple product as follows
\begin{equation}
[x,y,z]=\omega(x)\,[y,z]+\omega(y)\,[z,x]+\omega(z)\,[x,y], \quad x,y,z\in \frak g.
\label{ternary Lie bracket with omega}
\end{equation}
Obviously this triple product is symmetric with respect to cyclic permutations of $x,y,z$ and skew-symmetric with respect to non-cyclic permutations, i.e. it obeys the symmetries (\ref{symmetries of ternary commutator 2}). Straightforward computation of the left hand side and the right hand side of the Filippov-Jacobi identity
\begin{eqnarray}
 [x,y[u,v,t]] = [[x,y,u],v,t]+[u,[x,y,v],t]+[u,v,[x,y,t]],
\label{ternary Filippov identity 1}
\end{eqnarray}
shows that one can split all the terms into three groups, where two of them vanish because of binary Jacobi identity and skew-symmetry of binary Lie bracket. The third group can be split into subgroups of terms, where each subgroup has the same structure and it is determined by one of the commutators $[x,y],[u,v],[u,t],[v,t]$. For instance, if we collect all the terms containing the commutator $[x,y]$ then we get the expression
\begin{equation}
(\omega(u)\,\omega([v,t])+\omega(v)\,\omega([t,u])+\omega(t)\,\omega([u,v]))\,[x,y].
\label{sum of three terms}
\end{equation}
Hence the triple product (\ref{ternary Lie bracket with omega}) will satisfy the ternary Filippov-Jacobi identity if for any elements $u,v,t\in \frak g$ we require
\begin{equation}
\omega(u)\,\omega([v,t])+\omega(v)\,\omega([t,u])+\omega(t)\,\omega([u,v])=0.
\end{equation}
Now we consider $\omega$ as a $\mathbb C$-valued cochain of degree one of the Chevalley-Eilenberg complex of a Lie algebra $\frak g$. Making use of the coboundary operator $\delta:\wedge^k\frak g^\ast\to\wedge^{k+1}\frak g^\ast$ we obtain the degree two cochain $\delta\omega$, where $\delta\omega(x,y)=\omega([x,y])$. Finally we can form the wedge product of two cochains $\omega\wedge\delta\omega$, which is the cochain of degree three and
$$
\omega\wedge\delta\omega(u,v,t)=\omega(u)\,\omega([v,t])+\omega(v)\,\omega([t,u])+\omega(t)\,\omega([u,v]).
$$
Hence the third group of terms (\ref{sum of three terms}) of the Filippov-Jacobi identity vanishes if $\omega\in \frak g^\ast$ satisfies
\begin{equation}
\omega\wedge\delta\omega=0.
\label{condition for omega}
\end{equation}
Thus if an 1-cochain $\omega$ satisfies the equation (\ref{condition for omega}) then the triple product (\ref{ternary Lie bracket with omega}) is the ternary Lie bracket and by analogy with (\ref{triple commutator of square matrices}) we will call this ternary Lie bracket \emph{the quantum Nambu bracket induced by a 1-cochain}.

\vskip.3cm
\noindent
We can generalize the quantum Nambu bracket (\ref{ternary Lie bracket with omega}) if we consider a $\mathbb C$-valued 1-cochain of degree $n-2$, i.e. $\omega\in\wedge^{(n-2)}\frak g^\ast$, and define the $n$-ary product as follows
\begin{equation}
[x_1,x_2,\ldots,x_n]=\sum_{i<j}\,(-1)^{i+j+1}\;\omega(x_1,x_2,\ldots,\hat x_i,\ldots,\hat x_j,\ldots,x_n)\;[x_i,x_j].
\label{n-ary quantum Nambu bracket}
\end{equation}
\begin{thm}
Let $\frak g$ be a finite dimensional Lie algebra, $\frak g^\ast$ be its dual and $\omega$ be a cochain of degree $n-2$, i.e. $\omega\in\wedge^{n-2}\frak g^\ast$. The vector space of Lie algebra $\frak g$ equipped with the $n$-ary product (\ref{n-ary quantum Nambu bracket})
is the $n$-Lie algebra, i.e. the $n$-ary product (\ref{n-ary quantum Nambu bracket}) satisfies the Filippov-Jacobi identity (\ref{n-ary Filippov identity}) if and only if a $(n-2)$-cochain $\omega$ satisfies
\begin{equation}
\omega\wedge\delta\omega=0.
\label{condition for omega in the theorem}
\end{equation}
Particularly the vector space of a Lie algebra $\frak g$ endowed with the $n$-ary Lie bracket (\ref{n-ary quantum Nambu bracket}) is the $n$-Lie algebra if $\omega$ is an $(n-2)$-cocycle, i.e. $\delta\omega=0$.
\label{theorem 1}
\end{thm}
\noindent
Particularly from this theorem it immediately follows that the quantum Nambu bracket for matrices  (\ref{triple commutator of square matrices}) satisfies the ternary Filippov-Jacobi identity. Indeed in this case $\omega=\mbox{Tr}$ and this is 1-cocycle because for any two matrices $A,B$ we have
$$\delta\mbox{Tr}\,(A,B)=\mbox{Tr}\,([A,B])=0.$$
In \cite{Arnlind-Makhlouf-Silvestrov} it is proposed to call the 3-Lie algebra constructed by means of an analog of trace the 3-Lie algebra induced by a Lie algebra. In our approach we will use the similar terminology and call the $n$-Lie algebra constructed by means of a $(n-2)$-cochain $\omega$ (\ref{n-ary quantum Nambu bracket}) the \emph{$n$-Lie algebra induced by a Lie algebra $\frak g$ and an $(n-2)$-cochain} $\omega.$ The $n$-ary Lie bracket will be referred to as the \emph{$n$-ary quantum Nambu bracket induced by a $(n-2)$-cochain}.
\section{Quantum super Nambu bracket and 3-Lie superalgebra}
The aim of this section is to extend the approach developed in the previous section to 3-Lie superalgebras.

\vskip.3cm
\noindent
Let $\frak g=\frak g_0\oplus\frak g_1$ be a super vector space. We will denote the degree of a homogeneous element $x$ by $\bar x$. Thus $\bar x\in {\mathbb Z}_2$. A super vector space $\frak g$ is an \emph{$n$-Lie superalgebra} if it is endowed with an $n$-ary graded Lie bracket
$$
(x_1,x_2,\ldots,x_n)\in \frak g\times\frak g\times\ldots\times\frak g\;(n\; \mbox{factors}\,)\to [x_1,x_2,\ldots,x_n]\in\frak g,
$$
which is graded skew-symmetric and satisfies the graded Filippov-Jacobi identity or the graded fundamental identity (GFI)
\begin{eqnarray}
&&[x_1,x_2,\ldots,x_{n-1},[y_1,y_2,\ldots,y_n]]=\nonumber\\ &&\;\;\;\;\sum_{k=1}^n (-1)^{\alpha_k}
             [y_1,y_2,\ldots,y_{k-1},[x_1,x_2,\ldots,x_{n-1},y_k],y_{k+1},\ldots,y_n],
\end{eqnarray}
where
$$
\alpha_k=\sum_{i=1}^{n-1} \bar x_i\;\sum_{j=1}^{k-1} \bar y_j.
$$
Particularly a super vector space $\frak g =\frak g_0\oplus\frak g_1$ is said to be a 3-Lie superalgebra if it is endowed with a graded triple Lie bracket $[\;.\;,\;.\;,\;.\;]:\frak g\times\frak g\times \frak g\to \frak g$, i.e.
$$
[\mathfrak g_\alpha,\frak g_\beta,\frak g_\gamma]\subseteq \frak g_{\alpha+\beta+\gamma},\;
     \alpha,\beta,\gamma\in {\mathbb Z}_2,
$$
which is graded skew-symmetric
\begin{equation}
[x,y,z] = -(-1)^{\bar x\bar y}[y,x,z],\label{skew-symmetry 1}
\end{equation}
\begin{equation}
[x,y,z] = -(-1)^{\bar y\bar z}[x,z,y],\label{skew-symmetry 2}
\end{equation}
\begin{equation}
[x,y,z] = -(-1)^{\bar x\bar y+\bar y\bar z+\bar x\bar z}[z,y,x],\label{skew-symmetry 3}
\end{equation}
and satisfies the ternary graded Filippov-Jacobi identity
\begin{eqnarray}
&&[x,y,[z,v,w]]] = [[x,y,z],v,w]+(-1)^{\overline{xy}\,\bar z}[z,[x,y,v],w]\nonumber\\
       &&\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\qquad +(-1)^{\overline{xy}\,\overline{zv}}[z,v,[x,y,w]].
\label{graded Filippov identity}
\end{eqnarray}

\vskip.3cm
\noindent
Assume $\frak g=\frak g_0\oplus\frak g_1$ is a Lie superalgebra. Now we extend our formula for the quantum Nambu bracket induced by 1-cochain (\ref{ternary Lie bracket with omega}) to the case of a Lie superalgebra. We take $\omega\in\frak g^\ast$ and define the triple product of three elements of a Lie superalgebra by
\begin{equation}
[x,y,z]=\omega(x)\,[y,z]+(-1)^{\bar x\,\overline{yz}}\,\omega(y)\;[z,x]+(-1)^{\bar z\,\overline{xy}}\,\omega(z)\;[x,y],
\label{graded triple product with omega}
\end{equation}
where $[\;,\;]$ is the graded Lie bracket of a Lie superalgebra $\frak g$. It is easy to verify that the triple product is graded skew-symmetric, i.e. it satisfies the symmetries (\ref{skew-symmetry 1}) -- (\ref{skew-symmetry 3}).

\vskip.3cm
\noindent
Now we consider the ternary graded Filippov-Jacobi identity (\ref{graded Filippov identity}) in the case, where a graded skew-symmetric product is defined by the formula (\ref{graded triple product with omega}). By analogy with the case of the Filippov-Jacobi identity for a 3-Lie algebra considered in the previous section we might expect that there will be groups of terms which vanish by virtue of the binary graded Jacobi identity, pairs of terms which cancel by virtue of graded skew-symmetry of the triple product and finally groups of terms which will vanish if we impose a condition on a cochain $\omega$. The simplest case is when the pairs of terms with opposite signs (which arise by virtue of graded skew-symmetry) cancel, no additional conditions are required here. But in the case of terms which can be collected to the graded (binary) Jacobi identity, we have a situation different from the one considered in the previous section. For instance, at the left hand side of the graded Filippov-Jacobi identity we have the term
$$
\omega(x)\,\omega(z)\;[y,[v,w]].
$$
The terms on the right hand side of the graded Filippov-Jacobi identity which could complete the graded (binary) Jacobi identity for this term are
\begin{equation}
\omega(x)\,\omega(z)\;\big((-1)^{\bar z\,\overline{xy}}\;[[y,v],w]+(-1)^{\overline{xy}\,\overline{zv}}\;[v,[y,w]]\big).
\label{expression 1}
\end{equation}
Consequently we get the binary graded Jacobi identity (and then all three terms will vanish) if inside the parenthesis we have the expression
\begin{equation}
[[y,v],w]+(-1)^{\overline{yv}}\;[v,[y,w]].
\label{expression 2}
\end{equation}
It is easy to see that the expression in parenthesis in (\ref{expression 1}) gives the expression (\ref{expression 2}) only in one case, when $x,z$ are even elements, i.e. $\bar x=\bar z=\bar 0.$ Indeed then $\bar z\,\overline{xy}=\bar 0$ and $\overline{xy}\,\overline{zv}=\bar y\,\bar v$. In all other cases we have the improper coefficients in the expression (\ref{expression 1}). Consequently the product $\omega(x)\,\omega(z)$ must be zero as soon as at least one of elements $x,y$ is odd. This can be achieved if for any odd element $x$ of a Lie superalgebra we require $\omega(x)=0$. Hence the first condition for a cochain $\omega$ can be written in the form
\begin{equation}
\omega|_{\frak g_1}\equiv 0.
\label{condition 1 for omega}
\end{equation}
The remained terms in the graded Filippov-Jacobi identity are the following: at the left hand side there is the expression $\omega([z,v,w])\;[x,y]$ and the similar expressions at the right hand side are $\omega([x,y,z])\;[v,w], \omega([x,y,v])\;[z,w], \omega([x,y,w])\;[z,v]$ (all expressions are given up to a sign). The only way to get rid of these terms is to impose the condition
\begin{equation}
\omega([x,y,z])=0,
\label{condition 2 for omega}
\end{equation}
where $x,y,z$ are either all even elements or two of them are odd and one is even. Substituting the expression (\ref{graded triple product with omega}) into the condition (\ref{condition 2 for omega}) we can write this condition in the form
\begin{equation}
\omega(x)\,\omega([y,z])+(-1)^{\bar x\,\overline{yz}}\omega(y)\,\omega([z,x])+(-1)^{\bar z\,\overline{xy}}\omega(z)\,\omega([x,y])=0.
\label{condition 2prime for omega}
\end{equation}
This condition simplifies in the case of two odd degree elements and takes on the form
$$
\omega(z)\,\omega([x,y])=0,\quad z\in\frak g_0,\;x,y\in\frak g_1.
$$
Now $\delta\omega(x,y)=\omega([x,y])$ is the 2-cochain of a Lie superalgebra $\frak g$ and it is the element of the superexterior algebra $\wedge\frak g^\ast$. From the previous considerations it follows that $\delta\omega(x,y)=0$ if either $x\in \frak g_1,y\in\frak g_0$ or $x\in \frak g_0, y\in\frak g_1$. The condition (\ref{condition 2prime for omega}) can be written in the form
\begin{equation}
\omega\wedge\delta\omega=0,
\end{equation}
where we use the wedge product of superexterior algebra $\wedge \frak g^\ast$.
\begin{thm}
Let $\frak g=\frak g_0\oplus\frak g_1$ be a Lie superalgebra, $\frak g^\ast$ be its dual and $\omega\in \frak g^\ast$ be a 1-cochain. Define the graded triple product of elements of a Lie superalgebra $\frak g$ by
\begin{equation}
[x,y,z]=\omega(x)\,[y,z]+(-1)^{\bar x\,\overline{yz}}\,\omega(y)\;[z,x]+(-1)^{\bar z\,\overline{xy}}\,\omega(z)\;[x,y].
\label{graded triple Lie bracket with omega}
\end{equation}
This graded triple product satisfies the graded Filippov-Jacobi identity (graded fundamental identity) if and only if
\begin{enumerate}
\item[i)]
$\omega|_{\frak g_1}\equiv 0$,
\item[ii)]
$\omega\wedge\delta\omega=0$, where $\delta\omega(x,y)=\omega([x,y])$ is 2-cochain and $\wedge$ is the wdege product of superexterior algebra of $\frak g^\ast$.
\end{enumerate}
Thus the super vector space $\frak g$ endowed with the graded triple Lie bracket (\ref{graded triple Lie bracket with omega}) is the 3-Lie superalgebra.
\label{theorem about 3-Lie superalgebra}
\end{thm}
\noindent
In analogy with the terminology of the previous section the 3-Lie superalgebra constructed by means of the graded triple Lie bracket (\ref{graded triple Lie bracket with omega}) will be referred to as \emph{the 3-Lie superalgebra induced by a Lie superalgebra $\frak g$ and a cochain $\omega$}.

\vskip.3cm
\noindent
Theorem (\ref{theorem about 3-Lie superalgebra}) can be applied to matrix Lie superalgebras. Consider a matrix Lie superalgebra $\frak{gl}(m,n)$ of $(m,n)$-supermatrices. The super trace of a supermatrix is the linear $\mathbb C$-valued functional, which satisfies the conditions of Theorem (\ref{theorem about 3-Lie superalgebra}). Indeed for any odd degree matrix $X\in\frak{gl}_1(m,n)$ it holds
$$
\mbox{Str}\,X=0,
$$
and hence the first condition for $\omega$ is satisfied. The super trace is the 1-cocycle because for any two supermatrices we have
$$
\delta\,\mbox{Str}(X,Y)=\mbox{Str}([X,Y])=0,
$$
and the second condition is also satisfied. Thus if we endow a matrix Lie superalgebra $\frak{gl}(m,n)$ with the graded triple Lie bracket
\begin{eqnarray}
[X,Y,Z]&=&\mbox{Str}(X)\,[Y,Z]+(-1)^{x\,\overline{yz}}\mbox{Str}(Y)\,[Z,X]\nonumber\\
&&\qquad\qquad\qquad\qquad +(-1)^{z\,\overline{xy}}\mbox{Str}(Z)\,[X,Y],
\label{graded triple Lie bracket for super matrices}
\end{eqnarray}
where $x$ is the degree of a super matrix $X$ and $\overline{xy}=x+y$, then $\frak{gl}(m,n)$ becomes the 3-Lie superalgebra. This matrix 3-Lie superalgebra will be denoted by $\frak{glt}(m,n)$. By analogy with the terminology of the previous section the graded triple Lie bracket (\ref{graded triple Lie bracket for super matrices}) will be referred to as \emph{the quantum super Nambu bracket}.
\section{Basic structures of the calculus of cubic matrices}
The aim of this section is to remind some basic notions related to the structure of a 3-dimensional matrix which we will need later in this paper to construct a 3-Lie superalgebra of cubic supermatrices. We use the terminology and the notions of the calculus of space matrices developed in \cite{Sokolov}.

\vskip.3cm
\noindent
By a 3-dimensional $m\times n\times p$-matrix $A$ we mean a system of $mnp$ complex numbers equally spaced at the points with coordinates $(x,y,z)$, where $x=0,1,\ldots,m-1, y=0,1,\ldots,n-1$ and $z=0,1,\ldots,p-1$. We will assign to each element of 3-dimensional matrix three integers $i,j,k$ and denote a corresponding element by $A_{ijk}$. If an element $A_{ijk}$ of a 3-dimensional matrix $A$ is spaced at the point with coordinates $(x,y,z)$ then integers $i,j,k$ are expressed in terms of coordinates as follow $i=x+1,\;\;j=y+1,\;\;k=z+1$.
These relations show how we position a 3-dimensional matrix in the space relative to coordinate system. We will use the following directions in a 3-dimensional matrix: The directions $(i),(j),(k)$ of a cubic matrix coincide with the space directions of $X,Y,Z$ axes respectively. If $m=n=p$ then a 3-dimensional matrix $A$ will be called a \emph{cubic matrix} of $n$th order. The vector spaces of complex 3-dimensional $m\times n\times p$-matrices and complex cubic matrices of order $n$ will be denoted by $M^{(3)}_{m,n,p}$ and ${M}^{(3)}_n$ respectively. The vector space of rectangular complex $m\times n$-matrices will be denoted by ${M}_{m,n}$ and when $m=n$ the algebra of $n$th order square matrices will be denoted by ${M}_{n}$.

\vskip.3cm
\noindent
The elements $A_{ijk}$ of a 3-dimensional matrix $A$ with fixed first subscript $i$ are located in the plane perpendicular to $X$ axis and a collection of these elements of a 3-dimensional  matrix will be called a \emph{section of orientation} $(i)$. Similarly one can define sections of orientations $(j),(k)$. Geometrically one can get a section of a 3-dimensional matrix by cutting it by a plane perpendicular to one of the edges (or coordinate axes) of a cube. Evidently a section of any orientation can be viewed as a rectangular matrix and we will label each section of given orientation (considered as a rectangular matrix) with the help of fixed subscript. For instance, if we consider a section of orientation $(j)$ and it is determined by fixed second subscript $j=r$, where $r=1,2,\ldots,n$, then we will denote the corresponding rectangular matrix by $A^{(j)}_r$. Thus $A^{(j)}_r=(A_{irk})\in {M}_{m,p}$, where integer $r$ is fixed. A 3-dimensional matrix $A$ can be considered as the collection of $n$ rectangular $m\times p$-matrices $A^{(j)}_r$, where $r$ runs from 1 to $n$. Separating different sections by vertical lines and ordering them by section label we can map a 3-dimensional matrix $A$ to the plane of this page
\begin{equation}
\big(A^{(j)}_1\,|\,A^{(j)}_2\,|\,\ldots\,|\,A^{(j)}_n\big).
\label{collection of orientation j}
\end{equation}
Assume that $A$ is a cubic matrix of $n$th order. A section of any orientation of this matrix is a square matrix of $n$th order, and we can define the trace of a section of a cubic matrix as the trace of a square matrix. For instance given a section $A^{(j)}_r\in {M}_n$ of orientation $(j)$ we define its trace by
$$
\mbox{Tr}\,A^{(j)}_r=\sum_{i=1}^n A_{iri}.
$$

\vskip.3cm
\noindent
A collection of elements of a 3-dimensional matrix $A$ with two fixed subscripts $j,k$ is referred to as a \emph{double section of orientation} $(jk)$ or a \emph{row of direction} $(i)$. Obviously there are $np$ rows of direction $(i)$ in a 3-dimensional matrix $A$. Each row of direction $(i)$ can be considered as the row-matrix
$$
\big( A_{1jk}\;A_{2jk}\;\ldots\;A_{mjk}\big).
$$
Let $A$ be a cubic matrix of $n$th order. A collection of $n$ elements of this matrix is referred to as a \emph{transversal} of a cubic matrix if for any pair of elements of this collection there is no section of cubic matrix such that it contains these elements. For instance, if $A$ is a cubic matrix of 3rd order then three elements $A_{113}, A_{232}, A_{321}$ form the transversal of $A$. The elements of a cubic matrix equally spaced along straight lines passing through opposite vertices of a cube (diagonals of a cube) form the four transversals of a cubic matrix of $n$th order. One of these transversals $A_{111}, A_{222},\ldots, A_{nnn}$ is called the main diagonal of a cubic matrix and three others are called \emph{secondary diagonals}.

\vskip.3cm
\noindent
By analogy with the notion of transversal of a cubic matric one can define a notion of 2-dimensional transversal section of a cubic matrix. A collection of $n$ rows of direction $(i)$ is referred to as a \emph{2-dimensional transversal section of direction} $(i)$ if for any pair of rows of this collection there is no section of cubic matrix which passes through these rows. Similarly one can define a 2-dimensional transversal sections of directions $(j)$ and $(k)$.

\vskip.3cm
\noindent
The 2-dimensional transversal section of direction $(j)$ of a cubic matrix of $n$th order
\begin{eqnarray}
(A_{111}\;\;\;\,\;A_{121}\;\,\;\ldots\;\,\;A_{1n1}),\nonumber\\
(A_{212}\;\;\;\,\;A_{222}\;\,\;\ldots\;\,\;A_{2n2}),\nonumber\\
\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\nonumber\\
(A_{n1n}\;\;\,A_{n2n}\;\,\ldots\;\,A_{nnn}),\nonumber
\end{eqnarray}
will be called the \emph{main diagonal section of direction} $(j)$. Analogously one can define the main diagonal sections of a cubic matrix of directions $(i)$ and $(k)$.

\vskip.3cm
\noindent
We define the \emph{Hermitian adjoint relative to direction $(j)$} matrix $A^\dag$ as the complex conjugate of a matrix $A$ and transposed with respect to the main diagonal section of direction $(j).$ Hence if we denote an element of the Hermitian adjoint matrix by $b_{ijk}$ then $b_{ijk}={\bar a}_{kji}$.

\vskip.3cm
\noindent
Similarly to the trace of a square matrix one can define a trace of a cubic matrix by means of main diagonal section \cite{Awata-Li-Minic-Yaneya}. Since there are three main diagonal sections in a cubic matrix a notion of  trace depends on a direction of a cubic matrix. Thus we define the \emph{trace relative to direction} $(j)$ of a cubic matrix $A$ as the sum of all elements of the main diagonal section of direction $(j)$. The trace relative to direction $(j)$ of a cubic matrix $A$ will be denoted by $\mbox{Tr}^{(j)}A$ and
\begin{equation}
\mbox{Tr}^{(j)}A=\sum_{i,l=1}^{n}A_{ili}.
\label{trace of cubic matrix}
\end{equation}
We can define the traces of directions $(i)$ and $(k)$ of a cubic matrix $A$ in a similar manner and we get
$$
\mbox{Tr}^{(i)}A=\sum_{k,l=1}^{n}A_{lkk},\quad
   \mbox{Tr}^{(k)}A=\sum_{j,l=1}^{n}A_{jjl}.
$$
Since a cubic matrix can be viewed as the collection of sections of chosen orientation (\ref{collection of orientation j}) it is easy to see that the trace relative to chosen direction (let say $(j)$) of a cubic matrix is equal to the sum of traces of all sections of the orientation $(j)$. For example the trace of a cubic matrix relative to the direction $(j)$ is
$$
\mbox{Tr}^{(j)}A=\sum_{r=1}^n \mbox{Tr}\,A^{(j)}_r,
$$
where $A^{(j)}_r$ is an $n$th order matrix of section of orientation $(j)$. From this it follows that $\mbox{Tr}^{(j)}A^\dag=\overline{\mbox{Tr}^{(j)}A}$.
\section{Quantum Nambu bracket of cubic matrices}
In this section we consider a triple product of cubic matrices proposed in \cite{Awata-Li-Minic-Yaneya}. This triple product is defined with the help of the trace of a cubic matrix.

\vskip.3cm
\noindent
First we remind a notion of associativity in the case of an algebra with a ternary law of composition. Assume $\mathscr A$ is a vector space over $\mathbb C$ endowed with a triple product
$$
(a,b,c)\in \mathscr A\times\mathscr A\times\mathscr A\to (abc)\in \mathscr A.
$$
A triple product is said to be \emph{{lr}-associative} (left-right associative) \cite{Abramov2010} if it satisfies the relation
\begin{equation}
\big((abc)df\big)=\big(ab(cdf)\big).
\label{lr-associativity}
\end{equation}
A triple product is said to be an \emph{associative of the first kind} if in addition to lr-associativity (\ref{lr-associativity}) it satisfies the relation
$$
\big((abc)df\big)=\big(a(bcd)f\big).
$$
A triple product is said to be an \emph{associative of the second kind} if in addition to lr-associativity it satisfies the relation
$$
\big((abc)df\big)=\big(a(dcb)f\big).
$$

\vskip.3cm
\noindent
Let $A$ be a 3-dimensional $m\times n\times p$-matrix, $B$ be a 3-dimensional $p\times n\times q$-matrix. A \emph{product relative to direction $(j)$} of two 3-dimensional matrices $A\ast_{\small j}{\!} B$ is the 3-dimensional $m\times n\times q$-matrix whose $r$th section of orientation $(j)$ ($1\leq r\leq n$) is defined by
\begin{equation}
(A\ast_{\small j}{\!} B)^{(j)}_r=A^{(j)}_r\cdot B^{(j)}_r,
\label{binary product of 3-dim matrices}
\end{equation}
where $A^{(j)}_r$ is the $r$th section of orientation $(j)$ of a 3-dimensional matrix $A$ (considered as the $m\times p$-matrix), $B^{(j)}_r$ is the $r$th section of orientation $(j)$ of a 3-dimensional matrix $B$ (considered as the $p\times q$-matrix) and $A^{(j)}_r\cdot B^{(j)}_r$ stands for the product of two rectangular matrices. It follows from this definition that the product $A\ast B$ can be considered as the collection of $n$ sections of orientation $(j)$ (each section is an $m\times q$-matrix) and thus $A\ast_{\small j}{\!} B$ is the 3-dimensional $m\times n\times q$-matrix. If we denote this matrix by $C=A\ast_{\small j}{\!} B$ then its $r$th section of orientation $(j)$ can be expressed as follows
$$
C^{(j)}_r=A^{(j)}_r\cdot B^{(j)}_r.
$$
Similarly one can define product of 3-dimensional relative to the direction $(i)$ or $(k)$. In what follows we will use product of 3-dimensional matrices relative to direction $(j)$.

\vskip.3cm
\noindent
The rule for multiplication of two 3-dimensional matrices determined by the formula (\ref{binary product of 3-dim matrices}) can be described as follows: we take a section of orientation $(j)$ with label $r$ ($1\leq r\leq n$) in a first 3-dimensional matrix $A$ (this is a slice of a 3-dimensional matrix perpendicular to $Y$ axis), then we take a section of the same orientation $(j)$ with the same label $r$ in a second 3-dimensional matrix $B$, compute the product of these rectangular matrices and consider this product as the $r$th section of the orientation $(j)$ of the 3-dimensional matrix $C=A\ast_{j}\! B$. Virtually this multiplication of 3-dimensional matrices amounts to multiplication of rectangular matrices and consequently has properties very similar to ordinary multiplication of rectangular matrices. It is easy to see that the product defined in (\ref{binary product of 3-dim matrices}) is associative, i.e. $(A\ast_{j}\! B)\ast_{j}\! C=A\ast_{j}\! (B\ast_{j}\! C)$. Thus the multiplication of 3-dimensional matrices (\ref{binary product of 3-dim matrices}) induces the structure of unital associative algebra on the vector space of $n$th order cubic matrices $\cubic$, where the identity element of this algebra is the $n$th order cubic matrix $I$, whose any section of orientation $(j)$ is the $n$th order square unit matrix $E$, i.e. $I^{(j)}_r=E$ and $r=1,2,\ldots,n$. By other words each element of the main diagonal section of direction $(j)$ of the cubic matrix $I$ is equal to 1, $I_{iji}=1$.

\vskip.3cm
\noindent
Now $\cubic$ equipped with the binary multiplication of cubic matrices is the unital associative algebra and consequently the commutator of two cubic matrices $A,B\in \cubic$ defined as usual $[A,B]=A\ast_{\small j}{\!} B-B\ast_{\small j}{\!} A$ satisfies the Jacobi identity. Hence the unital associative algebra of cubic matrices $\cubic$ endowed with the commutator becomes the Lie algebra, which we will denote by ${\frak gl}^{(3)}_n$. It is easy to show that the trace of a cubic matrix introduced in the previous section (\ref{trace of cubic matrix}) vanishes on the commutator of two cubic matrices. Indeed we have
$$
\mbox{Tr}^{(j)}\,[A,B]=\sum_{r=1}^m \mbox{Tr}\,[A,B]^{(j)}_r
                      =\sum_{r=1}^m \mbox{Tr}\,[A^{(j)}_r,B^{(j)}_r]=0.
$$

\vskip.3cm
\noindent
In order to construct a 3-Lie algebra of cubic matrices we will need a triple product. Following \cite{Awata-Li-Minic-Yaneya} we consider the triple product of cubic matrices which is the combination of the binary product (\ref{binary product of 3-dim matrices}) of two cubic matrices and the trace of a cubic matrix
\begin{equation}
(ABC)=\mbox{Tr}^{(j)}B\,\;(A\ast_{j}\! C).
\label{triple product of cubic matrices}
\end{equation}
It is natural to call a triple product \emph{proper triple product} if it can not be reduced to a binary product, and \emph{binary generated triple product} if it is constructed by means of a binary product. Evidently the triple product (\ref{triple product of cubic matrices}) is binary generated triple product.
\begin{prop}
The triple product (\ref{triple product of cubic matrices}) of cubic matrices satisfies the following identities:
\begin{enumerate}
\item  it is lr-associative $((ABC)DF)=(AB(CDF))$,
\item $(ABC)^\dag=(C^\dag B^\dag A^\dag)$,
\item $(A(BCD)F)=(A(DCB)F)$,
\item $((ABC)DF)=((ADC)BF)$.
\end{enumerate}
\end{prop}
\noindent
We will prove only the lr-associativity of the triple product because the remained properties can be verified analogously. Making use of the definition of triple product we can write the left hand side of the lr-associativity as follows
$$
((ABC)DF)=\mbox{Tr}^{(j)}B\;\mbox{Tr}^{(j)}D\;(A\ast_{j}\! C)\ast_{j}\! F.
$$
Evidently there is the same product of scalar factors at the right hand side of the lr-associativity (because cubic matrices $B,D$ are still in the middle of the corresponding triple products), but the matrix part has the form $A\ast_{j}\! (C\ast_{j}\! F)$, which by virtue of the associativity of the binary product is equal to $(A\ast_{j}\! C)\ast_{j}\! F$.

\vskip.3cm
\noindent
Now in order to construct a 3-Lie algebra of cubic matrices we can consider the following quantum triple Nambu bracket
\begin{equation}
[A,B,C]=(ABC)+(BCA)+(CAB)-(CBA)-(BAC)-(ACB),
\label{ternary commutator}
\end{equation}
which plays the role of ternary Lie bracket for the triple product (\ref{triple product of cubic matrices}). The structure of quantum triple Nambu bracket is based on the well known representation of $S_3$, where the cyclic (even) permutations are taken with the sign "+" and the non-cyclic (odd) permutations are taken with sign "-". Evidently the quantum triple Nambu bracket (\ref{ternary commutator}) satisfies the symmetry properties of ternary Lie bracket (\ref{symmetries of ternary commutator 2}). Thus the quantum triple Nambu bracket (\ref{ternary commutator}) vanishes whenever there are two (or three) equal elements in (\ref{ternary commutator}). It is worth to mention that there is a different approach to notion of ternary commutator which is based on the group of cyclic permutations $\mathbb Z_3$ and its representation by cubic roots of unity \cite{Abramov1997}. In this approach a ternary commutator vanishes when all three elements are equal, but it is not necessarily zero if there are two equal elements.

\vskip.3cm
\noindent
Since the triple product of cubic matrices is binary generated we can write this product in terms of binary commutators as follows
\begin{equation}
[A,B,C]=\mbox{Tr}^{(j)}B\;[A,C]+\mbox{Tr}^{(j)}A\;[C,B]+\mbox{Tr}^{(j)}C\;[B,A],
\label{triple commutator expressed in binaries}
\end{equation}
and we see that the triple commutator of cubic matrices is analogous to the triple commutator of square matrices (\ref{triple commutator of square matrices}). The triple commutator (\ref{triple commutator expressed in binaries}) will be referred to as the \emph{quantum Nambu bracket for cubic matrices}. Now taking into account the property of the trace of a cubic matrix that the trace vanishes on any commutator of two cubic matrices we see that the condition (\ref{condition for omega in the theorem}) stated in Theorem (\ref{theorem 1}) is satisfied and consequently the triple commutator of cubic matrices (\ref{ternary commutator}) satisfies the Filippov-Jacobi identity. Hence the vector space of $n$th order cubic matrices $M^{(3)}_n$ equipped with the triple commutator (\ref{ternary commutator}) (or (\ref{triple commutator expressed in binaries})) is the 3-Lie algebra.

\section{Cubic supermatrix and quantum super Nambu bracket}
In this section we define the $\mathbb Z_2$-graded (super) structure of a cubic matrix by assigning parities to sections of two fixed orientations and defining the parity of the intersection of two sections (which is the row of direction different from two fixed orientations) as the sum of parities of intersecting sections. Then we define the super trace of a cubic supermatrix and show that it amounts to the sum of super traces of all sections of chosen orientation.  Making use of the super trace of a supermatrix we introduce the triple product of cubic supermatrices and find the identities for this triple product. Finally we define the quantum super Nambu bracket and show that it satisfies the graded Filippov-Jacobi identity and hence induces the structure of 3-Lie superalgebra of cubic supermatrices. The residue classes modulo 2 will be denoted by $\bar 0,\bar 1$, i.e. $\mathbb Z_2=\{\bar 0,\bar 1\}$.

\vskip.3cm
\noindent
Let $A=(A_{ijk})$ be a cubic matrix of $n$th order. We define a \emph{super structure relative to the direction $(j)$} of an $n$th order cubic matrix $A$ as follows. First we fix an integer $0<r<m$, choose $r$ sections of orientation $(i)$, $r$ sections of orientation $(k)$ and assign the parity $\bar 0$ to each of chosen sections. Naturally we assign the parity $\bar 1$ to each of remained $s=n-r$ sections of orientation $(i)$ and $s$ sections of orientation $(k)$. Accordingly a section will be called even if its parity is $\bar 0$ and odd if its parity is $\bar 1$. Each row of direction $(j)$ is the intersection of one section of orientation $(i)$ and one section of orientation $(k)$. Second we define the parity of each row of direction $(j)$ as the sum of the parities of sections whose intersection is the given row of direction $(j)$. Finally we define the parity of an element of a cubic matrix to be equal to the parity of the row of direction $(j)$ which passes through this element. Similarly one can define a super structure of a cubic matrix relative to direction $(i)$ or $(k)$.

\vskip.3cm
\noindent
In the described above super structure of a cubic matrix, the choice of $r$ even sections can be made arbitrarily, but it is more convenient to order even sections, placing them in the first place.
Hence we assign the parity $\bar 0$ to first $r$ sections of orientation $(i)$, the parity $\bar 1$ to the remaining $s=n-r$ sections of the same orientation and in the same way assign the parities to the sections of orientation $(k)$. Then the sections of orientation $(i)$ with the labels $1,2,\ldots,r$ are even sections, the section of orientation $(i)$ with the labels $r+1,r+2,\ldots,m$ are odd sections and exactly the same distribution of parities takes place for sections of orientation $(k)$. A cubic matrix endowed with a super structure will be called a \emph{cubic ($r,s$)-supermatrix}. We will denote the vector space of complex $n$th order cubic ($r,s$)-supermatrices by ${M}^{(3)}_{r,s}$, where $r+s=n$.

\vskip.3cm
\noindent
Consider a section of orientation $(j)$. This section is an $n$th order square matrix denoted by $A^{(j)}_r$. Each element of this square matrix is at the intersection of given section of orientation $(j)$ and the row of direction $(j)$. But each row of direction $(j)$ has the certain parity. Consequently each element of a square matrix $A^{(j)}_r$ has the certain parity and we see that the super structure of a cubic matrix induces the super structure of each section of orientation $(j)$. Hence each $A^{(j)}_r$ is an $n$th order square supermatrix with the usual super structure of square matrices: In the upper left corner and lower right corner there are the even blocks ($r$th order square matrix and $s$th order square matrix respectively) which will be denoted by $A^{r}_{00}, A^{r}_{11}$ (here $r$ is the label of a section and we do not show the orientation $(j)$ assuming it is fixed) and two odd blocks which are the rectangular matrices $A^r_{01},A^r_{10}$ of dimensions $r\times s$ and $s\times r$ respectively. The even blocks $A^r_{00}$, where $r=1,2,\ldots,n$, form the 3-dimensional $r\times n\times r$-matrix and the even blocks $A^r_{11}$ form the 3-dimensional ($s\times n\times s$)-matrix which will be denoted by $A^{(j)}_{00},A^{(j)}_{11}$ respectively and will be called the even blocks of a cubic supermatrix $A$. Analogously we define the odd blocks of a cubic supermatrix and denote them by $A^{(j)}_{01},A^{(j)}_{10}$, where $A^{(j)}_{01}$ is the 3-dimensional $r\times n\times s$-matrix and $A^{(j)}_{10}$ is the 3-dimensional $s\times n\times r$-matrix.

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\noindent
As usual a supermatrix $A$ is said to be an \emph{even degree supermatrix} if its odd blocks $A^{(j)}_{01},A^{(j)}_{10}$ are zero 3-dimensional matrices and $A$ is said to be a \emph{odd degree supermatrix} if its even blocks $A^{(j)}_{00},A^{(j)}_{11}$ are zero 3-dimensional matrices. The degree of a supermatrix $A$ will be denoted by $|A|$ (hence $|A|$ is either $\bar 0$ or $\bar 1$) and we will also use the small letter to denote degree of a supermatrix, for example the degree of a supermatrix $A$ will be denoted by $a$. Obviously the binary product (\ref{binary product of 3-dim matrices}) of two cubic ($r,s$)-supermatrices $A,B$ has the property
$$
|A\ast_{j}\! B|=|A|+|B|,
$$
and thus
the vector space of cubic ($r,s$)-supermatrices $M^{(3)}_{r,s}$ endowed with the binary product (\ref{binary product of 3-dim matrices}) is the superalgebra.

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\noindent
Next structure on the super vector space of cubic supermatrices arises if we consider the graded binary commutator of two cubic supermatrices
\begin{equation}
[A,B]=A\ast_{j}\! B - (-1)^{ab}\;B\ast_{j}\! A,
\end{equation}
where $a,b$ are the degrees of supermatrices $A,B$ respectively. Evidently this graded commutator satisfies the graded Jacobi identity and the super vector space of cubic supermatrices becomes the Lie superalgebra, which will be denoted by ${\frak gl}^{(3)}(r,s)$.

\vskip.3cm
\noindent
Now we can extend the notion of a trace relative to direction $(j)$ of a cubic matrix to cubic supermatrices. For this purpose we will use the formula (\ref{trace of cubic matrix}). Indeed we know that each section $A^{(j)}_l$ of orientation $(j)$ is an $n$th order square ($r,s$)-supermatrix and we can compute its supertrace. Thus we define the \emph{supertrace relative to direction $(j)$ of a cubic supermatrix $A$} as follows
\begin{equation}
\mbox{Str}^{(j)} A=\sum_{l=1}^n \mbox{Str}\; A^{(j)}_l.
\end{equation}
Obviously the supertrace of any odd degree cubic supermatrix vanishes.
It is easy to verify that the supertrace defined in this way vanishes on graded commutators of cubic supermatrices. Indeed
$$
\mbox{Str}^{(j)}\;[A,B]=\sum_{l=1}^n \;\mbox{Str}\;[A^{(j)}_l,B^{(j)}_l]=0.
$$
By analogy with the triple product of cubic matrices (\ref{triple product of cubic matrices}) we define the triple product of cubic supermatrices
\begin{equation}
(ABC)=\mbox{Str}^{(j)}A\;(B\ast_j\! C).
\label{triple product of cubic supermatrices}
\end{equation}
\begin{prop}
The triple product (\ref{triple product of cubic supermatrices}) satisfies the following identities:
\begin{enumerate}
\item $(AB(CDF))=(A(CBD)F)$,
\item $((ABC)DF)=(-1)^{bc}\;((ACB)DF)$,
\item $(AB(CDF))=(CB(ADF))$,
\end{enumerate}
where $b,c$ are the degrees of cubic supermatrices $B,C$ respectively.
\end{prop}
\noindent
Next we define the graded triple commutator of cubic supermatrices by the formula
\begin{eqnarray}
[A,B,C]\!\!\!&\!=\!&\!\!\!(ABC)+(-1)^{a\,\overline{bc}}\,(BCA)+(-1)^{c\,\overline{ab}}\,(CAB)\nonumber\\
             &&-(-1)^{ab}\,(BAC)-(-1)^{bc}\,(ACB)-(-1)^{ab+bc+ac}\,(CBA),\nonumber
\end{eqnarray}
where $\overline{ab}=a+b$.
Substituting the definition of the triple product (\ref{triple product of cubic supermatrices}) into this formula we get
\begin{equation}
[A,B,C]=\mbox{Str}^{(j)}A\;[B,C]+(-1)^{a\,\overline{bc}}\mbox{Str}^{(j)}B\;[C,A]+(-1)^{c\,\overline{ab}}\mbox{Str}^{(j)}C\;[A,B].
\label{viimane valem}
\end{equation}
Since the supertrace of a cubic supermatrix vanishes on the subspace of odd degree cubic supermatrices and it also vanishes on graded commutators of cubic supermatrices we see that the both conditions of Theorem (\ref{theorem about 3-Lie superalgebra}) are satisfied and we conclude that the super vector space of cubic supermatrices equipped with the graded triple commutator (\ref{viimane valem}) is the 3-Lie superalgebra. Hence the graded triple commutator (\ref{viimane valem}) satisfies the graded Filippov-Jacobi identity and by analogy with the previously used terminology it will be referred to as the \emph{quantum super Nambu bracket for cubic supermatrices.}
\subsection*{Acknowledgment}
The author gratefully acknowledges that this work was financially supported by the institutional funding IUT20-57 of the Estonian Ministry of Education and Research.

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