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Merrick Garland Must Follow Through With His Commitment to Hold All January 6 Perpetrators Accountable
By Joyce White Vance
Vance is distinguished professor of the practice of law at the University of Alabama, a former U.S. Attorney for the Northern District of Alabama and an NBC News and MSNBC legal analyst.
The House committee investigating the January 6 attack on the Capitol has been tearing it up. They worked straight through the holidays, including on Christmas and New Year's Day. They've interviewed 300-plus witnesses, pursuing those who disobey their subpoenas. In fact, the committee's work looks a lot like what we might expect a criminal investigation to look like—so much so that Trump's lawyers are trying to fend off the release of his presidential papers to the committee on the basis that the committee may use them to make criminal referrals to the Department of Justice.
In the meantime, before today's speech by Attorney General Merrick Garland, we haven't heard a lot about high-level January 6 investigations from DOJ. We know from the department's year-end press release that more than 725 cases are in progress. They range from misdemeanors to assaults against Capitol police officers to interference with Congress's certification of the 2020 election. Some defendants, including members of extremist groups, have been charged with conspiracy. But if you were looking for clear signs that DOJ is investigating those most responsible for the attempt to interfere with the smooth transfer of power, you'd have been hard-pressed to find any. No witnesses protesting grand-jury subpoenas, no leaks of information to the press. Some have suggested Garland's Justice Department wouldn't pursue those who are most culpable out of misplaced sentiment that such prosecutions would smack of politics, calling the attorney general feckless and the wrong man for the job.
Read More: How Donald Trump Turned Jan. 6 into a Windfall
There were some clues DOJ might be investigating. The same year-end press release noted, "The Department of Justice's resolve to hold accountable those who committed crimes on Jan. 6, 2021, has not, and will not, wane." DOJ speaks in a restrained voice when commenting on investigations, because investigations don't always culminate in prosecutions. That's how we should want our justice system to work—we want investigators to gather evidence and learn the truth before making decisions about whether the law and the facts warrant prosecution. DOJ conducts its criminal investigations, frustratingly for the public, in secrecy. This serves justice both by protecting the reputations of those who are never charged and maximizing investigators' ability to obtain evidence. Garland, who oversaw notoriously close-hold investigations including Oklahoma City and the Unabomber in his earlier stint in the department, knows this. So, while some DOJ watchers believe they would see signs of high-level investigations in process, that's not necessarily true.
Garland seemed to confirm investigation was in progress today, making an important commitment after recapping the department's prosecutions to date: "The actions we have taken thus far will not be our last. The Justice Department remains committed to holding all Jan. 6 perpetrators, at any level, accountable under law—whether they were present that day or were otherwise criminally responsible for the assault on our democracy. We will follow the facts wherever they lead." Garland is saying that no one is above the law, not even a former president. He's also answering concerns that his focus is limited to events on Jan. 6, not to the ongoing effort to prevent the transfer of power. This is a key point—Jan. 6 was the culmination of the effort to interfere with the outcome of the election, and it would be a swing and a miss if DOJ focused exclusively on that one day.
As for those who have criticized him for moving too slowly, Garland admonished that his work would take "as long as it takes and whatever it takes for justice to be done–consistent with the facts and the law." Prosecutors who have worked on complicated investigations, especially when the law requires proof of a defendant's state of mind, will understand this for the hard-nosed nod to reality that it is. These cases take time. Even when prosecutors are working at full speed, the evidence does not fall into place overnight. And yet given his discussion of voting rights, it's clear that Garland is mindful of the clock that is ticking on democracy. Prosecutors understand that justice unnecessarily delayed is justice denied.
Garland's job today was to close the trust gap that has developed between his Justice Department and the country, speaking to people who have lost faith that the rule of law has enough grit left to hold a former president, one with nothing but disdain for it, accountable. At the same time, Garland knows there is a significant percentage of the country that believes the lies Trump peddles about the election. Those people will believe that any investigation, let alone prosecution, is politically motivated.
Garland has undoubtedly struggled with that problem. But DOJ's mission is to do what is right, not what is politically expedient. Acting with reserve to avoid the appearance of political interference at the department is a laudable goal, but DOJ's moral imperative is to unflinchingly prosecute if–and it's still a big if–there is sufficient admissible evidence to prove serious crimes beyond a reasonable doubt. Failing to do so in the face of provable interference with a presidential election would end any claim that our legal system operates on the principle that no man is above the law. And Garland, it appears, is not willing to relegate the country to that fate.
Read More: Jan. 6 Changed How the Far Right Organize, Not What They Believe
Some of the facts are already clear. Democrats didn't steal the election, but Trump tried to. He is captured on tape cajoling and threatening Georgia's secretary of state to find him the 11,780 votes he needed for a victory in that state. He explicitly pushed the discredited narrative of the big lie on the country. And he sat in front of his television, doing nothing to end the violence, for more than three hours as the Capitol was overrun, police officers were attacked and our elected officials, including his own vice president, were at risk of harm. Those matters should all be investigated. And it should not fall on state and local prosecutors, or even the capable shoulders of the House committee investigating Jan. 6, to do the heavy lifting in the matter of a failed presidency.
But it's premature to ask the attorney general to promise indictments before all of the evidence is in. So Garland must navigate the challenge of restoring trust with a nation that is wary of DOJ after living through the excesses of the Trump Administration. He must find a path to convince people to have faith in him in the absence of a guarantee that DOJ will prosecute the people responsible for the near decimation of our democracy. That means he has the ongoing job of convincing them that, for now, investigation is enough. And that's not an easy sell.
Read More: For the Jan. 6 Rioters, Justice Is Still Coming
Yet if the past year has taught us anything, it's that neither law enforcement nor the public has a full understanding of complicated events immediately upon their conclusion. Not even when we watch the final act unfold before our eyes. As the House investigation proceeds, we are learning that there is a lot lurking beneath the surface. Hundreds of witnesses and thousands of pages of documents with new, but critical, details are still coming to light. DOJ may benefit from the evidence the committee is accumulating. DOJ can't decide to prosecute (or not) prematurely.
Criminal cases aren't easy. Prosecutors can't bring a case because "everyone knows what Trump did." It takes evidence that is admissible in court under strict federal rules. There must be proof beyond a reasonable doubt of each element of a crime. And prosecutors must be prepared not just to obtain a conviction at trial but to defend it on appeal. That means they must avoid stretching into novel legal theories that are vulnerable to being reversed. These difficulties don't mean DOJ can avoid its responsibilities. They mean it must perform them diligently and properly.
Read More: January 6th May Have Been Only the First Wave of Christian Nationalist Violence
Garland made a start on rebuilding trust with Americans today. But he will have to commit to and continue to work on the relationship. Although it's impossible and perhaps unfair to criticize and assess Garland's decisions until his work is concluded and we know the outcome, he's asking us to rely on him while we wait. We don't know exactly what he is and isn't doing. It's hard to avoid the concern the former president will, yet again, escape accountability, which would destroy many Americans' confidence in the rule of law.
An operating principle for prosecutors is that if you're going to shoot at the king, you'd best not miss. It's always much easier to be an armchair prosecutor than to prove a case in the courtroom and convince a panel of judges on appeal that the defendant's legal rights were all in place. DOJ must get the Jan. 6 prosecutions right because they are all that stands between our democracy and those who would take it away from us. So DOJ had best not miss. Garland said enough today to allay concerns that DOJ would miss because they weren't going to take the shot. Now it's up to him to convince us he'll take the right ones.
TIME Ideas hosts the world's leading voices, providing commentary on events in news, society, and culture. We welcome outside contributions. Opinions expressed do not necessarily reflect the views of TIME editors.
How Donald Trump Turned Jan. 6 into a Windfall
How We Can Learn to Live with COVID-19 After Vaccinations


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North Carolina Law Makes It Harder For Judges To Waive Fees And Fines
KELLY MCEVERS, HOST:
There has been a big effort across the country to reduce the number of people who go to jail because they can't afford to pay court fees or make bail. But a new law in North Carolina is doing the opposite. It took effect on Friday, and it is meant to make it harder for judges to waive court fines and fees.
Joe Neff of The Marshall Project writes it's thought to be the first law of its kind in the U.S. And the weird thing is there's no sponsor for this law. There's no name attached to it. Joe Neff joins us from member station WUNC in Durham. Hi there.
JOSEPH NEFF: Hello.
MCEVERS: Before we get into the specifics of this law, what do we know about where it came from? And how is it that there's no lawmaker attached to it?
NEFF: Well, it appeared in the budget, which is a way that laws in North Carolina and many other states just show up - in the budget. There was no bills filed. There is no amendment filed. There are no hearings in the judiciary committee, for example. It appeared in the budget.
MCEVERS: So as I said, this law makes it harder for judges to waive court fees and fines - how?
NEFF: Well, the Supreme Court has said you cannot send someone to jail just because they can't afford to pay. And so this law puts up an obstacle like a bureaucratic logjam. Before waiving a fine or a fee, a judge has to notify any public agency that might get a piece of it.
MCEVERS: Could you just give an example of how the fees work?
NEFF: OK. The lowest fee is a seatbelt violation. You go into court. You have to pay $154 general court fee and 25 bucks for not wearing your seatbelt. That $154 is split up anywhere from $2.50 to the law enforcement retirement fund, money to the local police officer for making arrest, money to the treasury to run the courthouse.
MCEVERS: So I guess I can imagine these agencies saying, hey, we want our money. I mean, is that the argument for this law?
NEFF: The state senator who talked to me about this law says, yes, these state agencies - the State Bureau of Investigation or the local jail or the school board - they deserve this money. So this is a revenue source for government.
MCEVERS: But if we're talking about people who can't afford to pay the fines in the first place, they're not going to get the revenue. I mean, how is this going to affect people who will be getting these fines?
NEFF: Poor people often have to make choices. Are they going to pay a $250 fine or pay part of it, or are they going to keep the lights on and keep rent paid? And if you don't pay a fine in North Carolina in 40 days you're reported to the motor vehicles, and then you can lose your license. And then you can lose your job or - so it just puts people in a real financial crunch.
MCEVERS: What are judges saying about the law?
NEFF: I haven't talked to any judges who like it, and I've talked to a number of them for this story. Judges don't like being told what to do by nature. But also, they're there watching a parade of people come into the courtroom every day in traffic court or misdemeanor court, and they get a sense that some of these people just can't pay. And a number of judges have told me that it's just wrong to fund the justice system on the backs of the poor since the poor are overrepresented in the justice system to begin with.
MCEVERS: As I said in the introduction, I mean, there's this, you know, broader reform of court systems so poor people aren't, you know, jailed because they can't afford fines. What's been the response to the North Carolina law nationally?
NEFF: Well, the National Center for State Courts flat-out said, you know, North Carolina is going in the wrong direction.
MCEVERS: Do you think - is there any sort of movement in North Carolina to see this law overturned?
NEFF: It's going to be hard to overturn it in the current legislature in North Carolina. The - both houses are run by Republicans. They have a veto-proof majority. And it's clear that they want this law in place.
MCEVERS: Joseph Neff is a staff writer for The Marshall Project. Thank you so much.
NEFF: Thank you. Transcript provided by NPR, Copyright NPR.

redpajama_set_name: RedPajamaCommonCrawl


Reviews of the Canon EOS M from around the web.
Canon hasn't exactly been quick out of the blocks in the race to launch compact system cameras and its first, the EOS M, comes four years after Panasonic kicked-off the whole shebang back in September 2008 with the launch of the Lumix G1.
The EOS M is Canon's long-awaited entry into the mirror-less compact system camera market. Announced in July 2012 it employs an 18 Megapixel APS-C sensor and a new EF-M lens mount with a 1.6x field-reduction factor. Canon launched the format with two native M-mount lenses, a 22mm f2 pancake prime (35mm equivalent) and an 18-55mm f3.5-5.6 stabilized zoom (29-88mm equivalent), both of which feature STM stepper motors for quick and quiet movie AF. Meanwhile, the entire Canon EF and EF-S lens catalogue is supported with autofocus and auto exposure using the optional EF-EOS M adapter.
Canon made its loyal customers wait quite a long time before it finally joined the mirrorless camera revolution, announcing the Canon EOS M back in June. The camera comes nearly four years after Panasonic kicked things off by "friending" Olympus - forming the Micro Four Thirds alliance - and introducing the Lumix DMC-G1, making Canon the last major DSLR maker to join the fray.


Mobiles – Blessing or a Curse We are Living?
This electronic gadget has different uses and capacities. It has turned into an essential a vital part of our life. Without mobile phones we for the most part feel fragmented. In vogue and sleek PDAs have turned into a piece of design now days. It has now turned into a multi-utility gadget, and not simply remain a specialized gadget.
Cell phones are the most advantageous and quickest methods for correspondence. You can call anybody from anyplace and whenever.
It is an extraordinary wellspring of excitement too. You can play diversions, tune in to music, converse with your companions, send messages, et cetera when you get exhausted.
It can be utilized for catching and putting away critical data.
It can work as a radio. You can listen music while driving and voyaging.
Mobiles can supplant watches to check the time and wake up timers that awakens you.
Your voice can likewise be recorded in your mobiles.
It can be utilized as a light when the light goes.
Mobiles with cameras are utilized to take photographs and making recordings. You can likewise solidify your significant minutes.
Mobiles likewise have logbooks to check days and dates.
You can download your main tunes, backdrops, ring tones, and motion pictures in your mobiles.
In the event that you lost the way, you can utilize your versatile and call for help.
Utilize cell phones instead of paper and pen. This spares paper, which thus spares trees.
Cell phones having the office of adding machine help to do mental computations.
This remote gadget influences you to advise the guardians when you to get late, illuminates your supervisor that you will be late as you stuck in the congested driving conditions, et cetera.
Mobiles having Internet office diminishes the workload of the business and office fill in as it influences you to work anyplace to and whenever.
It is compact gadget and you can convey it with you where you will go, so you go to all your essential calls.
Business arrangements should be possible on a solitary call.
Maintain a strategic distance from to utilize cell phone at whatever point conceivable.
Your discussion over cell phones ought to be succinct and to the point. Longer the discussion is, the more its effect on a human's wellbeing.
Fend off your mobiles from your body.
Try not to utilize versatile while driving, as it expands the danger of getting into auto crashes. In the event that it is a critical, stop your auto and after that go to your call.
Utilize a without hands set keeping in mind the end goal to lessen the measure of radiation. It diminishes the electronic field impact to 92%.
Utilization of cell phones by hoodlums to do their criminal exercises builds the wrongdoing rate.
Attempt to turn off your mobiles when you are in broad daylight places, religious spots, going to a gatherings or gatherings.
Disgraceful utilization of cell phones some of the time makes unsettling influence.
Keep your telephones on turn off mode while contemplating as it diverts your consideration.
Purchase telephones of present day computerized measures (GSM-1800 and 3-G) as it creates less effect on a man.
The emanation of radiations from mobiles may demonstrate unfavorable to your wellbeing as it might cause migraine, ear infection and obscuring vision. These unsafe radiations demolish the cells situated in head and ear area that makes harms the mind and nephrons in the head locale.
This electronic device will demonstrate an incredible assistance in the midst of crisis.
Cell phones clients additionally encounter power outage, sleepiness, poor fixation and expanded fatigability.
Try not to keep your portable in the front pocket when it is on dynamic mode. It might make extreme harm heart.
Ring tones ought not be noisy as it makes harm your eardrums.
German analysts found that discussion that went on for over 30 minutes prompt the expansion of circulatory strain of 5-10 mm on mercury segment.
Utilization of mobile phones is very aggravating in quiet and quiet places like libraries, galleries, motion picture lobbies, healing centers, and so forth.
Intemperate warming up of its battery brings about mobile phones impact. In spite of the fact that it happens very uncommon, however extremely deadly on the off chance that it happens.
Utilization of PDAs amid examinations as a methods for conning by the understudies makes an issues for them and it spoils their future.
Delayed utilization of mobiles can cause hearing imperfections in the person.
Cameras mobiles can be abused to take unlawful pictures and after that showed on the Internet. It can likewise be utilized for coercing, which thusly, can likewise destroy the individual existence of that individual.
Formation of cell phones is an awesome accomplishment of man. Man makes this electronic gadget for making our life more helpful and agreeable. It makes our lives simpler. Be that as it may, tragically numerous individuals are utilizing it for wrong purposes. It demonstrates a help for us in the event that it is utilized with extraordinary care and effectiveness.


Cast: Jim Carrey, Steve Carell, Will Arnett, Seth Rogen, Dan Fogler, Isla Fisher, Jonah Hill, Amy Poehler, Jaime Pressly, Carol Burnett, Charles Osgood.
Directed by Jimmy Hayward and Steve Martino, adapted from the books by Dr Seuss.
Story: Elephant talks to microscopic creatures.
That's a really difficult title to remember. Horton Who's Hearing a Hoover ... er ... Hairy Whores in Hoses ... it makes more sense when you see it.
Dr Seuss adapations have had something of a rocky ride in the past, with both The Cat in the Hat and The Grinch dividing opinion. Horton Hears a Who sticks more closely to the source material and is entertaining and funny to boot. Although primarily aimed at kids, the script has seriously profound implications, drawing clear parallels to religion without picking out any one doctrine for specific analysis. Big friendly elephant Horton with his big flappy ears is able to hear one of the microscopic villagers on a speck perched on a flower, but that's the only villager that can hear him talking back. Neither can see the other, so each has to persuade their respective races that the other exists, without any kind of evidence to back it up. See what I mean about religious parallels? The specky village is called Whoville and each little chap within is a "Who" - so now the title makes sense too!
Horton Hears a Who is fun for all the family. The kids will be entranced while the adults can ponder the metaphysical connotations if they like - or lap up the good clean fun with the rest of us.
To enjoy this film you should: take a kid along so you have an excuse to see it.
Horton Hears a Who was released in the UK on 21st March 2008.


Very nice San Lorenzo 100 with Hard Top. She is in a virtually new condition. Equipped with a Raised Pilothouse and a Hard Top. Commercially registered.
All interior in light contemporary oak in satin finish. Wooden parquette floors on main deck and cream carpet on lower deck. Galley with granit stone sids and floor.
4 en-suite cabins, master cabin on main deck with walk-in wardrobe, VIP cabin midships on lower deck with walk-in wardrobe as well, further forward a twin cabin to the port side and a double cabin to the starboard side. 3 crew cabins for 5 with 3 bathrooms. Crew mess with galley, dinette and laundry. Crew accommodation access through the main galley on main deck. Day head on main deck.


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From The Booth: Texans vs. Lions Radio Highlights
Voice of the Houston Texans and Andre Ware call the Texans vs. Lions Thanksgiving Day game on SportsRadio610.
Follow along on Gameday Central presented by Mattress Firm for more highlights.
Zach Cunningham is the NFL's leading tackler in 2020
Zach Cunningham will finish the 2020 regular season as the NFL's leading tackler with 159 total tackles (104 solo) through 16 starts for the Texans.
Sounds of the Game: Texans vs. Titans Reactions
Watch the press conferences following the Houston Texans Week 17 game against the Tennessee Titans.
Rapid Reaction: Texans fall 41-38 in a Week 17 shootout with Titans
The Houston Texans fell 41-38 to the Tennessee Titans in their final game of the 2020 regular season.
Deshaun Watson makes franchise history, clinches 2020 NFL passing title
QB Deshaun Watson is the 2020 NFL passing yards leader.
Brandin Cooks cracks 1,000 yards receiving in 2020 season
WR Brandin Cooks caught a 20-yard pass to start Sunday's game, and eclipsed the 1,000-yard mark in 2020.
Deshaun Watson sets brand-new all-time passing record for Texans
With one game remaining in 2020, Deshaun Watson has recorded 30 passing touchdowns in a single season, becoming the Houston Texans all-time franchise leader.
Pair of Texans playmakers have big days in loss to Bengals
WR Brandin Cooks and RB David Johnson exploded for big individual days in the Texans loss to Cincinnati.
Sounds of the Game: Texans vs. Bengals Reactions
Watch the press conferences following the Houston Texans Week 16 game against the Cincinnati Bengals.
Rapid Reaction: Texans fall 37-31 to Bengals in shootout
Despite an impressive offensive performance, the Houston Texans fell, 37-31, to the Cincinnati Bengals in Week 16.
David Johnson impacts offense as pass-catcher
RB David Johnson caught 11 passes for 106 yards in Sunday's loss at Indianapolis.
Sounds of the Game: Texans vs. Colts Reactions
Watch the press conferences following the Houston Texans Week 15 game against the Indianapolis Colts.
Deshaun Watson sets new single-season TD mark
With his second touchdown pass, Deshaun Watson set a record in Sunday's game against the Indianapolis Colts.


Q: On the night of June 15 Beijing time, the Office of US Trade Representative (USTR) published a list of Chinese goods subject to additional tariffs. What's your comment?
A: I have to stress that China and the US have conducted several rounds of negotiations in an attempt to resolve our differences and achieve win-win results. It is deeply regrettable that in disregard of the consensus between the two sides, the US has demonstrated flip-flops and ignited a trade war. This move not only hurts bilateral interests, but also undermines world trade order. The Chinese side firmly opposes that.
China doesn't want a trade war. However, confronted by such short-sighted act that hurts both US itself and others, China has no choice but to fight back forcefully, to firmly safeguard the interests of the nation and its people and uphold economic globalization and the multilateral trading system. We will immediately take tariff measures of the same scale and intensity. All the economic and trade outcomes of the previous talks will also lose effect.
Waging a trade war doesn't conform to global interests nowadays. We call on all countries to take collective actions to firmly curb such outdated and regressive move and steadfastly safeguard the common interests of the mankind.


we will build on in future reports.
is often made between these two regional blocs, which represent the two leading efforts to integrate their respective regions.
The EU's economic integration is significantly deeper and supported by a far more developed institutional and legal framework.
its Member States than is the case across the 10 members of ASEAN.
Read highlights of our journey in 2017 promoting a vibrant economy, better governance and national unity in Malaysia.
Learn about the values that shape our work and meet our distinctive community of thinkers, researchers and advocates.
Interested in funding us? Read our research and advocacy plans for open markets, good governance and progress in Malaysia.


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\section{Introduction}

For positive integers $k$ and $n$ with $k\le n$, a tensor $\mathcal{T}=(T_{i_1\dots i_k})$ of order $k$ and dimension $n$ refers to a multidimensional array
with complex entries $T_{i_1\dots i_k}$ for $i_j \in [n]:=\{1,\dots, n\}$ and  $j\in[k]$.
Obviously, a  vector is a  tensor of order $1$ and a matrix is a tensor of order $2$.

Let $\mathcal{M}$  be a  tensor of order $s\geq 2$  and  dimension $n$, and $\mathcal{N}$ a tensor of order $k\ge 1$ and  dimension $n$.
The product $\mathcal{M}\mathcal{N}$ is the tensor of order $(s-1)(k-1)+1$ and dimension $n$ with entries \cite{Sh}
\[
(\mathcal{M}\mathcal{N})_{ij_{1}\dots j_{s-1}}=\sum_{i_{2},\dots ,i_{s}\in [n]}M_{ii_{2}\dots i_{s}}N_{i_{2}j_{1}}\cdots N_{i_{s}j_{s-1}},
\]
with $i\in [n]$ and $j_{1},\dots ,j_{s-1}\in [n]^{k-1}$.

For a tensor $\mathcal{T}$ of order $k\ge 2$ and dimension $n$ and  a vector $x=(x_1,\dots, x_n)^\top$,  $\mathcal{T}x$ is an $n$-dimensional vector whose $i$-th entry is
\[(\mathcal{T}x)_i=\sum_{i_2,\dots ,i_k\in [n]} T_{ii_2\dots i_k}x_{i_2}\cdots x_{i_k},\]
where $i\in[n]$. Let $x^{[r]}=(x_1^r,\dots,x_n^r)^\top$.
For some complex $\rho$, if there is a nonzero $n$-dimensional vector $x$ such that
\[\mathcal{T}x=\rho x^{[k-1]},\]
then $\rho$ is called an eigenvalue of $\mathcal{T}$, and $x$ an eigenvector of $\mathcal{T}$ corresponding to $\rho$, see \cite{Q1,Qi}.
Let $\rho(\mathcal{T})$ be the largest modulus of the eigenvalues of $\mathcal{T}$.

Let $\mathcal{G}$ be a hypergraph with vertex set $V(\mathcal{G})=[n]$ and edge set $E(\mathcal{G})$, see~\cite{Be}.
If every edge of $\mathcal{G}$ has cardinality $k$,
then we say that $\mathcal{G}$ is a $k$-uniform hypergraph.
Throughout this paper, we consider $k$-uniform hypergraphs on $n$ vertices with $2\le k\le n$. A uniform hypergraph is a hypergraph that is $k$-uniform
for some $k$.
For $i\in [n]$,  $E_i$ denotes the set of edges of $\mathcal{G}$ containing $i$.
The degree of a vertex $i$ in $\mathcal{G}$ is defined as $d_i=|E_i|$. If $d_i=d$ for $i\in V(\mathcal{G})$, then $\mathcal{G}$ is called a regular hypergraph (of degree $d$).
For $i,j\in V(\mathcal{G})$, if there is a sequence of edges $e_1,\dots,e_r$ such that $i\in e_1$, $j\in e_r$ and $e_s\cap e_{s+1}\neq \emptyset $ for all $s\in[r-1]$, then we say that $i$ and $j$ are connected.  A hypergraph is  connected if every pair of different vertices of $\mathcal{G}$ is connected.

The adjacency tensor of a  $k$-uniform hypergraph $\mathcal{G}$ on $n$ vertices is defined as the  tensor $\mathcal{A}(\mathcal{G})$ of order $k$ and dimension $n$  whose $(i_1\dots i_k)$-entry is
\[A_{i_1 \dots i_k}= \begin{cases}
\frac{1}{(k-1)!} & \text{if } \{i_1, \dots, i_k\}\in E(\mathcal{G}),\\
0  & \text{otherwise}.
\end{cases}\]
Let $\mathcal{D}(\mathcal{G})$ be the diagonal tensor of order $k$ and dimension $n$ with its diagonal entry $D_{i\dots i}$ the degree of vertex $i$ for $i\in[n]$.
Then $\mathcal{Q}(\mathcal{G})=\mathcal{D}(\mathcal{G})+\mathcal{A}(\mathcal{G})$ is the signless Laplacian tensor of  $\mathcal{G}$.
We call $\rho(\mathcal{A}(\mathcal{G}))$ the (ordinary) spectral radius of $\mathcal{G}$, which is denoted by $\rho(\mathcal{G})$, and $\rho(\mathcal{Q}(\mathcal{G}))$ the signless Laplacian spectral radius of $\mathcal{G}$, which is denoted by $\mu (\mathcal{G})$.

For a nonnegative tensor $\mathcal{T}$ of order $k\ge 2$ and dimension $n$, the $i$-th row sum of $\mathcal{T}$ is $r_i(\mathcal{T})=\sum_{i_2,\dots,i_k\in [n]}T_{ii_2\dots i_k}$. If $r_i(\mathcal{T})>0$, then the $i$-th average $2$-row sum of $\mathcal{T}$ is defined as
\[m_i(\mathcal{T})=\frac{\sum_{i_2,\dots,i_k\in [n]}T_{ii_2\dots i_k}r_{i_2}(\mathcal{T})\cdots r_{i_k}(\mathcal{T})}{r_i^{k-1}(\mathcal{T})}.\]

Let $\mathcal{G}$ be a  $k$-uniform  hypergraph on $n$ vertices. Let $\mathcal{A}=\mathcal{A}(\mathcal{G})$. For $i\in V(\mathcal{G})$ with $d_i>0$,
\begin{eqnarray*}
m_i(\mathcal{A})&=&\frac{\sum_{i_2,\dots,i_k\in [n]}A_{ii_2\dots i_k}r_{i_2}(\mathcal{A})\cdots r_{i_k}(\mathcal{A})}{r_i^{k-1}(\mathcal{A})}\\
&=&\frac{\sum_{\{i,i_2,\dots,i_k\}\in E_i}d_{i_2}\cdots d_{i_k}}{d_i^{k-1}},
\end{eqnarray*}
which is called the average $2$-degree of vertex $i$ of $\mathcal{G}$ (average of degrees of vertices in $E_i$) \cite{YZL}

For a $k$-uniform hypergraph $\mathcal{G}$ with maximum degree $\Delta$, we know that $\rho(\mathcal{G})\le \Delta$ \cite{CD} and $\mu(\mathcal{G})\le 2\Delta$ \cite{Qi} with either equality when $\mathcal{G}$ is connected if and only if $\mathcal{G}$ is regular (see \cite{QSW}). Recently, upper bounds for $\rho(\mathcal{G})$ and $\mu(\mathcal{G})$ are given in \cite{YZL} using degree sequence. In this note, we present  sharp upper bounds for $\rho(\mathcal{G})$ and $\mu(\mathcal{G})$  using average $2$-degrees or degrees, and we also give a lower bound for $\rho(\mathcal{G})$. We compare these bounds with known bounds by examples.

\section{Preliminaries}

A nonnegative tensor $\mathcal{T}$ of order $k\ge 2$ dimension $n$ is called weakly irreducible if the associated  directed graph $D_\mathcal{T}$ of $\mathcal{T}$ is strongly connected, where $D_\mathcal{T}$ is the directed graph with vertex set $\{1, \dots, n\}$ and arc set $\{(i,j): a_{ii_2\dots i_k}\ne 0 \mbox{ for some } i_s=j \mbox{ with } s=2, \dots, k\}$ \cite{FGH,Qi}.

For an $n$-dimensional real vector $x$, let $\|x\|_k=(\sum_{i=1}^n|x_i|^k)^{\frac{1}{k}}$, and  if $\|x\|_k=1$, then we say that $x$ is a unit vector. Let $\mathbb{R}^n_{+}$ be the set of  $n$-dimensional nonnegative vectors.

\begin{Lemma}\label{irreducibility}\cite{FGH,YY}
Let $\mathcal{T}$ be a  nonnegative tensor. Then $\rho(\mathcal{T})$ is an eigenvalue of $\mathcal{T}$ and there is a unit nonnegative eigenvector corresponding to $\rho(\mathcal{T})$.
If furthermore $\mathcal{T}$ is weakly irreducible, then there is
a unique unit positive eigenvector corresponding to $\rho(\mathcal{T})$.

\end{Lemma}

\begin{Lemma} \cite{Qi} \label{QLQ}
Let $\mathcal{G}$ be a $k$-uniform hypergraph with $n$ vertices. Then
$\rho(\mathcal{G})=\max\{x^\top(\mathcal{A}(\mathcal{G})x): x\in\mathbb{R}^n_{+},\|x\|_{k}=1\}$.
\end{Lemma}

\begin{Lemma} \label{connec} \cite{PT,Qi} Let $\mathcal{G}$ be a $k$-uniform hypergraph. Then $\mathcal{A}(\mathcal{G})$ ($\mathcal{Q}(G)$, respectively) is weakly irreducible if and only if $\mathcal{G}$ is connected.
\end{Lemma}

A hypergraph $\mathcal{H}$ is a subhypergraph of $\mathcal{G}$ if $V(\mathcal{H})\subseteq V(\mathcal{G})$ and  $E(\mathcal{H})\subseteq E(\mathcal{G})$

\begin{Lemma} \cite{CD, FYZ} \label{subhypergraph}
Let $\mathcal{G}$ be a connected $k$-uniform hypergraph and $\mathcal{H}$ a subhypergraph of $\mathcal{G}$. Then
$\rho(\mathcal{H})\leq \rho(\mathcal{G})$ with equality if and only if $\mathcal{H}=\mathcal{G}$.
\end{Lemma}

For two tensors $\mathcal{M}$ and $\mathcal{N}$ of order $k\ge 2$ and dimension $n$, if  there is an $n\times n$ nonsingular diagonal matrix $U$ such that $\mathcal{N}=U^{-(k-1)}\mathcal{M}U$, then we say that $\mathcal{M}$ and $\mathcal{N}$ are  diagonal similar.

\begin{Lemma}\cite{Sh} \label{Tensor1}
Let $\mathcal{M}$ and $\mathcal{N}$ be two   diagonal similar tensors of order $k\ge 2$ and dimension $n$. Then $\mathcal{M}$  and $\mathcal{N}$
have the same  real eigenvalues.
\end{Lemma}

\begin{Lemma}\label{Tensor2} \cite{LCL,YY}
Let $\mathcal{T}$ be a nonnegative tensor of order $k\ge 2$ and dimension $n$. Then
\[\min_{1\leq i \leq n}r_i(\mathcal{T})\leq \rho(\mathcal{T})\leq \max_{1\leq i \leq n}r_i(\mathcal{T}).\]
Moreover, if $\mathcal{T}$ is weakly irreducible, then either equality holds if and only if $r_1(\mathcal{T})=\cdots=r_n(\mathcal{T})$.
\end{Lemma}

\begin{Proposition}\label{Tensor3}
Let $\mathcal{T}$ be a nonnegative tensor of order $k\ge 2$ and dimension $n$ with all row sums positive. Then
\[\min_{1\leq i \leq n}m_i(\mathcal{T})\leq \rho(\mathcal{T})\leq \max_{1\leq i \leq n}m_i(\mathcal{T}).\]
Moreover, if $\mathcal{T}$ is weakly irreducible, then either equality holds if and only if $m_1(\mathcal{T})=\cdots=m_n(\mathcal{T})$.
\end{Proposition}

\begin{proof} Let $U=diag (r_1(\mathcal{T}),\dots, r_n(\mathcal{T}))$ and $\mathcal{B}=U^{-(k-1)}\mathcal{T}U$.
Then $\mathcal{T}$ and $\mathcal{B}$ are diagonal similar, and thus we have by Lemma~\ref{Tensor1} that $\rho(\mathcal{T})=\rho(\mathcal{B})$.
Obviously,
\[B_{i_1\dots i_k}=\frac{T_{i_1\dots i_k}r_{i_2}(\mathcal{T})\cdots r_{i_k}(\mathcal{T})}{r_{i_1}^{k-1}(\mathcal{T})}\]
for $i_1,i_2,\dots, i_k\in[n]$.
Thus
\[r_i(\mathcal{B})=\frac{\sum_{i_2,\dots, i_k\in[n]}T_{ii_2\dots i_k}r_{i_2}(\mathcal{T})\cdots r_{i_k}(\mathcal{T})}{r_{i}^{k-1}(\mathcal{T})}=m_i(\mathcal{T})\]
for $i\in[n]$.
By Lemma~\ref{Tensor2}, we have \[\min_{1\leq i \leq n}m_i(\mathcal{T})=\min_{1\leq i \leq n}r_i(\mathcal{B})\leq \rho(\mathcal{T})=\rho(\mathcal{B})\leq \max_{1\leq i \leq n}r_i(\mathcal{B})=\max_{1\leq i \leq n}m_i(\mathcal{T}),\]
and if $\mathcal{T}$ is weakly irreducible, then since $D_\mathcal{T}=D_\mathcal{B}$, $\mathcal{B}$ is also weakly irreducible, and thus $\rho(\mathcal{T})=\min_{1\leq i \leq n}m_i(\mathcal{T})$ or $\rho(\mathcal{T})=\max_{1\leq i \leq n}m_i(\mathcal{T})$ if and only if $r_1(\mathcal{B})=\cdots=r_n(\mathcal{B})$, i.e., $m_1(\mathcal{T})=\cdots=m_n(\mathcal{T})$.
\end{proof}

For a  hypergraph $\mathcal{G}$, the blow-up of $\mathcal{G}$, denoted by $\mathcal{G}^1$, is the hypergraph obtained from $\mathcal{G}$ by adding a new common vertex $v$ to each edge. If $\mathcal{G}$ is a regular $(k-1)$-uniform hypergraph on $n-1$ vertices of degree $d$, then  $\mathcal{G}^1$ is a $k$-uniform hypergraph on $n$ vertices.

We use the techniques in \cite{YZL}.

\section{Main results}

Let $\mathcal{G}$ be a $k$-uniform hypergraph on $n$ vertices  without isolated vertices with average $2$-degrees $m_1\geq \dots \geq m_n$. By Proposition \ref{Tensor3},
$\rho(\mathcal{G})\leq m_1$. In the following, we give a upper bound for $\rho(\mathcal{G})$ using $m_1$ and $m_2$.

\begin{Theorem}\label{Tensor5}
Let $\mathcal{G}$ be a $k$-uniform hypergraph on $n$ vertices without isolated vertices with average $2$-degrees $m_1\geq  \dots \geq m_n$.
Then
\begin{equation} \label{HY}
\rho(\mathcal{G})\leq m_1^{\frac{1}{k}}m_{2}^{1-\frac{1}{k}}.
\end{equation}
Moreover, if $\mathcal{G}$ is connected,  then equality holds in (\ref{HY}) if and only  if each vertex of $\mathcal{G}$ has the same average $2$-degree
\end{Theorem}

\begin{proof} Let $\mathcal{A}=\mathcal{A}(\mathcal{G})$.

If $m_1=m_2$, then by Proposition~\ref{Tensor3}, we have \[\rho(\mathcal{G})\leq m_1^{\frac{1}{k}}m_{2}^{1-\frac{1}{k}}=m_1,\]
and when $\mathcal{G}$ is connected, $A$ is weakly irreducible, and thus
equality holds in (\ref{HY}) if and only  if each vertex of  $\mathcal{G}$ has the same average $2$-degree.

Suppose in the following that $m_1>m_2$.
Let $d_1, d_2,\dots, d_n$ be the degree sequence of $\mathcal{G}$.
Let $U$ be diagonal matrix $\mbox{diag}(td_1, d_2, \dots, d_n)$, where $t>1$ is a variable to be determined later.
Let  $\mathcal{T}=U^{-(k-1)}\mathcal{A}U$.
Then $\mathcal{A}$ and $\mathcal{T}$ are diagonal similar.
By Lemma~\ref{Tensor1}, $\mathcal{A}$ and $\mathcal{T}$ have the same real eigenvalues. By Lemma~\ref{irreducibility}, $\rho(\mathcal{A})$ is an eigenvalue of $\mathcal{A}$  and $\rho(\mathcal{T})$ is an eigenvalue of $\mathcal{T}$. Thus $\rho(\mathcal{G})=\rho(\mathcal{A})=\rho(\mathcal{T})$.
Obviously,
\[T_{i_1\dots i_k}=U^{-(k-1)}_{i_1i_1}A_{i_1\dots i_k}U_{i_2i_2}\cdots U_{i_ki_k}\]
for $i_1,\dots, i_k\in[n]$.
Then
\begin{eqnarray*}
r_1(\mathcal{T})&=& \sum_{i_2,\dots, i_k\in[n]}T_{ii_2\dots i_k}\\
&=&\sum_{i_2,\dots, i_k\in[n]}U^{-(k-1)}_{11}A_{1i_2\dots i_k}U_{i_2i_2}\cdots U_{i_ki_k}\\
&=& \sum_{i_2,\dots, i_k\in[n]\setminus\{1\}}(td_1)^{-(k-1)}A_{1i_2\dots i_k}d_{i_2}\cdots d_{i_k}\\
&=&\frac{\sum_{\{1,i_2,\dots, i_k\}\in E_1}d_{i_2}\cdots d_{i_k}}{(t d_1)^{k-1}}\\
&=& \frac{m_1}{t^{k-1}}.
\end{eqnarray*}
For $i=2, \dots, n$, let \[m_{1,i}=m_i-\frac{\sum_{1\notin\{i,i_2,\dots,i_k\}\in E_i}d_{i_2}\cdots d_{i_k}}{d_i^{k-1}},\]
and then
\begin{eqnarray*}
r_i(\mathcal{T})&=& \sum_{i_2,\dots, i_k\in[n]}T_{ii_2\dots i_k}\\
&=&\sum_{i_2,\dots, i_k\in[n]}U^{-(k-1)}_{ii}A_{ii_2\dots i_k}U_{i_2i_2}\cdots U_{i_ki_k}\\
&=&\sum_{i_2,\dots, i_k\in[n]\atop 1\in\{i_2,\dots, i_k\}}d_i^{-(k-1)}A_{ii_2\dots i_k}d_{i_2}\cdots d_{i_k}+\sum_{i_2,\dots, i_k\in[n]\atop 1\notin\{i_2,\dots, i_k\}}d_i^{-(k-1)}A_{ii_2\dots i_k}d_{i_2}\cdots d_{i_k}\\
&=&\frac{\sum_{i_2,\dots, i_k\in[n]\atop 1\in\{i_2,\dots, i_k\}}A_{ii_2\dots i_k}d_{i_2}d_{i_3}\cdots d_{i_k}}{d_i^{k-1}}+\frac{\sum_{i_2,\dots, i_k\in[n]\atop 1\notin\{i_2,\dots, i_k\}}A_{ii_2\dots i_k}d_{i_2}\cdots d_{i_k}}{d_i^{k-1}}\\
&=&\frac{\sum_{\{i,1,i_3,\dots, i_k\}\in E_i}(td_1)d_{i_3}\cdots d_{i_k}}{d_i^{k-1}}+\frac{\sum_{1\notin\{i,i_2,\dots,i_k\}\in E_i}d_{i_2}\cdots d_{i_k}}{d_i^{k-1}}\\
&=&\frac{t\sum_{\{i,1,i_3,\dots, i_k\}\in E_i} d_1d_{i_3}\cdots d_{i_k}}{d_i^{k-1}}+\frac{\sum_{1\notin\{i,i_2,\dots,i_k\}\in E_i}d_{i_2}\cdots d_{i_k}}{d_i^{k-1}}\\
&=& tm_{1,i}+m_i-m_{1,i}\\
&\leq& tm_i\\&\leq& tm_2,
\end{eqnarray*}
with equality if and only if $m_i=m_{1,i}$ and $m_i=m_2$.
Take $t=\left(\frac{m_1}{m_2}\right)^\frac{1}{k}$. Obviously, $t> 1$.
Then $r_1(\mathcal{T})=m_1^{\frac{1}{k}}m_{2}^{1-\frac{1}{k}}$ and
for $2\leq i\leq n$, $r_i(\mathcal{T})\leq tm_2=m_1^{\frac{1}{k}}m_{2}^{1-\frac{1}{k}}$. By Lemma~\ref{Tensor2},
 \[\rho(\mathcal{G})=\rho(\mathcal{T})\leq m_1^{\frac{1}{k}}m_{2}^{1-\frac{1}{k}}.\]

Now suppose that $\mathcal{G}$ is connected. By Lemma \ref{connec}, $\mathcal{A}(\mathcal{G})$ is weakly irreducible, and thus $\mathcal{T}$ is weakly irreducible since $D_{\mathcal{A}(\mathcal{G})}=D_{\mathcal{T}}$.

Suppose that  equality holds in (\ref{HY}). By Lemma \ref{Tensor2}, $r_1(\mathcal{T})=\cdots=r_n(\mathcal{T})$.
From the above argument, we have (i) $m_2=\cdots=m_n$, and (ii) $m_i=m_{1,i}$ for $i=2,\ldots,n$.  From (ii) and the definition of $m_{1,i}$, each edge of $\mathcal{G}$ contains vertex $1$, which implies that $d_1(k-1)=\sum_{i=2}^nd_i$. From (i),
\[m_1=\frac{\sum_{i=2}^n \frac{d^{k}_im_i}{d_1}}{d_1^{k-1}(k-1)}=\frac{\sum_{i=2}^n d^k_im_2}{d_1^k}=\frac{\sum_{i=2}^n d_im_2\left(\frac{d_i}{d_k}\right)^{k-1}}{d_1(k-1)}\leq\frac{\sum_{i=2}^nm_2d_i}{d_1(k-1)} =m_2,\]
a contradiction. Thus the inequality (\ref{HY})  is strict  if $m_1>m_2$.
\end{proof}

Let $\mathcal{G}$ be a connected $k$-uniform hypergraph on $n$ vertices with degree sequence $d_1\ge \cdots \ge d_n$. Then \cite{YZL}
\begin{equation} \label{old}
\rho(\mathcal{G})\le d_1^{\frac{1}{k}}d_2^{1-\frac{1}{k}}
\end{equation}
with equality if and only if $\mathcal{G}$ is a regular hypergraph or the blow-up hypergraph $\mathcal{H}^1$ of a regular $(k-1)$-uniform hypergraph $\mathcal{H}$ on $n-1$ vertices.

Obviously, if $\mathcal{G}$ is a regular hypergraph, then each vertex of $\mathcal{G}$ has the same average $2$-degree, and thus equality holds in (\ref{HY}) and (\ref{old}). For the blow-up hypergraph $\mathcal{H}^1$ of a regular $(k-1)$-uniform hypergraph $\mathcal{H}$ on $n-1$ vertices, the upper bound in (\ref{old}) is attained, while the upper bound in (\ref{HY}) is not attained. However,
in the following, we give two examples to show that
there are irregular hypergraphs for which each vertex has the same average $2$-degree. For such hypergraphs, the upper in (\ref{HY}) is attained, while the upper bound in (\ref{old}) can not be attained.

Let $\mathcal{H}_1$ be a $3$-uniform hypergraph with vertex set $V(\mathcal{H}_1)=[34]$ and $E(\mathcal{H}_1)=\{e_i:1\leq i \leq 51\}$, where
\begin{eqnarray*}
\begin{matrix}
e_1=\{1,2,5\}, & e_2=\{1,2,6\}, & e_3=\{1,2,7\}, &
e_4=\{1,2,8\},\\
e_5=\{1,2,9\}, & e_6=\{1,2,10\}, & e_7=\{1,2,11\}, & e_8=\{1,2,12\},\\
e_9=\{1,2,13\}, & e_{10}=\{3,4,5\}, & e_{11}=\{3,4,6\}, & e_{12}=\{3,4,7\},\\
e_{13}=\{3,4,8\}, & e_{14}=\{3,4,9\},& e_{15}=\{3,4,10\},& e_{16}=\{3,4,11\},\\
e_{17}=\{3,4,12\}, & e_{18}=\{3,4,13\}, & e_{19}=\{5,6,14\}, & e_{20}=\{6,7,15\}, \\
e_{21}=\{7,8,16\}, & e_{22}=\{8,9,17\}, & e_{23}=\{9,10,18\}, & e_{24}=\{10,11,19\}, \\
e_{25}=\{11,12,20\}, & e_{26}=\{12,13,21\}, & e_{27}=\{13,5,22\}, & e_{28}=\{5,23,24\}, \\
e_{29}=\{5,25,26\}, & e_{30}=\{6,27,28\}, & e_{31}=\{6,29,30\}, & e_{32}=\{7,31,32\}, \\
e_{33}=\{7,33,34\}, & e_{34}=\{8,23,24\}, & e_{35}=\{8,25,26\}, & e_{36}=\{9,27,28\}, \\
e_{37}=\{9,29,30\}, & e_{38}=\{10,31,32\}, & e_{39}=\{10,33,34\}, & e_{40}=\{11,23,24\}, \\
e_{41}=\{11,25,26\}, & e_{42}=\{12,27,28\}, & e_{43}=\{12,29,30\}, & e_{44}=\{13,31,32\}, \\
e_{45}=\{13,33,34\}, & e_{46}=\{14,15,16\}, & e_{47}=\{17,18,19\}, & e_{48}=\{20,21,22\}, \\
e_{49}=\{14,17,20\}, & e_{50}=\{15,18,21\}, & e_{51}=\{16,19,22\}.
\end{matrix}
\end{eqnarray*}
By direct calculation, we have
\[d_i= \begin{cases}
9& \text{if } 1\leq i\leq 4,\\
6 & \text{if } 5\leq i\leq 13,\\
3 & \text{if } 14\leq i\leq 34,
\end{cases}\]
and
\begin{eqnarray*}
m_i&=& \begin{cases}
\frac{(9\times 6)\times 9}{9\times9}& \text{if } 1\leq i\leq 4,\\
\frac{(9\times 9)\times 2+(6\times 3)\times 2+(3\times 3)\times 2}{6\times6} & \text{if } 5\leq i\leq 13,\\
\frac{6\times 6+(3\times 3)\times 2}{3\times3} & \text{if } 14\leq i\leq 22,\\
\frac{(6\times 3)\times 3}{3\times3} & \text{if } 23\leq i\leq 34
\end{cases}\\
&=&6.
\end{eqnarray*}
By Theorem~\ref{Tensor5}, we have $\rho(\mathcal{H}_1)=6$.

Let $\mathcal{H}_2$ be a $3$-uniform hypergraph with vertex set $V(\mathcal{H}_2)=[54]$ and $E(\mathcal{H}_2)=\{e_i:1\leq i \leq 64\}$, where
\begin{eqnarray*}
\begin{matrix}
e_1=\{1,2,3\}, & e_2=\{3,4,6\}, & e_3=\{3,4,5\}, & e_4=\{4,5,6\}, \\
e_5=\{1,2,6\}, & e_6=\{1,2,5\}, & e_7=\{1,3,4\}, & e_8=\{2,5,6\},\\
e_9=\{1,7,8\}, & e_{10}=\{1,8,9\}, & e_{11}=\{1,9,10\}, & e_{12}=\{1,10,11\},\\
e_{13}=\{2,11,12\}, & e_{14}=\{2,12,13\}, & e_{15}=\{2,13,14\}, & e_{16}=\{2,14,15\},\\
e_{17}=\{3,15,16\}, & e_{18}=\{3,16,17\}, & e_{19}=\{3,17,18\}, & e_{20}=\{3,18,19\}, \\
e_{21}=\{4,19,20\}, & e_{22}=\{4,20,21\}, & e_{23}=\{4,21,22\}, & e_{24}=\{4,22,23\}, \\
e_{25}=\{5,23,24\}, & e_{26}=\{5,24,25\}, & e_{27}=\{5,25,26\}, & e_{28}=\{5,26,27\}, \\
e_{29}=\{6,27,28\}, & e_{30}=\{6,28,29\}, & e_{31}=\{6,29,30\}, & e_{32}=\{6,30,7\}, \\
e_{33}=\{7,8,31\}, & e_{34}=\{7,8,32\}, & e_{35}=\{9,10,33\}, & e_{36}=\{9,10,34\}, \\
e_{37}=\{11,12,35\}, & e_{38}=\{11,12,36\}, & e_{39}=\{13,14,37\}, & e_{40}=\{13,14,38\},\\
e_{41}=\{15,16,39\}, & e_{42}=\{15,16,40\}, & e_{43}=\{17,18,41\}, & e_{44}=\{17,18,42\},\\
e_{45}=\{19,20,43\}, & e_{46}=\{19,20,44\}, & e_{47}=\{21,22,45\}, & e_{48}=\{21,22,46\},\\
e_{49}=\{23,24,47\}, & e_{50}=\{23,24,48\}, & e_{51}=\{25,26,49\}, & e_{52}=\{25,26,50\},\\
e_{53}=\{27,28,51\}, & e_{54}=\{27,28,52\}, & e_{55}=\{29,30,53\}, & e_{56}=\{29,30,54\}, \\
e_{57}=\{31,32,33\}, & e_{58}=\{34,35,36\}, & e_{59}=\{37,38,39\}, & e_{60}=\{40,41,42\}, \\
e_{61}=\{43,44,45\}, & e_{62}=\{46,47,48\}, & e_{63}=\{49,50,51\}, & e_{64}=\{52,53,54\}.
\end{matrix}
\end{eqnarray*}
By direct calculation, we have
\[d_i= \begin{cases}
8& \text{if } 1\leq i\leq 6,\\
4 & \text{if } 7\leq i\leq 30,\\
2 & \text{if } 31\leq i\leq 54,
\end{cases}\]
and
\begin{eqnarray*}
m_i&=& \begin{cases}
\frac{(8\times 8)\times 4+(4\times 4)\times 4}{8\times8}& \text{if } 1\leq i\leq 6,\\
\frac{(8\times 4)\times 2+(4\times 3)\times 2}{4\times4} & \text{if } 7\leq i\leq 30,\\
\frac{4\times 4+2\times 2}{2\times2} & \text{if } 31\leq i\leq 54
\end{cases}\\
&=&5.
\end{eqnarray*}
By Theorem~\ref{Tensor5}, we have $\rho(\mathcal{H}_2)=5$.

Let $\mathcal{G}$ be a $k$-uniform hypergraph of order $n$  without isolated vertices with maximum degree $\Delta$ and  average $2$-degrees $m_1\geq \dots \geq m_n$. Note that $\mu(\mathcal{G})\le \Delta+\rho(\mathcal{G})$. By Theorem~\ref{Tensor5}, we have
Then $\mu(\mathcal{G})\leq m_1^{\frac{1}{k}}m_{2}^{1-\frac{1}{k}}+\Delta$.

If we  take $U=\mbox{diag} (d_1, \ldots, d_{n-1},yd_n)$
with $y=\left(\frac{m_n}{m_{n-1}}\right)^\frac{1}{k}$ in the proof of Theorem~\ref{Tensor5}, then $\rho(\mathcal{G})\geq m_n^{\frac{1}{k}}m_{n-1}^{1-\frac{1}{k}}$,
and if $\mathcal{G}$ is connected,  then equality holds if and only  if each vertex of $\mathcal{G}$ has the same average $2$-degree.

For a $k$-uniform hypergraph $\mathcal{G}$, if there is a disjoint partition of $V(\mathcal{G})$ as $V(\mathcal{G})=V_0\cup V_1\cup\dots\cup V_d$, where $|V_0|=1,|V_1|=\dots=|V_d|=k-1$, and $E(\mathcal{G})=\{V_0\cup V_i:i\in[d]\}$, then $\mathcal{G}$ is called a hyperstar, denoted by $\mathcal{S}_d^k$. The  vertex (of degree $d$) in $V_0$  is called the heart. Obviously, it is an isolated vertex if $d=0$.

For positive integers $d_1,\gamma$ and nonnegative integer $d_2$,  let $\mathcal{G}_{d_1,d_2,\gamma}$ be the $k$-uniform hypergraph obtained vertex-disjoint $\mathcal{S}_{d_1}^k$ and $\mathcal{S}_{d_2}^k$ by adding $\gamma(k-2)$ new vertices $v_{1,1},\dots ,v_{1,k-2},\dots,$ $v_{\gamma,1},\dots ,v_{\gamma,k-2} $ and $\gamma$ new edges $e_1,\dots,e_{\gamma}$,
 where $e_i=\{u,v,v_{i,1},\dots ,v_{i,k-2}\}$ for $ i\in[\gamma]$, and $u,v$ are the hearts of  $\mathcal{S}_{d_1}^k$ and $\mathcal{S}_{d_2}^k$, respectively. Obviously, if $d_2=0$ and $\gamma=1$, then $\mathcal{G}_{d_1,d_2,\gamma}\cong \mathcal{S}_{d_1+1}^k$.

Next we give a lower bound for $\rho(\mathcal{G})$ of a $k$-uniform hypergraph $\mathcal{G}$.

\begin{Theorem} \label{Tensor9}  Let $\mathcal{G}$ be a $k$-uniform  hypergraph  with $u\in V(\mathcal{G})$ of maximum degree $\Delta\ge 1$. Let $v$ be a  neighbor of $u$ with maximum degree. Then
\begin{equation} \label{star}
\rho(\mathcal{G})\geq \left(\frac{\Delta+\delta-2\gamma+\gamma^2+\sqrt{(\Delta-\delta)^2+\gamma^4+2(\Delta +\delta-2\gamma)\gamma^2}}{2}\right)^{\frac{1}{k}},
\end{equation}
where $\delta$ is the degree of $v$, and  $\gamma$ is the number of edges containing $u$ and $v$.
Moreover, if $\mathcal{G}$ is connected, then equality holds in (\ref{star})  if and only if $\mathcal{G}\cong \mathcal{G}_{\Delta,\delta,\gamma}$.
\end{Theorem}

\begin{proof}
Let $e_1, \dots, e_{\Delta}$ be the $\Delta$ edges of $\mathcal{G}$ containing  $u$. Among these edges, $\gamma$ of them, say  $e_1, \dots, e_{\gamma}$, contain $v$.
Let $e_{\Delta+1}, \dots, e_{\Delta+\delta-\gamma}$ be the $\delta-\gamma$ edges of $\mathcal{G}$ containing  $v$  different from $e_1, \dots, e_{\gamma}$.
Let $\mathcal{G}_1$ be the  subhypergraph of $\mathcal{G}$ induced by $\{e_1, \dots, e_{\Delta+\delta-\gamma}\}$. Then $V(\mathcal{G}_1)=\cup_{i=1}^{\Delta+\delta-\gamma}e_{i}$.
For $1\leq i\leq \Delta+\delta-\gamma$, let
$e_{i}=\{v_{i,1},\dots,v_{i,k}\}$, where $v_{i,1}=u$ and $v_{i,2}=v$ if  $1\leq i\leq \gamma$, $v_{i,1}=u$ if $\gamma+1\leq i\leq \Delta$, and $v_{i,1}=v$ if $\Delta+1\leq i\leq \Delta+\delta-\gamma$.
Note that maybe some of $v_{i,s}$ and $v_{j,t}$ for $1\leq s, t\leq k$ and  $1\leq i<j\leq \Delta+\delta-\gamma$  with $v_{i,s}, v_{j,t}\ne u,v$
represent the same vertex.

Let $\mathcal{G}'_1$ be a new hypergraph such that $V(\mathcal{G}'_1)=\cup_{i=1}^{\Delta+\delta-\gamma}e'_{i}$ and
$E(\mathcal{G}'_1)=\{e'_1, \dots, e'_{\Delta+\delta-\gamma}\}$, where
 $e'_{i}=\{v'_{i,1},\dots,v'_{i,k}\}$ with  $v'_{i,1}=u$ and $v'_{i,2}=v$ if $i=1, \dots, \gamma$,
$v'_{i,1}=u$ if $\gamma+1\leq i\leq \Delta$, and $v'_{i,1}=v$ if $\Delta+1\leq i\leq \Delta+\delta-\gamma$.
Note that $v\notin e_i$ for  $\gamma+1\leq i\leq \Delta$, $u\notin e_i$ for $\Delta+1\leq i\leq \Delta+\delta-\gamma$, and
$v'_{i,s}$  and $v'_{j,t}$ for $1\leq s, t\leq k$ and  $1\le i<j\le \Delta+\delta-\gamma$ with $v'_{i,s}, v'_{j,t}\ne u,v$ are different vertices.
Obviously, $\mathcal{G}'_1\cong \mathcal{G}_{\Delta,\delta,\gamma}$

By Lemma~\ref{irreducibility}, there is a unit positive eigenvector $x$ of $\mathcal{A}(\mathcal{G}'_1)$ corresponding to $\rho(\mathcal{G}'_1)$, in which the entry at $v'_{i,s}$ is denoted by $x_{i,s}$, where $1\leq i\leq \Delta+\delta-\gamma$ and $1\leq s\leq k$.  Then  $\rho(\mathcal{G}'_1)=x^\top (\mathcal{A}(\mathcal{G}'_1)x)$.
Let $w$ be any vertex of $\cup_{i=\Delta-\gamma+1}^{\Delta}e_{i}\setminus\{u\}$.
Since $\rho(\mathcal{G}'_1)x_w^{k-1}=x_u x^{k-2}_w$, we have $x_w=\frac{ x_u}{\rho(\mathcal{G}'_1)}$.
Thus the entry of $x$ at each vertex of $\cup_{i=\Delta-\gamma+1}^{\Delta}e_{i}\setminus\{u\}$ is the same, denoted by $a$.
Similarly, the entry of $x'$ at each vertex of $\cup_{i=1}^{\gamma}e_{i}\setminus\{u,v\}$ is the same, denoted by $b$,
and the entry of $x'$ at each vertex of
$\cup_{i=\Delta+1}^{\Delta+\delta-\gamma}e_{i}\setminus\{v\}$ is the same, denoted by $c$.
Then
\begin{eqnarray*
\rho(\mathcal{G}'_1)a^{k-1}&=&  x_u a^{k-2},\\
\rho(\mathcal{G}'_1)x^{k-1}_u&=&  (\Delta-\gamma) a^{k-1}+\gamma b^{k-2}x_v,\\
\rho(\mathcal{G}'_1)b^{k-1}&=&  x_ux_vb^{k-3},\\
\rho(\mathcal{G}'_1)x^{k-1}_v&=& (\delta-\gamma) c^{k-1}+\gamma b^{k-2}x_u, \\
\rho(\mathcal{G}'_1)c^{k-1}&=& x_vc^{k-2}.
\end{eqnarray*}
Thus $\rho(\mathcal{G}'_1)$ is the largest root of the equation $f(\rho)=0$, where $f(\rho)=(\rho^k-\Delta+\gamma)\left(\rho^{2k}-(\Delta+\delta-2\gamma+\gamma^2)\rho^k+(\Delta-\gamma)(\delta-\gamma)\right)$.
It follows that \[\rho(\mathcal{G}'_1)=\left(\frac{\Delta+\delta-2\gamma+\gamma^2+\sqrt{(\Delta-\delta)^2+\gamma^4+2(\Delta +\delta-2\gamma)\gamma^2}}{2}\right)^{\frac{1}{k}}.\]

Construct a surjection $\sigma$ from $V(\mathcal{G}'_1)$ to $V(\mathcal{G}_1)$ such that $\sigma(v'_{i,s})=v_{i,s}$ for $1\leq i\leq \Delta +\delta-\gamma$ and  $1\leq s\leq k$.  Let $y=(y_1,\dots,y_{|V(\mathcal{G}_1)|})^{\top}$ such that $y_{i}=\max_{v'_{j,s}\in\sigma^{-1}(i)}\{x_{j,s}\}$ for $1\leq i\leq |V(\mathcal{G}_1)|$.
Obviously, $\|y\|_{k}\leq \|x\|_{k}=1$. Let $z=\frac{y}{\|y\|_{k}}$. Then $\|z\|_{k}=1$.
By Lemma~\ref{QLQ},
\begin{equation} \label{med}
\rho(\mathcal{G}_1)\geq z^\top (\mathcal{A}(\mathcal{G}_1)z) =\frac{y^\top (\mathcal{A}(\mathcal{G}_1)y)}{\|y\|_{k}^k}\geq\frac{x^\top (\mathcal{A}(\mathcal{G}'_1)x)}{\|x\|_{k}^k}= x^\top (\mathcal{A}(\mathcal{G}'_1)x)
=\rho(\mathcal{G}'_1).
\end{equation}
Since $\mathcal{G}_1$ is a subhypergraph of $\mathcal{G}$, we have by Lemma~\ref{subhypergraph} that $\rho(\mathcal{G})\geq\rho(\mathcal{G}_1)$.
Thus
 \[\rho(\mathcal{G})\geq\rho(\mathcal{G}'_1)= \left(\frac{\Delta+\delta-2\gamma+\gamma^2+\sqrt{(\Delta-\delta)^2+\gamma^4+2(\Delta +\delta-2\gamma)\gamma^2}}{2}\right)^{\frac{1}{k}}.\]

If $\mathcal{G}\cong \mathcal{G}_{\Delta,\delta,\gamma}$, then by the above proof, equality holds in (\ref{star}).

Suppose that $\mathcal{G}$ is connected and  equality holds in (\ref{star}). Then all equalities hold in (\ref{med}) and $\rho(\mathcal{G})=\rho(\mathcal{G}_1)$.
Thus by the construction of $\mathcal{G}'_1$, we have $\mathcal{G}_1 \cong \mathcal{G}'_1$. Otherwise, $|V(\mathcal{G}_1)| <|V(\mathcal{G}'_1)|$, and then $\|y\|_{k}< \|x\|_{k}=1$, a contradiction.
By Lemma~\ref{subhypergraph}, we have $\mathcal{G}=\mathcal{G}_1$. Thus $\mathcal{G}\cong \mathcal{G}_{\Delta,\delta,\gamma}$.
\end{proof}

Let $\mathcal{G}$ be a $k$-uniform  hypergraph  with maximum degree $\Delta\ge 1$. Let $f(\Delta, \delta, \gamma)$ be the  lower bound in (\ref{star}). For $\gamma\le \delta\le \Delta$, $f(\Delta, \delta, \gamma)$ is a increasing  function at $\delta$. Note that $\gamma\ge 1$.
By  Theorem~\ref{Tensor9},
$\rho(\mathcal{G})\geq f(\Delta, 1,1)=\Delta^{\frac{1}{k}}$,
and  if $\mathcal{G}$ is connected, then equality holds if and only if $\mathcal{G}$ is a hyperstar. Moreover, if $\mathcal{G}$ is connected and is not a hyperstar, then $\rho(\mathcal{G})\geq f(\Delta, 2,1)= \left(\frac{\Delta+1+\sqrt{\Delta^2-2\Delta+5}}{2}\right)^{\frac{1}{k}}$ with equality if and only if $\mathcal{G}\cong \mathcal{G}_{\Delta,2,1}$.

In the following, we give upper bounds for $\mu(\mathcal{G})$ of a $k$-uniform hypergraph.

\begin{Theorem} \label{Mo} Let $\mathcal{G}$ be a $k$-uniform  hypergraph on $n$ vertices with degree sequence $d_{1}\geq \dots \geq d_{n}$.  Let $d^*=1$ if $d_1=d_2$ and $d^*$ be  a root of $h(t)=0$ in $((\frac{d_{1}}{d_{2}})^{\frac{1}{k}},\frac{d_{1}}{d_{2}})$ if $d_1>d_2$, where $h(t)=d_{2}t^{k}+(d_{2}-d_{1})t^{k-1}-d_{1}$. Then
\begin{equation} \label{MZh}
\mu(\mathcal{G})\leq d_{1}+d_{1}\left(\frac{1}{d^*}\right)^{k-1}.
\end{equation}
 Moreover, if $\mathcal{G}$ is connected, then equality holds in (\ref{MZh}) if and only if $\mathcal{G}$ is a regular hypergraph or the blow-up hypergraph of a regular $(k-1)$-uniform hypergraph on $n-1$ vertices.
\end{Theorem}

\begin{proof} Let $\mathcal{Q}=\mathcal{Q}(\mathcal{\mathcal{G}})$, $\mathcal{A}=\mathcal{A}(\mathcal{\mathcal{G}})$, and $\mathcal{D}=\mathcal{D}(\mathcal{\mathcal{G}})$.

If $d_{1}=d_{2}$, then $d^*=1$, and by Lemma~\ref{Tensor2}, we have
\[
\mu(\mathcal{G})=\rho(\mathcal{Q})\leq\max_{1\leq i\leq n}r_{i}(\mathcal{Q})=\max_{1\leq i\leq n}2d_{i}=2d_{1}= d_{1}+d_{1}\left(\frac{1}{d^*}\right)^{k-1},
\]
and when $\mathcal{G}$ is connected, we have by Lemma~\ref{connec} that $\mathcal{Q}$ is weakly irreducible, and thus equality holds  if and only if $r_1(\mathcal{Q})=\dots =r_n(\mathcal{Q})$, i.e., $\mathcal{G}$ is a regular hypergraph.

Suppose in the following that $d_{1}> d_{2}$. Let $U=\mbox{diag}(t,1,\dots,1)$ be an $n\times n$ diagonal matrix, where $t>1$ is a variable to be determined later. Let $\mathcal{T}=U^{-(k-1)}\mathcal{Q}U$. By Lemma~\ref{Tensor1}, $\mathcal{Q}$ and $\mathcal{T}$ have the same real eigenvalues. By Lemma~\ref{irreducibility}, $\rho(\mathcal{Q})$ is an eigenvalue of $\mathcal{Q}$  and $\rho(\mathcal{T})$ is an eigenvalue of $\mathcal{T}$. Thus $\mu(\mathcal{G})=\rho(\mathcal{Q})=\rho(\mathcal{T})$.
We have
\begin{eqnarray*}
r_{1}(\mathcal{T})&=&\sum_{i_{2},\ldots,i_{k}\in[n]}T_{1i_{2}\dots i_{k}}\\
&=&\sum_{i_{2},\ldots,i_{k}\in[n]}U_{11}^{-(k-1)}A_{1i_{2}\ldots i_{k}}U_{i_{2}i_{2}}\cdots U_{i_{k}i_{k}}+D_{1\dots 1}\\
&=&\sum_{i_{2},\ldots,i_{k}\in[n]\setminus\{1\}}\frac{1}{t^{k-1}}A_{1i_{2}\ldots i_{k}}+d_{1}\\
&=&\frac{d_{1}}{t^{k-1}}+d_{1}.
\end{eqnarray*}
For $i\in [n]\setminus \{1\}$, let
$d_{1,i}=|\{e: 1,i\in e\in E(\mathcal{G})\}|$. Obviously, $d_{1,i}\le d_i$.
For $2\leq i\leq n$, we have
\begin{eqnarray*}
r_{i}(\mathcal{T})&=&\sum_{i_{2},\dots,i_{k}\in[n]}T_{ii_{2}\dots i_{k}}\\
&=&\sum_{i_{2},\dots,i_{k}\in[n]}U_{ii}^{-(k-1)}A_{ii_{2}\ldots i_{k}}U_{i_{2}i_{2}}\cdots U_{i_{k}i_{k}}+D_{i\dots i}\\
&=&\sum_{i_{2},\dots,i_{k}\in[n]\atop 1\in\{i_{2},\ldots,i_{k}\}}U_{ii}^{-(k-1)}A_{ii_{2}\ldots i_{k}}U_{i_{2}i_{2}}\cdots U_{i_{k}i_{k}}\\
&&+\sum_{i_{2},\dots,i_{k}\in[n]\atop 1\not\in\{i_{2},\ldots,i_{k}\}}U_{ii}^{-(k-1)}A_{ii_{2}\ldots i_{k}}U_{i_{2}i_{2}}\cdots U_{i_{k}i_{k}}+d_{i}\\
&=&\sum_{i_{2},\dots,i_{k}\in[n]\atop 1\in\{i_{2},\ldots,i_{k}\}}A_{ii_{2}\ldots i_{k}}t+\sum_{i_{2},\dots,i_{k}\in[n]\atop 1\not\in\{i_{2},\ldots,i_{k}\}}A_{ii_{2}\ldots i_{k}}+d_{i}\\
&=&td_{1,i}+d_{i}-d_{1,i}+d_i\\
&\leq& (t+1)d_{i}\\
&\leq& (t+1)d_{2}
\end{eqnarray*}
with equality if and only if $d_i=d_{1,i}$ and $d_i=d_2$.

Note that $h((\frac{d_{1}}{d_{2}})^{\frac{1}{k}})=(d_2-d_1)\left(\frac{d_1}{d_2}\right)^{1-\frac{1}{k}}<0$ and $h((\frac{d_{1}}{d_{2}}))=d_1\left(\left(\frac{d_1}{d_2}\right)^{k-2}-1\right)>0$. Thus  $h(t)=0$ does have a root  $d^*$ in $((\frac{d_{1}}{d_{2}})^{\frac{1}{k}},\frac{d_{1}}{d_{2}})$. Let  $t=d^*$. Then  $t>1$. We have
\[
r_{1}(\mathcal{T})=d_{1}+d_{1}\left(\frac{1}{d^*}\right)^{k-1},
\]
and for $2\leq i\leq n$,
\[
r_{i}(\mathcal{T})\leq d_{1}+d_{1}\left(\frac{1}{d^*}\right)^{k-1}.
\]
Thus  by Lemma~\ref{Tensor2},
\[
\mu(\mathcal{G})=\rho(\mathcal{T})\leq \max_{1\leq i\leq n}r_{i}(\mathcal{T})=d_{1}+d_{1}\left(\frac{1}{d^*}\right)^{k-1}.
\]
This proves (\ref{MZh}).

Now suppose that $\mathcal{G}$ is connected. By Lemma~\ref{connec}, $\mathcal{Q}$ is weakly irreducible, and so is $\mathcal{T}$.

If  equality holds in (\ref{MZh}), then by Lemma~\ref{Tensor2}, $r_1(\mathcal{T})=\cdots=r_n(\mathcal{T})$, and thus  from the above arguments,  we have $d_{1,i}=d_{i}$ for $i=2, \dots, n$ (implying that each edge of $\mathcal{G}$ contains vertex $1$), $d_{2}=\cdots=d_{n}$, and thus  $\mathcal{G}$ is a blow-up hypergraph of a regular $(k-1)$-uniform hypergraph on $n-1$ vertices of degree $d_2$.

Conversely,  if $\mathcal{G}=\mathcal{H}^1$, where $\mathcal{H}$ is a regular $(k-1)$-uniform hypergraph on $n-1$ vertices of degree $d_2$, then by the above arguments, we have   $r_{i}(\mathcal{T})=d_{1}+d_{1}(\frac{1}{d^*})^{k-1}$ for $1\leq i\leq n$, and thus by Lemma~\ref{Tensor2},
 $\mu(\mathcal{G})=\rho(\mathcal{Q})=\rho(\mathcal{T})=d_{1}+d_{1}\left(\frac{1}{d^*}\right)^{k-1}$.
\end{proof}

 Let $\mathcal{G}$ be a  $k$-uniform hypergraph on $n$ vertices with degree sequence $d_1\ge \cdots \ge d_n$. If $d_1>d_2$, then $d_{1}+d_{1}(\frac{1}{d^*})^{k-1}<d_{1}+d_{1}(\frac{d_2}{d_1})^{1-\frac{1}{k}}
=d_{1}+d_{1}^{\frac{1}{k}}d_{2}^{1-\frac{1}{k}}$.  By Theorem~\ref{Mo}, we have
\[
\mu(\mathcal{G})\le d_1+d_1^{\frac{1}{k}}d_2^{1-\frac{1}{k}},
\]
and if $\mathcal{G}$ is connected, then equality holds  if and only if $\mathcal{G}$ is a regular hypergraph, see  \cite{YZL}.

\begin{Theorem}\label{Tensor7}
Let $\mathcal{G}$ be a $k$-uniform hypergraph on $n$ vertices  without isolated vertices with average $2$-degrees $m_1\geq \dots \geq m_n$,  and degree sequence $d_1, \dots, d_n$.
Then
\begin{equation} \label{Final}
\mu(\mathcal{G})\leq \min_{1\leq j\leq n}\max\left\{m_1^{\frac{1}{k}}m_{j}^{1-\frac{1}{k}}+d_1, \theta_j\right\},
\end{equation}
where \[
\theta_j =\max\left\{ m_1^{\frac{1}{k}}m_{i}m_j^{-\frac{1}{k}}+d_i :2\leq i\leq n\right\}.
\]

\end{Theorem}
\begin{proof} Let $\mathcal{Q}=\mathcal{Q}(\mathcal{G})$.
Let $U$ be a diagonal matrix $\mbox{diag} (td_1, d_2,\dots, d_n)$, where $t\geq1$ is a variable to be determined later.
Let $\mathcal{T}=U^{-(k-1)}\mathcal{Q}U$.
Then $\mathcal{Q}$ and $\mathcal{T}$ are diagonal similar. By Lemma~\ref{Tensor1}, $\mu(\mathcal{G})=\rho(\mathcal{T})$.
Obviously,
\[T_{i_1\dots i_k}=U^{-(k-1)}_{i_1i_1}Q_{i_1\dots i_k}U_{i_2i_2}\cdots U_{i_ki_k}.\]
for $i_1,\dots, i_k\in[n]$.
Then
\begin{eqnarray*}
r_1(\mathcal{T}
&=&\sum_{i_2,\dots, i_k\in[n]}U^{-(k-1)}_{11}Q_{1i_2\dots i_k}U_{i_2i_2}\cdots U_{i_ki_k}\\
&=&\frac{\sum_{i_2,\dots, i_k\in[n]\setminus\{1\}}Q_{1i_2\dots i_k}d_{i_2}\cdots d_{i_k}}{(t d_1)^{k-1}}\\
&=&\frac{D_{1\dots 1}(td_1)^{k-1}}{(t d_1)^{k-1}}+\frac{\sum_{i_2,\dots, i_k\in[n]}A_{1i_2\dots i_k}d_{i_2}\cdots d_{i_k}}{(t d_1)^{k-1}}\\
&=&d_1+\frac{\sum_{\{1,i_2,\dots, i_k\}\in E_1}d_{i_2}\cdots d_{i_k}}{(t d_1)^{k-1}}\\
&=&d_1+\frac{m_1}{t^{k-1}}.
\end{eqnarray*}
For $i=2, \dots, n$, let \[m_{1,i}=m_i-\frac{\sum_{1\notin\{i,i_2,\dots,i_k\}\in E_i}d_{i_2}\dots d_{i_k}}{d_i^{k-1}},\]
and then
\begin{eqnarray*}
r_i(\mathcal{T}
&=&\sum_{i_2,\dots, i_k\in[n]}U^{-(k-1)}_{ii}Q_{ii_2\dots i_k}U_{i_2i_2}\cdots U_{i_ki_k}\\
&=&\sum_{i_2,\dots, i_k\in[n]\atop 1\in\{i_2,\dots, i_k\}}d_i^{-(k-1)}Q_{ii_2\dots i_k}d_{i_2}\cdots d_{i_k}\\
&&+\sum_{i_2,\dots, i_k\in[n]\atop 1\notin \{i_2,\ldots , i_k\}}d_i^{-(k-1)}Q_{ii_2\dots i_k}d_{i_2}\cdots d_{i_k}\\
&=&\frac{D_{i\dots i}d_i^{k-1}}{ d_i^{k-1}}+\frac{\sum_{i_2,\dots, i_k\in[n]\atop 1\in\{i_2,\dots, i_k\}}A_{ii_2\dots i_k}d_{i_2}\cdots d_{i_k}}{d_i^{k-1}}\\
&&+\frac{\sum_{i_2,\dots, i_k\in[n]\atop 1\notin\{i_2,\dots, i_k\}}A_{ii_2\dots i_k}d_{i_2}d_{i_2}\cdots d_{i_k}}{d_i^{k-1}}\\
&=&d_i+\frac{\sum_{\{i,1,i_3,\dots, i_k\}\in E_i}(td_1)d_{i_3}\cdots d_{i_k}}{d_i^{k-1}}\\
&&+\frac{\sum_{1\notin\{i,i_2,\dots,i_k\}\in E_i}d_{i_2}\cdots d_{i_k}}{d_i^{k-1}}\\
&=&d_i+ tm_{1,i}+m_i-m_{1,i}\\
&\leq&d_i+ tm_i.
\end{eqnarray*}
For an arbitrary fixed $j$ with  $1\leq j\leq n$, let $t=\left(\frac{m_1}{m_j}\right)^\frac{1}{k}$. Obviously, $t\geq 1$.
Then \[r_1(\mathcal{T})=m_1^{\frac{1}{k}}m_{j}^{1-\frac{1}{k}}+d_1,\]
for $2\leq i\leq n$, \[r_i(\mathcal{T})\leq tm_i+d_i=m_1^{\frac{1}{k}}m_{i}m_j^{-\frac{1}{k}}+d_i.\]

Let $\theta_j =\max\left\{ m_1^{\frac{1}{k}}m_{i}m_j^{-\frac{1}{k}}+d_i :2\leq i\leq n\right\}$.
Thus for $1\leq i\leq n$, we have  \[r_i(\mathcal{T})\leq\max\left\{m_1^{\frac{1}{k}}m_{j}^{1-\frac{1}{k}}+d_1, \theta_j\right\}\]
Thus \[r_i(\mathcal{T})\leq \min_{1\leq j\leq n}\max\left\{m_1^{\frac{1}{k}}m_{j}^{1-\frac{1}{k}}+d_1, \theta_j\right\}.\]
Now the result follows from   Lemma~\ref{Tensor2}.
\end{proof}

If we  take $U=\mbox{diag} (d_1,\ldots,d_{n-1}, yd_n)$
with $y=\left(\frac{m_n}{m_j}\right)^\frac{1}{k}$ for an arbitrary  fixed $j$ in the above proof, then we have
\[\mu(\mathcal{G})\ge \max_{1\leq j\leq n}\min\left\{m_n^{\frac{1}{k}}m_{j}^{1-\frac{1}{k}}+d_n, \gamma_j\right\},
\]
where $\gamma_j =\min\left\{ m_n^{\frac{1}{k}}m_{i}m_j^{-\frac{1}{k}}+d_i :2\leq i\leq n\right\}$ for $1\leq j\leq n$.

Consider $4$-uniform hypergraph $\mathcal{G}_1$ with vertex set $[25]$ and edge set $E(\mathcal{G}_1)=\{e_1,\ldots,e_{14}\}$, where \begin{eqnarray*}
\begin{matrix}
e_1=\{1,2,3,4\}, & e_2=\{5,6,7,8\}, & e_3=\{9,10,11,12\},\\
e_4=\{13,14,15,16\}, &
e_5=\{17,18,19,20\}, & e_6=\{21,22,23,24\}, \\
e_7=\{1,2,3,25\}, & e_8=\{4,5,6,25\}, &
e_9=\{7,8,9,25\},\\
 e_{10}=\{10,11,12,25\}, & e_{11}=\{13,14,15,25\}, & e_{12}=\{16,17,18,25\}\\
e_{13}=\{19,20,21,25\}, & e_{14}=\{22,23,24,25\}.
\end{matrix}
\end{eqnarray*}
In notation of Theorem~\ref{Tensor7}, we have
\[
d_1=\dots=d_{24}=2, d_{25}=8,
\]
and
\[
m_1=\dots=m_{24}=5, m_{25}= 0.125,
\]
implying that $\theta_1=\dots=\theta_{24}=14.5743$, $\theta_{25}\approx 8.31436$,  and $m_1^{\frac{1}{4}}m_j^{\frac{3}{4}}+d_1=7$ for $1\leq j\leq 24$ and $m_1^{\frac{1}{4}}m_{25}^{\frac{3}{4}}+d_1\approx 2.31436$. Thus $\mu(\mathcal{G}_1)\leq 8.125$.
Note that $8 =d_1>d_2=\dots=d_{25}=2$ in notation of Theorem \ref{Mo} and  that $h(d^*)=d_{1}+d_{1}\left(\frac{1}{d^*}\right)^{k-1}$ is a decreasing function for $d^*\in((\frac{d_{1}}{d_{2}})^{\frac{1}{k}},\frac{d_{1}}{d_{2}})$. Then
\[
d_{1}+d_{1}\left(\frac{1}{d^{\ast}}\right)^{k-1}
> d_{1}+d_{1}\left(\frac{d_2}{d_1}\right)^{k-1}
= d_{1}+\frac{d_2^{k-1}}{d_1^{k-2}}
= 8+\frac{2^3}{8^2}
=8.125.
\]
For $\mathcal{G}_1$, the upper bound in (\ref{Final}) is smaller than the one in (\ref{MZh}). Obviously,
the blow-up hypergraph of a regular $(k-1)$-uniform hypergraph on $n-1$ vertices, the upper bound in (\ref{MZh}) is smaller than the one in (\ref{Final}).

For a $k$-uniform hypergraph $\mathcal{G}$,
let $d_1\geq \dots\geq d_n$ be the degree sequence of $\mathcal{G}$ and $m_1,\dots, m_n$ be the average $2$-degrees of $\mathcal{G}$. In~\cite{YZL}, the following upper bounds for $\mu(\mathcal{G})$ are given.
\begin{eqnarray}\label{TT2}
\mu(\mathcal{G})\leq \max_{e\in E(\mathcal{G})}\max_{\{i,j\}\in e}(d_i+d_j),
\end{eqnarray}
\begin{eqnarray}\label{TT3}
\mu(\mathcal{G})\leq \max_{e\in E(\mathcal{G})}\max_{\{i,j\}\in e}\frac{d_i+d_j+\sqrt{(d_i-d_j)^2+4m_im_j}}{2}.
\end{eqnarray}

Consider $3$-uniform hypergraph $\mathcal{G}_2$ with vertex set $[9]$ and edge set $E(\mathcal{G}_2)=\{e_1,\ldots,e_4\}$, where
\[
e_1=\{1,2,9\}, e_2=\{3,4,8\},e_3=\{5,6,7\},e_4=\{7,8,9\}.
\]
In notation of Theorem~\ref{Tensor7}, we have
\[
d_1=\dots=d_6=1, d_7=d_8=d_9=2,
\]
and
\[
m_1=\dots =m_6=2, m_7=m_8=m_9=\frac{5}{4},
\]
implying that $\theta_1=\dots=\theta_6=3.25$, $\theta_7=\theta_8=\theta_9\approx3.462$, and $m_1^{\frac{1}{3}}m_j^{\frac{2}{3}}+d_1=3$ when $1\leq j\leq 6$ and $m_1^{\frac{1}{3}}m_j^{\frac{2}{3}}+d_1\approx2.462$ when $7\leq j\leq 9$, and thus $\mu(\mathcal{G}_2)\leq 3.25$.
By direct calculation, the bounds in (\ref{TT2}) and (\ref{TT3}) are  $4$ and $3.25$, respectively.
For $\mathcal{G}_2$, the upper bound in (\ref{Final}) is smaller than the upper bound in (\ref{TT2}).

Consider $4$-uniform hypergraph $\mathcal{G}_3$ with vertex set $[7]$ and edge set $E(\mathcal{G}_3)=\{e_1,\ldots,e_{8}\}$, where \begin{eqnarray*}
\begin{matrix}
e_1=\{1,2,3,4\}, & e_2=\{1,5,6,7\}, & e_3=\{2,3,4,5\}, \\
e_4=\{3,4,5,6\}, & e_5=\{4,5,6,7\}, & e_6=\{5,6,7,2\}, \\
e_7=\{6,7,2,3\}, & e_8=\{7,2,3,4\}.
\end{matrix}
\end{eqnarray*}
In notation of Theorem~\ref{Tensor7}, we have
\[
d_1=2, d_2=\dots=d_7=5,
\]
and
\[
m_1=31.25, m_2=\dots=m_7=4.4,
\]
implying that $\theta_1=9.4$, and $\theta_2=\dots=\theta_7\approx 12.18294$, and $m_1^{\frac{1}{4}}m_j^{\frac{3}{4}}+d_1=33.25$ for  $j=1$ and $m_1^{\frac{1}{4}}m_j^{\frac{3}{4}}+d_1\approx9.18294$ for $2\leq j\leq 7$. Thus $\mu(\mathcal{G}_3)\leq 12.18294$. It is easily seen that  the bounds in (\ref{TT2}) and (\ref{TT3}) are $10$ and $15.32159$, respectively.
For $\mathcal{G}_3$, the upper bound in Theorem~\ref{Tensor7} is smaller than the one in  (\ref{TT3}) but larger than the one in  (\ref{TT2}).

\vspace{5mm}

\noindent {\bf Acknowledgement.}  We thank the referees for their valuable and constructive comments and suggestions.

redpajama_set_name: RedPajamaArXiv


text:
1952 - Table of Contents/Authority for 1952
Approved State Plans for Enforcement of State Standards
Table of Contents/Authority for 1952
PART 1952 -- APPROVED STATE PLANS FOR ENFORCEMENT OF STATE STANDARDS
Subpart A -- List of Approved State Plans for Private-Sector and State and Local Government Employees
1952.1 South Carolina.
1952.2 Oregon.
1952.3 Utah.
1952.4 Washington.
1952.5 North Carolina.
1952.6 Iowa.
1952.7 California.
1952.8 Minnesota.
1952.9 Maryland.
1952.10 Tennessee.
1952.11 Kentucky.
1952.12 Alaska.
1952.13 Michigan.
1952.14 Vermont.
1952.15 Nevada.
1952.16 Hawaii.
1952.17 Indiana.
1952.18 Wyoming.
1952.19 Arizona.
1952.20 New Mexico.
1952.21 Virginia.
1952.22 Puerto Rico.
Subpart B -- List of Approved State Plans for State and Local Government Employees
1952.23 Connecticut.
1952.24 New York.
1952.25 New Jersey.
1952.26 The Virgin Islands.
1952.27 Illinois.
1952.28 Maine.
Authority: Sec. 18, 84 Stat. 1608 (29 U.S.C. 667); 29 CFR part 1902; Secretary of Labor's Order No. 1–2012 (77 FR 3912, Jan. 25, 2012).
[59 FR 2295, Jan. 14, 1994; 59 FR 14554, March 29, 1994; 59 FR 32649, June 24, 1994; 59 FR 39257, Aug. 2, 1994; 59 FR 42493, Aug. 18, 1994; 59 FR 42495, Aug. 18, 1994; 60 FR 12416, March 7, 1995; 60 FR 12417, March 7, 1995; 60 FR 20191, April 25, 1995; 60 FR 43969, Aug. 24, 1995; 60 FR 56950, Nov. 13, 1995; 61 FR 28053, June 4,1996; 61 FR 36824, July 15, 1996; 61 FR 55098, Oct. 24, 1996; 61 FR 66593, Dec. 18, 1996; 62 FR 2558, Jan. 17, 1997; 62 FR 49908, Sept. 24, 1997; 63 FR 53281, Oct. 5, 1998; 65 FR 36619, June 9, 2000; 66 FR 6135, Jan. 19, 2001; 67 FR 60128, Sept. 25, 2002; 68 FR 43460, July 23, 2003; 69 FR 20827 and 20829, April 19, 2004; 70 FR 24954, May 12, 2005; 71 FR 2886, Jan. 18, 2006; 71 FR 36990, June 29, 2006; 76 FR 63190, Oct. 12, 2011; 76 FR 63191, Oct. 12, 2011; 77 FR 58590, Sept. 21, 2012; 80 FR 49904-49909, August 18, 2015; 81 FR 6178, February 5, 2016]


text:
Reinforcing Joyful Noise
by Worship Facility Staff | Dec 22, 2022 | Audio, Case Studies, Production
Christ the King Lutheran Church, known as "The Music Church" by locals, has been a small but significant presence in southwest Houston, TX since 1944. With an appreciation of history, tradition, and classical, faithful music, the church was recently faced with a dilemma. Leadership was focused on an audio upgrade they knew was necessary to improve the quality of spoken word for services, but they were concerned about the impact that any new loudspeakers would have on music reproduction as well as the visual of the nave. It was a quest that ultimately led them to an approach incorporating Renkus-Heinz digitally steerable loudspeakers.
"I've worked with many houses of worship," says Michael Austin, president of Native Media, the Renkus-Heinz sales representative for Texas and other states in the region. "But this church, acoustically speaking for musical applications, is stunningly gorgeous to the ear. I've never heard any nave or sanctuary that sounded better for music, but when it came to the spoken word, there were big acoustic challenges – people couldn't understand what was being said just eight feet in front of them."
The modest nave provides seating for about 300 in a setting of sharply angled hardwood ceilings, stone floors, a complement of stained-glass windows, and a large pipe organ within the rear wall. The space isn't a typical design; it's an in-the-round layout, with the altar and services taking place in the center of the nave.
BUILT FOR BACH
Over the years, the church has invested significantly in music. The pipe organ was designed by the legendary Fritz Noack in
the spirit of instruments cherished by Johann Sebastian Bach. The church also has a Bach-favorite, 1719-style harpsichord, and two top-quality grand pianos, a Blüthner and a Bösendorfer. With such a musical foundation, the Bach Society at Christ
the King Lutheran Church was founded in 1982 as an interest group within the congregation.
"The liturgy we provide on Sundays is a traditional Lutheran Eucharist with vested clergy, fully robed choir, featured vocalists, and accompanying live music tailored to each service – we even use incense on special Holy days," explains Rachel McWhirter, who serves as the church's director of digital ministry and media producer for the Bach Society. "Our church is one of only two Lutheran churches in the Houston area that specialize in this liturgy style. Our congregation has been preserving the history and integrity of the liturgy and the worship space, and people were nervous about adding a sound system. The congregation was asking, 'Are we going to change our tradition for the sake of technological advancement?'"
Faced with the challenge to improve the audio quality and intelligibility of the spoken word within the nave, the church began a system upgrade project with strict guidelines: a discreet system that provided high intelligibility without compromising live music or aesthetics. It had to be as visually discreet as possible to preserve the architecture and history of the space, yet provide enough audio control to accommodate the in-the-round configuration.
"When the church contacted us to help design and install their new system, we immediately thought of Renkus-Heinz, so
I reached out to Michael at Native Media, for his advice," says Danny Elizondo, project manager for integration firm Conference Technologies Inc. (CTI). "This was a unique project and we had to make sure we had the best sound solution possible."
After evaluating the space, the CTI team worked with Austin to set up a temporary Renkus-Heinz Iconyx loudspeaker to demonstrate the system's beam steering capabilities. The technology would allow the church to focus the spoken-word audio where it needed to be – on the congregation – while avoiding unwanted reflections from walls, ceilings, floors or other architectural features.
"We set up the demo system, dialed in the beam steering to fire over the in-the-round altar space to cross over and cover
the pews on the opposite sides," Austin notes. "We usually don't want to do this because the pastors are walking around with
wireless microphones, so it's an invitation for feedback. But with the beam steering, we were able to position the acoustic center over the presenter area to avoid any trouble and perfectly cover the congregation with intelligible audio. We left the temporary system in place for a few months and everyone was elated and ready to move forward for the permanent install."
THE RIGHT BALANCE
Ultimately, the team opted for a set of four ICLive X modules in stacked ICLXL/ICLLX pairs mounted on vertical wooden beams that rise in the center and sides of the nave. The placement allows the sound to be directed over the worship space in the center of the room to cover the pews on each side of the nave. Further back, toward the Bach organ and the far end, f ourDC12/2 column loudspeakers are mounted on columns to deliver the spoken word to the choir and overflow seating. All
loudspeakers are factory color-matched to blend in with their surroundings.
"Now, for the first time, our entire congregation can easily understand our pastors without sacrificing aesthetics or natural
acoustics required for the organ," states McWhirter.
When the presiding bishop visited the church late last year, she expressed high praise for the sound quality and discreet
appearance within the worship space. Since the installation, the congregation has shown interest in upgrading the audio
systems to Renkus-Heinz in additional parts of the church, including at its Christ the King west campus.
"It's interesting to note that the pedal tower of our Bach organ has a Latin inscription from Psalm 100 that perfectly captures our church's spirit: 'Jubilate Deo omnis terra, servite Dominum'," McWhirter concludes. "In Latin it translates as 'Make a joyful noise unto the Lord, all ye lands. Serve the Lord with gladness' and Renkus-Heinz is allowing us to do exactly that."
This appeared previously in Church Sound magazine.