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Month: May 2020
Message from His Excellency David Granger, President of the Co-operative Republic of Guyana, at the PAHO/WHO virtual concert for Public Health workers 2020.05.31
Guyana and the world are enduring the excruciating effects of the Novel CoronavirusDisease 2019 (COVID-19). Twelve of our citizens have died. The competence of our public health workerscontributed to the recovery of sixty-seven persons and to the continued care of one hundred and fifty confirmed cases. I join the Pan American Health Organization (PAHO) and …
Presentation of His Excellency David Granger President of the Co-operative Republic of Guyana to the High-Level Event on Financing for Development in the Era of Novel Coronavirus Disease 2019 (COVID-19) 2020.05.28
Excellencies, Guyana records its appreciation to the Prime Ministers of Canada and Jamaica and the Secretary-General of the United Nations for convening this important high-level event. Guyana, one of the smallest states on the continent of South America, borders countries which have a high incidence of Novel Coronavirus Disease 2019 (COVID-19). Migrants, seeking refuge, enter …
Address of His Excellency David Granger President of the Co-operative Republic of Guyana on the occasion of Guyana's54th Independence Anniversary 2020.05.26
Guyanese, We celebrate the 54th Anniversary of Independence recalling the sacrifice, suffering and struggle of our working people who struck the first blows against colonial oppression one hundred and fifteen years ago in Georgetown. We remember the militancy of the workers who were shot by the colonial police in the Ruimveldt Riots of 1905, an …
Flooded Gunns/ Masekenari village benefit from relief efforts
Georgetown, Guyana – (May 25, 2020) A team from the Civil Defence Commission (CDC), today, travelled to Gunns/Masekenari Village in the Konashen District, Upper Takutu-Upper Essequibo (Region Nine) to conduct a disaster risk assessment and handed over 60 food hampers to the Toshao, Mr. Paul Chikema, as residents continue to experience flooding in the village. …
Message of His Excellency David Granger President of the Co-operative Republic of Guyana on the occasion of Eid-ul-Fitr 2020.05.24
Eid Mubarak!I extend greetings to all Muslims and to all other Guyanese on the joyous occasion of Eid-ul-Fitr. This sacred festival commemorates the conclusion of the holy month of Ramadan during which Muslims performed acts of abstinence and engaged in ritual fasting and prayer. Eid-ul-Fitr reminds us of how divine worship is connected to social …
MOTP condemns falsehoods in Nascimento, Guyana Times letter – calls for immediate apology and retraction
Georgetown, Guyana (May 23, 2020): The Ministry of the Presidency firmly rejects the falsehoods contained in a fallacious letter written by Mr. Kit Nascimento titled, "Granger is not innocent of attempt to force rigged elections on Guyana" and published in today's edition of the Guyana Times. Mr. Nascimento, in his Letter to the Editor, asserts …
Update: REOI Provision of Marketing Services for the Cooperative Republic of Guyana's Oil Entitlement from the Lisa Destiny FPSO Vessel
The Department of Energy hereby notifies the Applicants who submitted Expressions of Interest on April 26, 2020, for the "Provision of Marketing Services for the Cooperative Republic of Guyana's Oil Entitlement from the Lisa Destiny FPSO Vessel", that the preparation of the Shortlist of Firms to be invited to submit full Technical and Financial Proposals …
Government launches data collection mechanism to facilitate COVID-19 public assistance
Georgetown, Guyana – (May 20, 2020) The Government of Guyana, today, launched a data collection mechanism, which is aimed at identifying and providing tangible support for vulnerable households affected by the COVID-19 pandemic and its emergency measures. The initiative is made possible through the collaborative efforts of the National COVID-19 Task Force Executive Secretariat, Civil …
Declaration by Chair of Elections Commission will be accepted – President Granger
Georgetown, Guyana – (May 17, 2020): President David Granger, this morning, reiterated his position that he will abide by whatever declaration is made by Chairman of the Elections Commission, Justice (ret'd) Claudette Singh on the outcome of the March 2, 2020 General and Regional Elections. The Head of State was at the time responding to …
Message from His Excellency David Granger President of the Co-operative Republic of Guyana on the occasion of International Nurses Day 2020.05.12
Nursing Guyanese to good health Guyanese, International Nurses Day acknowledges the contribution that nurses make to society and to humanity around the world. The Day commemorates the birth of Florence Nightingale, considered the founder of modern nursing. She was recognized for her service in the 19th century and is credited with implementing innovative methods of …


12 Deep Buffet Cabinet Tags : Delightful Large Dining Room Table Fantastic Square Dining Table. Adorable Pedestal Dining Table You Probably Did Not Know That Will Blow Your Mind.
Malaya Roberts Dining Room, 2018-09-13 22:31:55. Your dining tables, for example, can be accessorized with table leaves or linen of a creamy hue. Walnut furniture looks fantastic with lighter shades that are slightly creamier in texture. Beiges will not really work in this situation. All in all, consult a professional designer who will guide you through the process and will offer you suggestions and advice on how to match all your furniture so that your dining room area doesn't look too cluttered. âEUR¨Walnut dining furniture is best for those who want to be able to throw intimate, smart-casual get-togethers. The walnut furniture is perfect for that air of chic sophistication without looking too snooty. If you decide to only get one piece of walnut-based furniture, these have to be your dining room tables, as they will probably be the centerpiece of the room. Make sure that you get one in good quality and know how to care for it well so that it will stand the test of time and will look polished and brand-new.


I started this "Healthy Food for Healthy Kids Event Series" in Jan- 2012 and now after the successful completion of one year, I think it is time to reward my lovely friends who take so much effort in linking up their entries for the event. I have decided to randomly select A Winner Every Month for the Giveaway, giving equal chances of winning for all the participants.
As always, there will be a "TOP CONTRIBUTOR AWARD" and "PARTICIPATION AWARD" badges.
One more giveaway, of course a prestigious one "OVERALL WINNER OF THE EVENT SERIES" for the year will be picked up from the Top Contributors. This will be announced at the end of every year and he/she will be rewarded with a gift and badge.
Please note, that this giveaway is not a sponsored one. I am doing this on my own.
The overall Winner for the Event - 2012 will be announced soon:-) Wait till next post..
Hop on to Jaleela's Space to link your entries..Who knows, you may be the giveaway winner! So hurry..
That is great!!! Looking forward to host and also participate!!!


Helping educate tenants in regards to their rights as a tenant.
Ok, we are moving right along here.
Dr. Lilian Ma (the Chair of the Landlord and Tenant Board) seems to be taking my case seriously, and is not only having my court case reviewed, but she is having Ieva Martin investigated.
Ieva Martin is a member of the Landlord and Tenant board who I felt was not only in favour of the landlord right from day 1, but went out of her way to make me feel like a criminal, stupid, and that everything I said was a lie.
Once my book is available to be read, you will be amazed at how much evidence I had against the landlord, and what her end ruling was.
Are you a tenant within Ontario?
Do you rent only a portion of a house – basement or top floor?


Refinancing your mortgage to take advantage of beneficial rates or features can save you lots of money in the long term – however, there are a few negligible upfront costs you need to consider before making the decision. These small fees are normally offset by huge savings.
For example: An interest rate reduction of .5% on a $700,000 loan can save you $3500 a year.
It's not advisable to refinance your loan too regularly. Every two to three years is generally the sweet spot. To see how much you can save if you refinance use our free mortgage assessment tool.
When you refinance, your new lender may charge a range of upfront fees (see below). Not all lenders charge all of these fees, and some may be willing to negotiate or waive some costs. Some lenders also offer 'cash back' when you switch your mortgage to them, which could offset some or all of the costs of refinancing your mortgage.
Settlement fee – your lender may charge a fee for the pay out of your current mortgage.
Always ask your lender if they'll negotiate these fees. You never know until you try and in today's competitive market, you might save yourself some real money. A good mortgage broker will also know what deals are currently in the market to help you save or recoup costs.
Lender's Mortgage Insurance protects the lender against default. If you're borrowing more than 80 per cent of your home's value, your loan-to-value ratio (LVR) will be over 80 and you may be asked to take out LMI.
This will involve a one-off premium. You might be able to capitalise it (that is, add it to the amount you're borrowing) but this means you'll be paying interest on it, which will add to its cost in the long term.
Trying to keep under the 80 per cent LVR will save you LMI costs, which can run into thousands of dollars and end up costing you more than you bargained for. This may be more relevant for people looking to live in their home than investors.
Exit fees on new loans were abolished by Federal Parliament from 1 July 2011. If you took out your loan after that date, then you don't have to worry about exit fees. But if your loan pre-dates the change, then you should check what fees – which can go by different names, like 'early repayment fees' or 'deferred establishment fees' – you're liable to pay.
Exit fees vary widely though they usually only apply if you have had your current loan for less than five years.
Exit fees are becoming a thing of the past but can still catch homeowners out, so make sure you do your homework.
Some states charge stamp duty on your mortgage if you increase the size of your loan, and they may also charge a 'mortgage transfer fee'. This lets the various State Titles Offices know you've changed either your lender or your loan.
We can show you the mortgage duty and any transfer fees that apply for your area.
If you'd like to see if you're eligible for refinancing your mortgage, and for an obligation free chat to one of our friendly brokers, click here to take our quiz. It just takes two minutes.


Here are Realtor.com's hot housing market picks that outperform the national forecasted price growth of 2.2%. Lakeland, Florida tops the list with 7.4% forecasted price growth and 5% in home sales growth. Compare that to Realtor.com's national number on home sales growth of -2%. Grand Rapids, Michigan is in the second spot looking at an 8.2% price growth and 4% sales growth. Ranking at No. 3 is El Paso, Texas, with 2.5% in price growth, offset by 7.9% in sales growth. Other areas in the top ten include Chattanooga, Tennessee; Phoenix, Arizona; Bridgeport, Connecticut; Las Vegas, Nevada; Boise City, Idaho; Miami, Florida; and Boston, Massachusetts.
Realtor.com's chief economist Danielle Hale shares her take on the numbers: "These local markets are a direct reflection of the local economies. If they are thriving, so is housing. These are also markets that appeal to younger buyers." The criteria Realtor.com looked at included a strong economy with job growth outperforming the forecasted national average of 1.3%. Population growth numbers for the top ten are at 2% compared to 0.65% nationally were included in the number crunching. Increasing new home construction in those areas looked good thanks to available land.
Let's look at just what you can buy in the top three markets. In Lakeland, Florida, for $209,900 you can move into a three-bedroom lakefront home with a screened in back porch. In Grand Rapids, you'll find your money goes far compared to other Millennial hot spots. For $240,986, you can get a 1,746-square-foot updated brick ranch on the desirable northeast side of town. Moving over to El Paso, a 1,708-square-foot home with a pool, loft and cathedral ceilings is listed for $158,000.
Millennials are expected to take out 45% of the country's mortgages this year, according to Realtor.com. It's clear they already like Grand Rapids, Michigan, with 49% of the area's mortgages going to Millennials. Thanks to a continuing wave of craft breweries opening, Grand Rapids, according to Realtor.com, was recently named, "Brew City." El Paso, Texas came in with 45.7% of all mortgages going to Millennials. Chattanooga's Millennials had 43.9% of all mortgages. What this latest research shows is there is affordable housing around the country in areas with jobs and good economies that are attracting Millennial buyers. It's no surprise none of the markets on this list are in California.


The formation of more than one embryo in an ovule. These may develop by division of the fertilized zygote (see cleavage polyembryony ) and thus be a product of sexual reproduction. They may also arise, alongside a zygote, from somatic tissue (see adventive embryony ). Embryos derived in this manner will have the same genetic constitution as the maternal parent. See also apomixis , parthenogenesis.


text:
\section{Introduction}
In \cite{Ree}, Ree introduced the shuffle algebra which has been
studied extensively during the last fifty years. The shuffle
product is carried out on the tensor space $T(V)$ of a vector
space $V$ by using the shuffle rule. Its natural generalization is
the quasi-shuffle product where $V$ is moreover an associative
algebra and the new product on $T(V)$ involves both of the shuffle
product and the multiplication of $V$. Quasi-shuffle algebras
first arose in the work of Newman and Radford \cite{NR} for the
study of cofree irreducible Hopf algebras built on associative
algebras. Later, they were discussed by many other mathematicians
for various motivations, such as multiple zeta values (\cite{Hof1}
and \cite{IKZ}), Rota-Baxter algebras (\cite{GK},\cite{G} and
\cite{EG}), and commutative tridendriform algebras (\cite{Lod}).

For both of physical and mathematical considerations, people wants
to deform or quantize some important algebra structures. The most
famous example is absolutely the quantum group introduced by
Drinfeld \cite{D} and Jimbo \cite{J}. To our surprise, there is an
implicit but significant connection between quantum groups and
shuffle algebras. Rosso \cite{Ro} constructed the quantization of
shuffle algebras. This is a new kind of quantized algebras and
leads to an intrinsic understanding of the quantum group. Since
shuffle algebras are special quasi-shuffle algebras, and the
importance of the later ones, people would expect to find out what
the quantization of quasi-shuffle algebras is and whether it can
bring us some useful information. Hoffman's q-deformation of
quasi-shuffles (\cite{Hof2}) is such an attempt. His idea is to
deform the multiplication formula according to a special case of
Rosso's quantum shuffles. After replacing the shuffle part by its
quantized version in the formula of quasi-shuffles, Hoffman tried
to deform the mixed term by multiplying a power of q, and found
that there is only one way making the new product to be
associative. This construction is more or less experiential as he
said. The general construction of quantized quasi-shuffles is due
to Rosso (\cite{Ro2}) in the spirit of his quantum shuffles. We
describe Rosso's idea as follows. Let $M$ be
 a Hopf bimodule over a Hopf algebra $H$. In addition, if $M$ is an algebra and the multiplication is
compatible with the braiding coming from the Hopf bimodule
structure in some sense, then one can construct a new algebra
structure on the cotensor coalgebra $T^c_H(M)$ by using its
universal property. In general, given a braided algebra
$(A,m,\sigma)$, one can construct an analogue of the quasi-shuffle
algebra in the braided category, where the action of the usual
flip is replaced by that of the braiding. The resulting algebra is
called a quantum quasi-shuffle algebra. In particular, Hoffman's
q-deformation of quasi-shuffle products is a special case of
Rosso's quantum quasi-shuffle. In \cite{JR}, the construction of
quantum quasi-shuffles appears, as a special braided cofree Hopf
algebra, in the framework of quantum multi-brace algebras.

 Some interesting properties of the quantum
quasi-shuffle algebras have been studied in \cite{JRZ}, including
the commutativity, universal property, and etc. This paper
continues the trip. We establish some explicit results concerning
this new subject. We start by reformulating the product.
Originally, the quantum quasi-shuffle algebra is constructed by
using the universal property of connected coalgebras (\cite{JR}).
Later, it is defined by an inductive formula (\cite{JRZ}). But
neither of these two constructions can provide an explicit
formula. To know more about this new subject, a more clear form of
the multiplication formula is definitely helpful. Here, we use the
notion of mixable shuffles introduced in \cite{GK} to establish a
complete description of the quantum quasi-shuffle product.
Contrast to the quantum symmetric algebra, we describe the
subalgebra of the quantum quasi-shuffle bialgebra generated by the
primitive elements. On the other hand, for the reason that the
universal property of connected coalgebras is not so familiar by
non-algebraists, we use the universal property of tensor algebras
to construct a braided coalgebra structure on $T(C)$ for a braided
coalgebra $C$, and show that its dual is the quantum quasi-shuffle
algebra. This enables one to study the quantum quasi-shuffle
algebra through its dual. Finally, as applications, we provide
examples of Rota-Baxter algebras and tridendriform algebras by
using quantum quasi-shuffles. Rota-Baxter algebras and
tridendriform algebras are important subjects in mathematics and
physics. So our constructions not only enlarge their families, but
also demonstrate the value of the quantum quasi-shuffles.

This paper is organized as follows. In Section 2, several concrete
examples of braided algebras are provided. In Section 3, we recall
the construction of quantum quasi-shuffle algebras and established
explicit formulas for the new product and the subalgebra generated
by the primitive elements. In Section 4, we construct the dual
coalgebra of the quantum quasi-shuffle algebra. Finally, in
Section 5, it contains some applications of the quantum
quasi-shuffle algebra involving examples of Rota-Baxter algebras
and tridendriform algebras.

\noindent\textbf{Notation.} In this paper, we denote by
$\mathbb{K}$ a ground field of characteristic 0. All the objects
we discuss are defined over $\mathbb{K}$. For a vector space $V$,
we denote by $\otimes$ the tensor product within $T(V)$, and by
$\underline{\otimes}$ the one between $T(V)$ and $T(V)$.

We denote by $\mathfrak{S}_{n}$ the symmetric group of the set
$\{1,2,\ldots,n\}$ and by $s_{i}$, $1\leq i\leq n-1$, the standard
generators of $\mathfrak{S}_{n}$ permuting $i$ and $i+1$. For
fixed $k,n\in \mathbb{N}$, we define the shift map
$\mathrm{shift}_k:\mathfrak{S}_{n}\rightarrow\mathfrak{S}_{n+k}$
by $\mathrm{shift}_k(s_i)=s_{i+k}$ for any $1\leq i\leq n-1$. For
the reason of intuition and the simplicity of notation, we denote
$1_{\mathfrak{S}_{k}}\times w=\mathrm{shift}_k(w)$ for any $w\in
\mathfrak{S}_{n}$. The notations $w\times 1_{\mathfrak{S}_{k}}$,
$1_{\mathfrak{S}_{k}}\times w\times 1_{\mathfrak{S}_{l}}$ and
others are understood similarly.

A braiding $\sigma$ on a vector space $V$ is an invertible linear
map in $\mathrm{End}(V\otimes V)$ satisfying the quantum
Yang-Baxter equation on $V^{\otimes 3}$: $$(\sigma\otimes
\mathrm{id}_{V})(\mathrm{id}_{V}\otimes \sigma)(\sigma\otimes
\mathrm{id}_{V})=(\mathrm{id}_{V}\otimes \sigma)(\sigma\otimes
\mathrm{id}_{V})(\mathrm{id}_{V}\otimes \sigma).$$ A braided
vector space $(V,\sigma)$ is a vector space $V$ equipped with a
braiding $\sigma$. For any $n\in \mathbb{N}$ and $1\leq i\leq
n-1$, we denote by $\sigma_i$ the operator $\mathrm{id}_V^{\otimes
i-1}\otimes \sigma\otimes \mathrm{id}_V^{\otimes n-i-1}\in
\mathrm{End}(V^{\otimes n})$. For any $w\in \mathfrak{S}_{n}$, we
denote by $T^\sigma_w$ the corresponding lift of $w$ in the braid
group $B_n$, defined as follows: if $w=s_{i_1}\cdots s_{i_l}$ is
any reduced expression of $w$, then $T^\sigma_w=\sigma_{i_1}\cdots
\sigma_{i_l}$. This definition is well-defined (see, e.g., Theorem
4.12 in \cite{KT}).

We define $\beta:T(V)\underline{\otimes} T(V)\rightarrow
T(V)\underline{\otimes} T(V)$ by requiring that the restriction of
$\beta$ on $V^{\otimes i}\underline{\otimes} V^{\otimes j}$,
denoted by $\beta_{ij}$, is $T^\sigma_{\chi_{ij}}$ , where
\[\chi_{ij}=\left(\begin{array}{cccccccc}
1&2&\cdots&i&i+1&i+2&\cdots & i+j\\
j+1&j+2&\cdots&j+i&1& 2 &\cdots & j
\end{array}\right)\in \mathfrak{S}_{i+j},\] for any $i,j\geq 1$. For convenience, we denote by
$\beta_{0i}$ and $\beta_{i0}$ the usual flip.

\section{Braided algebras}
We start by recalling the notion of braided algebras which is the
correspondent object of associative algebras in braided
categories. In the following, all algebras are assumed to be
associative but not necessarily unital.
\begin{definition}Let $A=(A,m)$ be an algebra with product $m$, and $\sigma$ be a braiding on $A$. We call the triple $(A,m,\sigma)$ a \emph{braided algebra} if it satisfies the following conditions:
\[\left\{
\begin{array}{lll}
(\mathrm{id}_A\otimes m)\sigma_1\sigma_2&=&\sigma( m\otimes
\mathrm{id}_A),\\[3pt] ( m\otimes
\mathrm{id}_A)\sigma_2\sigma_1&=&\sigma(\mathrm{id}_A\otimes m).
\end{array} \right.
\]
Moreover, if $A$ is unital and its unit $1_A$ satisfies that for
any $a\in A$,
\[\left\{
\begin{array}{lll}
\sigma(a\otimes 1_A)&=&1_A\otimes a,\\[3pt] \sigma(1_A\otimes a)&=&a\otimes
1_A,
\end{array} \right.
\]then $A$ is called a \emph{unital braided algebra}.
\end{definition}

\begin{remark}1. For any braided vector space $(V,\sigma)$, there is a trivial braided algebra structure on it whose multiplication is the trivial one $m=0$.

2. The braided algebra structure is very crucial. Given a braided
vector space $(V,\sigma)$, it is not reasonable that there should
be a non-trivial braided algebra structure on $V$. For instance,
let $V$ be a vector space with basis $\{e_1,e_2\}$. We define two
braidings $\sigma_1$ and $\sigma_2$ on $V$ respectively by

\[\left\{
\begin{array}{lll}
\sigma_1(e_{1}\otimes e_{1})&=&e_{1}\otimes e_{1}, \\
\sigma_1(e_{i}\otimes e_{2})&=&qe_{2}\otimes e_{1},\\
\sigma_1(e_{2}\otimes e_{1})&=&qe_{1}\otimes
e_{2}+(1-q^{2})e_{2}\otimes e_{1},\\
\sigma_1(e_{2}\otimes e_{2})&=&e_{2}\otimes e_{2},
\end{array} \right.
\]
and
$$\sigma_2(e_{i}\otimes e_{j})=qe_{j}\otimes e_{i},\ \ \ \forall i,j,$$
where $q\in\mathbb{K}$ is nonzero and not equal to $\pm 1$.

Then by an easy argument, one can show that the only product on
$V$ which is compatible with $\sigma_1$ is just the trivial one.
The case is the same for $\sigma_2$.

3. For any braided algebra $(A,m,\sigma)$, one can embed it into a
unital braided algebra
$(\widetilde{A},\widetilde{m},\widetilde{\sigma})$ in the
following way. First of all, we set
$\widetilde{A}=\mathbb{K}\oplus A$. Then we define the
multiplication $\widetilde{m}$ and the braiding
$\widetilde{\sigma}$ by: for any $\lambda,\mu\in \mathbb{K}$ and
$a,b\in A$
$$\widetilde{m}\big((\lambda+a)\otimes (\mu+b)\big)=\lambda\mu+\lambda\cdot b+\mu\cdot
a+m(a\otimes b), $$ and
$$\widetilde{\sigma}\big((\lambda+a)\otimes (\mu+b)\big)=\mu\otimes
\lambda+b\otimes \lambda+\mu\otimes a+\sigma(a\otimes b).
$$It is easy to verify that $(\widetilde{A},\widetilde{m},\widetilde{\sigma})$ is a braided algebra with unit $1\in \mathbb{K}$.
\end{remark}

Because all the constructions in this paper are based on braided
algebras, we provide several concrete examples which will either
be used in our later discussion or afford the reader some
illustrations. Some of them may be known, while some may be new.
For more examples, one can see \cite{B} and \cite{JR}.

\begin{example}Let $V$ be a vector space with basis $\{e_i\}$ which is at most countable. We provide a braided algebra structure on $V$. The braiding $\sigma$ on $V$ is given by $\sigma(e_i\otimes
e_j)=q_{ij}e_j\otimes e_i$, where $q_{ij}$'s are nonzero scalars
in $\mathbb{K}$ such that $q_{ij}q_{ik}=q_{i\ j+k}$ and
$q_{ik}q_{jk}=q_{i+j\ k}$ for any $i,j,k$. For instance, let $q$
be a nonzero scalar in $\mathbb{K}$ and $q_{ij}=q^{ij}$. The
multiplication $\cdot$ on $V$ which is compatible with the
braiding $\sigma$ is given as follows.

Case 1. If $V$ is a finite-dimensional vector space with basis
$\{e_1,e_2,\ldots,e_N\}$, then we define

\[e_i\cdot e_j= \left\{
\begin{array}{lll}
e_{i+j},&& \mathrm{if\ }i+j\leq N,\\[3pt]
0,&&\mathrm{otherwise}.
\end{array} \right.
\]

Case 2. If $V$ is a vector space with basis $\{e_i\}_{i\in
\mathbb{N}}$, then we define $e_i\cdot e_j=e_{i+j}$ for any
$i,j\in \mathbb{N}$.

It is evident that $\cdot$ is an associative algebra structure on
$V$ in both cases. Notice that
\begin{eqnarray*}(\mathrm{id}_V\otimes
\cdot)\sigma_1\sigma_2 (e_i\otimes e_j\otimes
e_k)&=&q_{jk}q_{ik}e_k\otimes e_{i+j}\\[3pt]
&=&q_{i+j\ k}e_k\otimes e_{i+j}\\[3pt]&=&\sigma(
\cdot\otimes\mathrm{id}_V)(e_i\otimes e_j\otimes
e_k),\end{eqnarray*}
 and similarly
$(
\cdot\otimes\mathrm{id}_V)\sigma_2\sigma_1=\sigma(\mathrm{id}_V\otimes
\cdot)$. Therefore $(V,\cdot,\sigma)$ is a braided algebra.
\end{example}

\begin{example}All notions of this example can be found in \cite{Ka}. Let $q\neq 1$ be a invertible scalar in $\mathbb{K}$, and $x,y$ be two indeterminates. Denote by $\mathbb{K}_q[x,y]$ the quantum plane, i.e., the algebra generated by $x,y$ with the relation $yx=qxy$. It has a linear basis $\{x^iy^j\}_{i,j\geq 0}$. Define two algebra automorphisms $\omega_x$ and $\omega_y$ of $\mathbb{K}_q[x,y]$ by requiring that
$$\omega_x(x)=qx,\omega_x(y),\omega_y(x)=x,\omega_y(y)=qy,$$ and define two endomorphisms $\partial_q/\partial x$ and $\partial_q/\partial y$ by requiring that $$\frac{\partial_q(x^my^n)}{\partial x}=[m]x^{m-1}y^n,\frac{\partial_q(x^my^n)}{\partial y}=x^m[n]y^{n-1},$$ where $[k]=\frac{q^k-q^{-k}}{q-q^{-1}}$ for any $k\in \mathbb{N}$.

Let $U_q \mathfrak{sl}_2$ be the quantized algebra associated to
$\mathfrak{sl}_2$, i.e., the algebra generated by $E,F,K,K^{-1}$
with the relations $$KK^{-1}=K^{-1}K=1,$$
$$KE=q^2EK, \ KF=q^{-2}FK,$$  $$EF-FE=\frac{K-K^{-1}}{q-q^{-1}}.$$ It is well-known that $U_q \mathfrak{sl}_2$ is a quasi-triangular Hopf algebra.

By Theorem VII 3.3 in \cite{Ka}, $\mathbb{K}_q[x,y]$ is a $U_q
\mathfrak{sl}_2$-module-algebra with the following module
structure: for any $P\in \mathbb{K}_q[x,y] $,
$$EP=x\frac{\partial_q(P)}{\partial y}, FP=\frac{\partial_q(P)}{\partial x}y, $$ $$KP=(\omega_x\omega_y^{-1})(P), K^{-1}P=(\omega_y\omega_x^{-1})(P).$$
Set $V=\mathrm{Span}_\mathbb{K}\{x,y\}$. It is not hard to see
that the above action restricting on $V$ is the standard
2-dimensional simple $U_q \mathfrak{sl}_2$-module structure. We
know that (Theorem 2.7 in \cite{JR}) every module-algebra over a
quasi-triangular Hopf algebra has a braided algebra structure. So
$\mathbb{K}_q[x,y]$ is a braided algebra.\end{example}

\begin{example}Let $(V,\sigma)$ be a braided vector space.  It
is know that $(T(V),m,\beta)$ is a braided algebra, where $m$ is
the concatenation product. Let $M_n: V^{\otimes n}\rightarrow
T(V)$ be a linear map such that $\beta(M_n\otimes
\mathrm{id}_V)=(\mathrm{id}_V\otimes M_n)\beta_{n1}$. If we denote
by $\mathcal{I}$ the ideal of $T(V)$ generated by
$\mathrm{Im}M_n$, the image of $M_n$, then
$\beta\big(T(V)\underline{\otimes}\mathcal{I}+\mathcal{I}\underline{\otimes}T(V)\big)\subset
T(V)\underline{\otimes}I+I\underline{\otimes}T(V)$. So the
quotient algebra $T(V)/\mathcal{I}$ is also a braided algebra. For
instance, if $M_2=\mathrm{id}_V^{\otimes 2}-\sigma$, then the
quotient algebra is the $r$-symmetric algebra defined in \cite{B}.
\end{example}

\begin{example}Let $H$ be a finite dimensional quasi-triangular Hopf algebra. By a result of Majid (Theorem 3.3 in \cite{M}), the quantum double $\mathcal{D}(H)$ of $H$ is a braided algebra (according to a discussion in \cite{JR} for Radford's work \cite{Ra}).\end{example}

\section{Quantum quasi-shuffle algebras}
For any algebra $A$, it is a braided algebra with respect to the
flip map switching the two factors of $A\otimes A$. One can
construct an algebra structure on $T(A)$ which combines the the
multiplication of $A$ and the shuffle product of $T(A)$ (see
\cite{NR}). This structure is the the so-called quasi-shuffle
algebra. In fact, if the flip map is replaced by a braiding, one
can construct a quantized quasi-shuffle product by assuming some
compatibilities between the multiplication of $A$ and the braiding
(for more details, one can see \cite{JR} and \cite{JRZ}). Given a
braided algebra $(A, m,\sigma)$, the \emph{quantum quasi-shuffle
product} $\Join_\sigma$ on $T(A)$ is given by the following
inductive formula: for $i,j>1$ and any $a_1,\ldots,
a_i,b_1,\ldots, b_j\in A$,
\begin{eqnarray*}
\lefteqn{(a_1\otimes\cdots\otimes a_i)\Join_\sigma (b_1\otimes\cdots\otimes b_j)}\\
&=&a_1\otimes \big((a_2\otimes\cdots\otimes a_i)\Join_\sigma (b_1\otimes\cdots\otimes b_{j})\big)\\
&&+(\mathrm{id}_A\otimes \Join_{\sigma (i,j-1)})(\beta_{i,1}\otimes \mathrm{id}_A^{\otimes j-1})(a_1\otimes\cdots\otimes a_i\otimes b_1\otimes\cdots\otimes b_j)\\
&&+(m\otimes\Join_{\sigma (i-1,j-1)} )(\mathrm{id}_A\otimes
\beta_{i-1,1}\otimes \mathrm{id}_A^{\otimes
j-1})(a_1\otimes\cdots\otimes a_i\otimes b_1\otimes\cdots\otimes
b_j),
\end{eqnarray*}
where  $\Join_{\sigma (i,j)}$ denotes the restriction of
$\Join_\sigma$ on $V^{\otimes i}\underline{\otimes}V^{\otimes j}$.

\begin{remark}1. Given a braided algebra $(A, m,\sigma)$, $T_{\sigma}^{qsh}(A)=(T(A),\Join_\sigma)$ is a an associative algebra with unit $1\in \mathbb{K}$, and called the \emph{quantum quasi-shuffle algebra} built on $(A,
m,\sigma)$. Furthermore, the algebra $T_{\sigma}^{qsh}(A)$,
together with the braiding $\beta$ and the deconcatenation
coproduct $\delta$, forms a braided bialgebra in the sense of
\cite{Ta} (see \cite{JR}). The set of primitive elements is
exactly $A$.

2. By using Example 2.3, Proposition 17 in \cite{JRZ} can be
applied to any vector space whose basis is at most countable. In
other words, for any vector space $V$ with at most countable
basis, one can provide a linear basis of $T(V)$ by combining the
quantum quasi-shuffle product with Lyndon words.\end{remark}

\begin{example}[Hoffman's q-deformation]In \cite{Hof2}, Hoffman defined his q-deformation of
quasi-shuffles. It is an attempt to deform the quasi-shuffle
product according to the quantum shuffle product. Now we give an
explanation of the q-deformation from a point of view of the
quantum quasi-shuffles. Let $X$ be a locally finite set, i.e., $X$
is a disjoint union of finite set $X_n$, whose elements are called
letters of degree n, for $n\geq 1$. We denote by $\mathfrak{X}$
the vector space spanned by $X$. The elements in
$T(\mathfrak{X})$, which are of the form $a_1\otimes a_2\otimes
\cdots \otimes a_m$ with $a_i\in X$, are called words. Let $[ ,]$
be a graded associative product on $\mathfrak{X}$. Hoffman defined
an associative product $\ast_q$ on $T(\mathfrak{X})$: for any
words $w_1,w_2\in T(\mathfrak{X})$ and letters $a,b\in X $,
\begin{eqnarray*}\lefteqn{(a\otimes
w_1)\ast_q (b\otimes w_2)}\\
&=&a\otimes \big(w_1\ast_q(b\otimes w_2)\big)+q^{|a\otimes
w_1||b|}b\otimes \big((a\otimes w_1)\ast_q w_2
\big)+q^{|w_1||b|}[a,b](w_1\ast_q w_2),\end{eqnarray*}where $|w|$
is the degree of a word, i.e., the sum of degrees of its factors.

We define a braiding $\sigma$ on $\mathfrak{X}$ as follows: for
$x\in X_i$ and $y\in X_j$, $\sigma(x\otimes y)=q^{ij}y\otimes x$,
where $q\in K$ is a nonzero scalar. Since the product $[ ,]$
preserves the grading, it is an easy exercise to verify that
$(\mathfrak{X},[,],\sigma)$ is a braided algebra. By comparing
their reductive formulas, one can see that the quantum
quasi-shuffle algebra built on $(\mathfrak{X},[,],\sigma)$ is just
Hoffman's q-deformation.\end{example}

In order to give a more explicit description of the quantum
quasi-shuffle product, we need to recall some terminologies
introduced in \cite{GK}. An $(i,j)$-\emph{shuffle} is an element
$w\in \mathfrak{S}_{i+j}$ such that $w (1) < \cdots <w (i)$ and $w
(i+1) < \cdots <w (i+j)$. We denote by $\mathfrak{S}_{i,j}$ the
set of all $(i,j)$-shuffles. Given an $(i,j)$-shuffle $w$, a pair
$(k,k+1)$, where $1\leq k < i+j$, is called an \emph{admissible
pair} for $w$ if $w^{-1}(k)\leq i<w^{-1}(k+1)$. We denote by
$\mathcal{T}^w$ the set of all admissible pairs for $w$. For any
subset $S$ of $\mathcal{T}^w$, the pair $(w,S)$ is called a
\emph{mixable $(i,j)$-shuffle}. We denote by
$\overline{\mathfrak{S}}_{i,j}$ the set of all mixable
$(i,j)$-shuffles, i.e.,
$$\overline{\mathfrak{S}}_{i,j}=\{(w,S)|w\in \mathfrak{S}_{i,j}, S\subset \mathcal{T}^w\}.$$

Let $(A, m,\sigma)$ be a braided algebra. Define $m^k:A^{\otimes
k+1}\rightarrow A$ recursively by $m^0=\mathrm{id}_A$, $m^1=m$ and
$m^k=m(\mathrm{id}_A\otimes m^{k-1})$ for $k\geq 2$. Given
$n\in\mathbb{N}$, we denote
$$\mathcal{C}(n)=\{I=(i_1,\ldots,i_k)\in
\mathbb{N}^k|i_1+\cdots+i_k=n\}.$$
 The elements in $\mathcal{C}(n)$ are called \emph{compositions} of
 $n$. For any $I=(i_1,\ldots,i_k)\in \mathcal{C}(n)$, we define $m_I=m^{i_1-1}\otimes\cdots\otimes
 m^{i_k-1}$. For any $(w,S)\in\overline{\mathfrak{S}}_{i,j}$, we
 associate to $S$ a composition $\textrm{cp}(S)$ of $i+j$ as follows: if
$S=\{(k_1,k_1+1),\ldots, (k_s,k_s+1)\}$ with $k_1<\cdots <k_s$,
set  $$\textrm{cp}(S)=(\underbrace{1,\ldots\ldots,1}_{k_1-1\
\textrm{copies}},
2,\underbrace{1,\ldots\ldots\ldots\ldots,1}_{k_2-k_1-2\
\textrm{copies}},2,\ldots,2,\underbrace{1,\ldots\ldots\ldots,1}_{i+j-k_s-1\
\textrm{copies}}).$$  By convention, we set
$\textrm{cp}(\emptyset)=(1,1,\ldots,1)$. Denote
$T_{(w,S)}^\sigma=m_{\textrm{cp}(S)}\circ T_w^\sigma$.

\begin{theorem}Let $(A, m,\sigma)$ be a braided algebra. Then for any $a_1,\ldots,a_{i+j}\in A$,  $$(a_1\otimes\cdots\otimes a_i)\Join_\sigma (a_{i+1}\otimes\cdots\otimes a_{i+j})=\sum_{(w,S)\in\overline{\mathfrak{S}}_{i,j}}T_{(w,S)}^\sigma(a_1\otimes \cdots\otimes a_{i+j}).$$\end{theorem}
\begin{proof}We use induction on $i+j$.

When $i=j=1$,
$$a_1\Join_\sigma a_2=m(a_1\otimes a_2)+a_1\otimes a_2+\sigma(a_1\otimes a_2).$$
On the other hand,
$\overline{\mathfrak{S}}_{1,1}=\{(1_{\mathfrak{S}_2},\emptyset),
(1_{\mathfrak{S}_2},\{(1,2)\}), (s_1,\emptyset)\}$, where $s_1$ is
the generator of $\mathfrak{S}_2$. So the formula holds.

We assume that the formula is true in the case $\leq i+j$. By the
inductive formula of quantum quasi-shuffles, we have that
\begin{eqnarray*}\lefteqn{(a_1\otimes\cdots\otimes a_{i+1})\Join_\sigma (a_{i+2}\otimes\cdots\otimes a_{i+j+1})}\\[3pt]
&=&a_1\otimes \big((a_2\otimes\cdots\otimes a_{i+1})\Join_\sigma (a_{i+2}\otimes\cdots\otimes a_{i+j+1})\big)\\
&&+(\mathrm{id}_A\otimes \Join_{\sigma (i+1,j-1)})(\beta_{i+1,1}\otimes \mathrm{id}_A^{\otimes j-1})(a_1\otimes \cdots\otimes a_{i+j+1})\nonumber\\
&&+(m\otimes\Join_{\sigma (i,j-1)} )(\mathrm{id}_A\otimes
\beta_{i,1}\otimes \mathrm{id}_A^{\otimes
j-1})(a_1\otimes \cdots\otimes a_{i+j+1})\\[3pt]
&=&\sum_{(w,S)\in\overline{\mathfrak{S}}_{i,j}}(\mathrm{id}_A\otimes T_{(w,S)}^\sigma)(a_1\otimes \cdots\otimes a_{i+j+1})\\
&&+\sum_{(w,S)\in\overline{\mathfrak{S}}_{i+1,j-1}}(\mathrm{id}_A\otimes T_{(w,S)}^\sigma)(\beta_{i+1,1}\otimes \mathrm{id}_A^{\otimes j-1})(a_1\otimes \cdots\otimes a_{i+j+1})\nonumber\\
&&+\sum_{(w,S)\in\overline{\mathfrak{S}}_{i,j-1}}(m\otimes
T_{(w,S)}^\sigma )(\mathrm{id}_A\otimes \beta_{i,1}\otimes
\mathrm{id}_A^{\otimes j-1})(a_1\otimes \cdots\otimes
a_{i+j+1})\\[3pt]
&=&\sum_{(w,S)\in\overline{\mathfrak{S}}_{i,j}}(\mathrm{id}_A\otimes m_{\textrm{cp}(S)})T_{1_{\mathfrak{S}_1}\times w}^\sigma(a_1\otimes \cdots\otimes a_{i+j+1})\\
&&+\sum_{(w,S)\in\overline{\mathfrak{S}}_{i+1,j-1}}(\mathrm{id}_A\otimes m_{\textrm{cp}(S)})T_{(1_{\mathfrak{S}_1}\times w)\circ (\chi_{i+1,1}\times 1_{\mathfrak{S}_{j-1}})}^\sigma(a_1\otimes \cdots\otimes a_{i+j+1})\nonumber\\
&&+\sum_{(w,S)\in\overline{\mathfrak{S}}_{i,j-1}}(m\otimes
m_{\textrm{cp}(S)} )T^\sigma_{(1_{\mathfrak{S}_2}\times w)\circ
(1_{\mathfrak{S}_1}\times \chi_{i,1}\times
1_{\mathfrak{S}_{j-1}})}(a_1\otimes \cdots\otimes
a_{i+j+1}),\end{eqnarray*} where the third equality follows from
the fact that all the expressions of the permutations being lifted
are reduced.

Denote $$\mathcal{S}_1=\{(w,S)\in
\overline{\mathfrak{S}}_{i+1,j}|(1,2)\notin S, w(1)=1\},$$
$$\mathcal{S}_2=\{(w,S)\in \overline{\mathfrak{S}}_{i+1,j}|(1,2)\notin S, w(i+2)=1\},$$and
$$\mathcal{S}_3=\{(w,S)\in \overline{\mathfrak{S}}_{i+1,j}|(1,2)\in S\}.$$For any $(i+1,j)$-shuffle $w$, one has either $w(1)=1$ or $w(i+2)=1$.
Therefore $\mathcal{S}_1$, $\mathcal{S}_2$ and $ \mathcal{S}_3$
are mutually disjoint,  and
$\mathfrak{S}_{i+1,j}=\mathcal{S}_1\cup\mathcal{S}_2\cup\mathcal{S}_3$.

We make a further observation. It is easy to see that there is a
one-to-one correspondence between $\mathfrak{S}_{i,j}$ and $\{w\in
\mathfrak{S}_{i+1,j}|w(1)=1\}$ given by $w\mapsto
1_{\mathfrak{S}_1}\times w$ for any $w\in \mathfrak{S}_{i,j}$. So
$$\mathcal{S}_1=\{(1_{\mathfrak{S}_1}\times w,S)|w\in \mathfrak{S}_{i,j}, S\subset\mathcal{T}^{1_{\mathfrak{S}_1}\times w},(1,2)\notin S\}.$$
There is a one-to-one correspondence between
$\mathfrak{S}_{i+1,j-1}$ and $\{w\in
\mathfrak{S}_{i+1,j}|w(i+2)=1\}$ given by  $w\mapsto
\widetilde{w}=(1_{\mathfrak{S}_1}\times w)\circ
(\chi_{i+1,1}\times 1_{\mathfrak{S}_{j-1}})$ for any $w\in
\mathfrak{S}_{i+1,j-1}$. Consequently,
$$\mathcal{S}_2=\{(\widetilde{w},S)|w\in \mathfrak{S}_{i+1,j-1},S\subset\mathcal{T}^{\widetilde{w}},(1,2)\notin S\}.$$
Finally, for any $(w,S)\in \mathcal{S}_3$, we must have that
$w(1)=1$ and $w(i+2)=2$. There is a one-to-one correspondence
between $\mathfrak{S}_{i,j-1}$ and $\{w\in
\mathfrak{S}_{i+1,j}|w(1)=1,w(i+2)=2\}$ given by $w\mapsto
\overline{w}=(1_{\mathfrak{S}_2}\times w)\circ
(1_{\mathfrak{S}_1}\times \chi_{i,1}\times
1_{\mathfrak{S}_{j-1}})$ for any $w\in \mathfrak{S}_{i,j-1}$. So
$$\mathcal{S}_3=\{(\overline{w},S)\in \overline{\mathfrak{S}}_{i+1,j}|w\in \mathfrak{S}_{i,j-1},(1,2)\in S\}.$$

 As a conclusion, the three terms in the final step of the preceding
 computation come from $\mathcal{S}_1$, $\mathcal{S}_2$ and
 $\mathcal{S}_3$ respectively. So we have that $$(a_1\otimes\cdots\otimes a_{i+1})\Join_\sigma
(a_{i+2}\otimes\cdots\otimes
v_{i+j+1})=\sum_{(w,S)\in\overline{\mathfrak{S}}_{i+1,j}}T_{(w,S)}^\sigma(a_1\otimes
\cdots\otimes a_{i+j+1}),$$ which completes the induction.
 \end{proof}

\begin{remark}Let $(A,m)$ be an algebra and $\lambda$ be a scalar in $\mathbb{K}$. Then $(A, \lambda m)$ becomes a braided algebra with respect to the usual flip map. In this case, the formula in Theorem 3.3 coincides with the one of mixable shuffle product introduced in \cite{GK}.\end{remark}

Assume again that $(A,m,\sigma)$ is a braided algebra. We denote
by $S^{qsh}_\sigma(A)$ the subalgebra of $T^{qsh}_\sigma(A)$
generated by $A$. To describe this subalgebra, we need to
introduce some notations.

For a fixed $n\in \mathbb{N}$ and any $w\in \mathfrak{S}_n$, we
denote $$\mathcal{S}^w=\{(k,k+1)|1\leq k<n,
w^{-1}(k)<w^{-1}(k+1)\},$$and
$$\overline{\mathfrak{S}}_n=\{(w,S)|w\in \mathfrak{S}_n, S\subset
\mathcal{S}^w\}.$$For any $(w,S)\in \overline{\mathfrak{S}}_n$, we
associate to $S$ a composition $\textrm{cp}(S) $ of $n$ as
follows. Let $S=\{(k_1,k_1+1), \ldots, (k_s,k_s+1)\}$ with
$k_1<\cdots<k_s$. We divide $\{k_1,\ldots,k_s\}$ into several
subsets
$$\{k_1,\ldots,k_{i_1}\}, \{k_{i_1+1},\ldots,k_{i_1+i_2}\},
\ldots,\{k_{i_1+\cdots+i_{r-1}+1},\ldots,k_s\},$$ which obey the
rule that:
\[\left\{
\begin{array}{lll}
k_1+1=k_2,k_2+1=k_3,\ldots,k_{i_1-1}+1=k_{i_1},&&\\
k_{i_1+1}+1=k_{i_1+2},k_{i_1+2}+1=k_{i_1+3},\ldots,k_{i_1+i_2-1}+1=k_{i_2},&&\\
\ldots,&&\\
k_{i_1+\cdots+i_{r-1}+1}+1=k_{i_1+\cdots+i_{r-1}+2},\ldots,
k_{s-1}+1=k_{s},&&
\end{array} \right.
\]
 but $k_{i_1}+1<k_{i_1+1},
k_{i_1+i_2}+1<k_{i_1+i_2+1},\ldots,
k_{i_1+\cdots+i_{r-1}}+1<k_{i_1+\cdots+i_{r-1}+1}$. Denote
$i_r=s-i_1-\cdots-i_{r-1}$. Then we write
 $$\textrm{cp}(S)=(\underbrace{1,\ldots\ldots,1}_{k_1-1\ \textrm{copies}}, i_1+1,\underbrace{1,\ldots\ldots\ldots\ldots,1}_{k_{i_1+1}-k_{i_1}-1\ \textrm{copies}},i_2+1,\ldots,i_r+1,\underbrace{1,\ldots\ldots\ldots,1}_{n-k_{i_r}-1\ \textrm{copies}}).$$
We define as before the map
$T_{(w,S)}^\sigma=m_{\textrm{cp}(S)}\circ T_w^\sigma$ for any
$(w,S)\in\overline{\mathfrak{S}}_n$.

Now we provide a decomposition of $\overline{\mathfrak{S}}_{n+1}$
which will be used later. For any $1\leq i\leq n+1$, we denote
$\mathfrak{S}_{n+1}(i)=\{w\in \mathfrak{S}_{n+1}|w(1)=i\}$. It is
clear that $\mathfrak{S}_{n+1}$ is the disjoint union of all
$\mathfrak{S}_{n+1}(i)$'s, and for each $i$ there is a one-to-one
correspondence between $\mathfrak{S}_{n}$ and
$\mathfrak{S}_{n+1}(i)$ given by $w\mapsto
L(w,i)=(\chi_{1,i-1}\times
1_{\mathfrak{S}_{n+1-i}})\circ(1_{\mathfrak{S}_1}\times w) $ for
any $w\in \mathfrak{S}_{n-1}$. So
\begin{eqnarray*}\mathfrak{S}_{n+1}&=&\bigcup_{i=1}^{n+1}\mathfrak{S}_{n+1}(i)\\[3pt]
&=&\bigcup_{i=1}^{n+1}\bigcup_{w\in
\mathfrak{S}_{n}}\{L(w,i)\}\\[3pt]
&=&\bigcup_{w\in
\mathfrak{S}_{n}}\bigcup_{i=1}^{n+1}\{L(w,i)\}.\end{eqnarray*}
Then we have that
\begin{eqnarray*}\overline{\mathfrak{S}}_{n+1}&=&\bigcup_{w\in
\mathfrak{S}_{n}}\bigcup_{i=1}^{n+1}\{(L(w,i),S)|S\subset \mathcal{S}^{L(w,i)}\}\\[3pt]
&=&\big(\bigcup_{w\in
\mathfrak{S}_{n}}\bigcup_{i=1}^{n+1}\{(L(w,i),S)|S\subset
\mathcal{S}^{L(w,i)},(i,i+1)\notin
S\}\big)\\[3pt]
&&\cup\big(\bigcup_{w\in
\mathfrak{S}_{n}}\bigcup_{i=1}^{n+1}\{(L(w,i),S)|S\subset
\mathcal{S}^{L(w,i)},(i,i+1)\in S\}\big).\end{eqnarray*} All the
unions above are disjoint.

Given $w\in \mathfrak{S}_n$ and $S\subset \mathcal{S}^w$ with
$\mathrm{cp}(S)=(i_1,\ldots,i_s)$. For any $0\leq k\leq s$, we
denote
$$\mathrm{cp}(S)_k=(i_1,\ldots,i_k,1,i_{k+1},\ldots,i_s),$$
$$\mathrm{cp}(S)^k=(i_1,\ldots,i_{k-1},i_k+1,i_{k+1},\ldots,i_s),$$
and $$I_k=(\underbrace{1,\ldots,1}_{k\
\textrm{copies}},2,\underbrace{1,\ldots\ldots\ldots,1}_{n-1-|S|-k\
\textrm{copies}}).$$Here, $|S|$ denotes the cardinality of the set
$S$.

\begin{lemma}Under the assumptions above, we have \[\left\{
\begin{array}{lll}
(\beta_{1,k}\otimes \mathrm{id}_A^{\otimes
n-|S|-k})(\mathrm{id}_A\otimes
T_{(w,S)}^\sigma)&=&m_{\mathrm{cp}(S)_k}T_{L(w,i_1+\cdots+i_k+1)},\\[5pt]
m_{I_k}(\beta_{1,k}\otimes \mathrm{id}_A^{\otimes
n-|S|-k})(\mathrm{id}_A\otimes
T_{(w,S)}^\sigma)&=&m_{\mathrm{cp}(S)^k}T_{L(w,i_1+\cdots+i_k+1)}.
\end{array} \right.
\]\end{lemma}
\begin{proof}Since $\sigma(\mathrm{id}_A\otimes m^l)=(m^l\otimes
\mathrm{id}_A)\beta_{1,l+1}$ for any $l$ (see Lemma 2 in
\cite{B}), \begin{eqnarray*}\lefteqn{(\beta_{1,k}\otimes
\mathrm{id}_A^{\otimes n-|S|-k})(\mathrm{id}_A\otimes
T_{(w,S)}^\sigma)}\\[3pt]
&=&(\beta_{1,k}\otimes \mathrm{id}_A^{\otimes
n-|S|-k})(\mathrm{id}_A\otimes
m_{\textrm{cp}(S)})(\mathrm{id}_A\otimes T_w^\sigma)\\[3pt]
&=&m_{\textrm{cp}(S)_k}(\beta_{1,i_1+\cdots+i_k}\otimes
\mathrm{id}_A^{\otimes i_{k+1}+\cdots+i_s})(\mathrm{id}_A\otimes T_w^\sigma)\\[3pt]
&=&m_{\textrm{cp}(S)_k}T_{(\chi_{1,i_1+\cdots+i_k}\times
1_{\mathfrak{S}_{i_{k+1}+\cdots+i_s}})\circ(1_{\mathfrak{S}_1}\times
w)}^\sigma\\[3pt]
&=&m_{\mathrm{cp}(S)_k}T_{L(w,i_1+\cdots+i_k+1)}.\end{eqnarray*}

The second equality is a consequence of the first one.\end{proof}

\begin{theorem}Let $(A,m,\sigma)$ be a braided algebra. For any $a_1,\ldots,a_n\in A$, we have that$$a_1\Join_\sigma\cdots\Join_\sigma a_n=\sum_{(w,S)\in\overline{\mathfrak{S}}_n}T_{(w,S)}^\sigma(a_1\otimes \cdots\otimes a_{n}).$$ Therefore $S^{qsh}_\sigma(A)=\sum_{n\geq 0}\mathrm{Im}(\sum_{(w,S)\in\overline{\mathfrak{S}}_n}T_{(w,S)}^\sigma)$.\end{theorem}
\begin{proof}We use induction on $n$.

When $n=2$, it is trivial since
$$\overline{\mathfrak{S}}_2=\{(1_{\mathfrak{S}_2},\emptyset),(1_{\mathfrak{S}_2},\{(1,2)\}),(s_1,\emptyset)
\}.$$ By Theorem 3.5, we have that for any $a_1,\ldots,a_{r+1}\in
A$,
\begin{eqnarray*}\lefteqn{a_1\Join_\sigma (a_2\otimes\cdots\otimes a_{r+1})}\\[3pt]
&=&\sum_{k=0}^r(\beta_{1,k}\otimes \mathrm{id}_A^{\otimes r-k})(a_1\otimes \cdots\otimes a_{r+1})\\[3pt]
&&+\sum_{k=0}^{r-1}(\mathrm{id}_A^{\otimes k}\otimes m\otimes
\mathrm{id}_A^{\otimes r-k-1})(\beta_{1,k}\otimes
\mathrm{id}_A^{\otimes r-k})(a_1\otimes \cdots\otimes
a_{r+1}).\end{eqnarray*} Therefore,
\begin{eqnarray*}\lefteqn{a_1\Join_\sigma\cdots\Join_\sigma
a_{n+1}}\\[3pt]
&=&a_1\Join_\sigma\big(\sum_{(w,S)\in\overline{\mathfrak{S}}_n}T_{(w,S)}^\sigma(a_2\otimes
\cdots\otimes a_{n+1})\big)\\[3pt]
&=&\sum_{(w,S)\in\overline{\mathfrak{S}}_n}\sum_{k=0}^{n-|S|}(\beta_{1,k}\otimes
\mathrm{id}_A^{\otimes n-|S|-k})\big(a_1\otimes
T_{(w,S)}^\sigma(a_2\otimes
\cdots\otimes a_{n+1})\big)\\[3pt]
&&+\sum_{(w,S)\in\overline{\mathfrak{S}}_n}\sum_{k=0}^{n-|S|-1}(\mathrm{id}_A^{\otimes
k}\otimes m\otimes \mathrm{id}_A^{\otimes
n-|S|-k-1})\\[3pt]
&&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \
\circ(\beta_{1,k}\otimes \mathrm{id}_A^{\otimes
n-|S|-k})\big(a_1\otimes T_{(w,S)}^\sigma(a_2\otimes \cdots\otimes
a_{n+1})\big)\\[3pt]
&=&\sum_{(w,S)\in\overline{\mathfrak{S}}_n}\sum_{k=0}^{n-|S|}m_{\textrm{cp}(S)_k}T_{L(w,i_1+\cdots+i_k+1)}^\sigma(a_1\otimes
\cdots\otimes a_{n+1})\\[3pt]
&&+
\sum_{(w,S)\in\overline{\mathfrak{S}}_n}\sum_{k=0}^{n-|S|-1}m_{\textrm{cp}(S)^k}T_{L(w,i_1+\cdots+i_k+1)}^\sigma(a_1\otimes
\cdots\otimes a_{n+1}),\end{eqnarray*} where the last equality
follows from the preceding lemma.

On the other hand, by the decomposition of
$\overline{\mathfrak{S}}_{n+1}$ mentioned before,
\begin{eqnarray*}\lefteqn{\sum_{(w,S)\in\overline{\mathfrak{S}}_{n+1}}T_{(w,S)}^\sigma(a_1\otimes
\cdots\otimes a_{n+1})}\\[3pt]
&=&\sum_{w\in\mathfrak{S}_n}\sum_{i=1}^{n+1}\sum_{\substack{S\subset
\mathcal{S}^{L(w,i)}\\(i,i+1)\notin S}}T_{(L(w,i),S)}^\sigma
(a_1\otimes
\cdots\otimes a_{n+1})\\[3pt]
&&+\sum_{w\in\mathfrak{S}_n}\sum_{i=1}^{n+1}\sum_{\substack{S\subset
\mathcal{S}^{L(w,i)}\\(i,i+1)\in
S}}T_{(L(w,i),S)}^\sigma(a_1\otimes \cdots\otimes
a_{n+1}).\end{eqnarray*}We compare the terms in these two
expressions. Notice that for a fixed $1\leq i\leq n+1$ and
$S\subset \mathcal{S}^{L(w,i)}$ with $(i,i+1)\notin S$, there is a
unique $S'\in \mathcal{S}^w$ such that
$\textrm{cp}(S)=\textrm{cp}(S')_i$. Indeed, we can write down
$S^\prime$ explicitly: if $S=\{(k_1,k_1+1), \ldots, (k_s,k_s+1)\}$
with $k_1<\cdots<k_l<i<k_{l+1}<\cdots<k_s$, then
$S'=\{(k_1,k_1+1), \ldots,
(k_l,k_l+1),(k_{l+1}-1,k_{l+1}),\ldots,(k_s-1,k_s)\}$. Similarly,
if $(i,i+1)\in S$, there is a unique $S''\in \mathcal{S}^w$ such
that $\textrm{cp}(S)=\textrm{cp}(S'')^i$. It follows that every
term in
$\sum_{(w,S)\in\overline{\mathfrak{S}}_{n+1}}T_{(w,S)}^\sigma(a_1\otimes
\cdots\otimes a_{n+1})$ is from exactly one term in the formula of
$a_1\Join_\sigma\cdots\Join_\sigma a_{n+1}$. The converse is also
true. Since all terms in each formula are mutually distinct, we
get the conclusion.
\end{proof}

\begin{remark}Consider Example 3.2 and let $(\mathfrak{X},[,],\sigma)$ be the braided algebra introduced there. For any $w\in \mathfrak{S}_n$, we denote $\iota(w)=\{(i,j)|1\leq i<j\leq n, w(i)>w(j)\}$. Then for any $a_1,\cdots, a_n\in X$, $$T_w^\sigma(a_1\otimes\cdots\otimes a_n)=q^{\sum_{(i,j)\in \iota(w)}|a_i||a_j|}a_{w^{-1}(1)}\otimes \cdots \otimes a_{w^{-1}(n)}.$$

For any two compositions $I=(i_1,\ldots,i_k)$ and
$J=(j_1,\ldots,j_l)$ of $n$, we say $I$ is a \emph{refinement} of
$J$, written by $I\succeq J$, if there are $r_1,\ldots,r_l\in
\mathbb{N}$ such that $r_1+\cdots +r_l=k$ and
 $$i_1+\cdots+i_{r_1}=j_1, i_{r_1+1}+\cdots+i_{r_1+r_2}=j_2,\ldots, i_{r_1+\cdots+r_{l-1}+1}+\cdots+i_k=j_l.$$
 For instance, $(1,2,2,3)\succeq (3,2,3)$. For any $w\in \mathfrak{S}_n$, let $C(w)$ be the composition $(i_1,\ldots,i_k)$ of $n$ such that $$\{i_1,i_1+i_2,\ldots, i_1+\cdots+i_{k-1}\}=\{l|1\leq l\leq n-1, w(l)>w(l+1)\}.$$
For any $I=(i_1,\ldots,i_l)\in \mathcal{C}(n)$, we write
$I[a_1\otimes \cdots a_n]=[,]_I(a_1\otimes \cdots a_n)$.

Observing that for any $w\in \mathfrak{S}_n$ there is a one-to-one
correspondence between $S\subset \mathcal{S}^w$ and $I\in
\mathcal{C}(n)$ with $I\succeq C(w)$, one has immediately that
$$a_1\ast_q\cdots\ast_q
a_n=\sum_{w\in\mathfrak{S}_n}q^{\sum_{(i,j)\in
\iota(w)}|a_i||a_j|}\sum_{I\succeq C(w)}I[a_{w^{-1}(1)}\otimes
\cdots \otimes a_{w^{-1}(n)}].$$ This formula is given by Hoffman
when $[,]$ is commutative (see Lemma 5.2 in \cite{Hof2}).
\end{remark}

We conclude this section by an interesting formula. Let $V$ be a
vector space with basis $\{e_i\}_{i\in \mathbb{N}}$, and
$(V,m,\sigma)$ be the braided algebra structure given in Example
2.3. So we have that $\sigma(e_i\otimes e_j)=q_{ij}e_j\otimes e_i$
and $m(e_i\otimes e_j)=e_{i+j}$.

For fixed $i$, we denote $(0)_{q_{ii}}=1$ and
$(n)_{q_{ii}}=1+q_{ii}+q_{ii}^2+\cdots+q_{ii}^{n-1}=\frac{1-q_{ii}^n}{1-q_{ii}}$
when $n\in \mathbb{N}$. We denote
$(n)_{q_{ii}}!=(1)_{q_{ii}}\cdots (n)_{q_{ii}}$.

\begin{proposition}For nay $i,k\in \mathbb{N}$, we have that $$e_i^{\Join_\sigma k}=\sum_{n=1}^k\sum_{\substack{l_1+\cdots+l_n=k\\1\leq l_1,\ldots,l_n\leq k}}\frac{(k)_{q_{ii}}!}{(l_1)_{q_{ii}}!\cdots (l_n)_{q_{ii}}!}e_{l_1i}\otimes \cdots \otimes e_{l_ni}.$$\end{proposition}
\begin{proof}It is a direct verification by using induction and Theorem 3.6 or Hoffman's formula.\end{proof}

\section{The dual construction}
In this section, we give a dual construction of the quantum
quasi-shuffle algebra by using the universal property of tensor
algebras. First of all, we study a special coalgebra structure on
$T(C)$ which is a generation of the quantized cofree coalgebra
structure. For coalgebras, we adopt Sweedler's notation. That
means for a coalgebra $(C,\bigtriangleup,\varepsilon)$ and any
$c\in C$, we denote
$$\bigtriangleup(c)=\sum_{(c)}c_{(1)}\otimes c_{(2)},$$
or simply, $\bigtriangleup(c)=c_{(1)}\otimes c_{(2)}$.

To extend algebra structures to quantized case, one needs the
notion of braided algebras. By contrast, in the case of
coalgebras, one needs the so-called braided coalgebras.

\begin{definition}Let $C=(C,\bigtriangleup,\varepsilon)$ be a coalgebra with coproduct $\bigtriangleup$ and counit $\varepsilon$, and $\sigma$ be a braiding on $C$. We call $(C,\bigtriangleup,\sigma)$ a \emph{braided coalgebra} if it satisfies the following conditions:
\[\left\{
\begin{array}{lll}
(\mathrm{id}_C\otimes \bigtriangleup)\sigma&=&\sigma_1\sigma_2(
\bigtriangleup\otimes
\mathrm{id}_C),\\[3pt] ( \bigtriangleup\otimes
\mathrm{id}_C)\sigma&=&\sigma_2\sigma_1(\mathrm{id}_C\otimes
\bigtriangleup).
\end{array} \right.
\]

\end{definition}

In order to get new braided algebras and coalgebras from old ones,
we need the proposition below.
\begin{proposition}[\cite{HH}, Proposition 4.2]1.
For a braided algebra $(A,\mu,\sigma)$ and any $i\in \mathbb{N}$,
$(A^{\otimes i},\mu_{\sigma,i}, \beta_{ii})$ becomes a braided
algebra with product $\mu_{\sigma,i}=\mu^{\otimes i}\circ
T^\sigma_{w_i}$, where $w_i\in \mathfrak{S}_{2i}$ is given by
\[w_{i}=\left(\begin{array}{ccccccccc}
1&2&3&\cdots&i&i+1&i+2&\cdots & 2i\\
1&3&5&\cdots&2i-1&2& 4 &\cdots & 2i
\end{array}\right).\]

2. For a braided coalgebra $(C, \bigtriangleup,\sigma)$,
$(C^{\otimes i},\bigtriangleup_{\sigma,i}, \beta_{ii})$ becomes a
braided coalgebra with coproduct $\bigtriangleup_{\sigma,i}=
T^\sigma_{w_i^{-1}}\circ\bigtriangleup^{\otimes i}$ and counit
$\varepsilon^{\otimes i}:C^{\otimes i}\rightarrow
\mathbb{K}^{\otimes i}\simeq \mathbb{K}$.\end{proposition}

Let $(C,\bigtriangleup,\sigma)$ be a braided coalgebra. Consider
the tensor algebra $T(C)$ with the concatenation product $m$. Then
by
 the above proposition, $\mathscr{T} _\beta^2(C)=(T(C)\underline{\otimes}
 T(C),m_{\beta,2})$ and $\mathscr{T} _\beta^3=(T(C)\underline{\otimes}
 T(C)\underline{\otimes}
 T(C),m_{\beta,3})$ are associative algebras.

 We define
\[\begin{array}{cccc}
\phi_1: &C&\rightarrow& T(C)\underline{\otimes}
 T(C)
,\\[6pt]
&c&\mapsto&1\underline{\otimes} c+ c\underline{\otimes}1.
\end{array}\]

By the universal property of tensor algebras, there exists an
algebra map $\Phi_1: T(C)\rightarrow \mathscr{T} _\beta^2(C)$
whose restriction on $C$ is $\phi_1$. Moreover $\Phi_1$ is
coassociative and is the dual of quantum shuffle product (see,
e.g., \cite{FG}).

Now we define
\[\begin{array}{cccc}
\phi_2: &C&\rightarrow& T(C)\underline{\otimes}
 T(C)
,\\[6pt]
&c&\mapsto&\sum c_{(1)}\underline{\otimes} c_{(2)}.
\end{array}\]

By the universal property of tensor algebras again, there exists
an algebra map $\Phi_2: T(C)\rightarrow \mathscr{T} _\beta^2(C)$
whose restriction on $C$ is $\phi_2$.

\begin{proposition}For any $i\in \mathbb{N}$, we have that $\Phi_2\mid_{C^{\otimes
i}}=\bigtriangleup_{\sigma,i}$. So $(T(C), \Phi_2, \beta)$ is a
braided coalgebra.\end{proposition}
\begin{proof}We use induction on $i$. When $i=1$, it is trivial. We assume the equality holds for the case $i< n$.
Then for any $c_1,\ldots,c_n\in C$, we have
\begin{eqnarray*}\lefteqn{\Phi_2(c_1\otimes \cdots \otimes
c_n)}\\[3pt]
&=&m_{\beta,2}\big(\Phi_2(c_1)\otimes\Phi_2(c_2\otimes \cdots \otimes c_n)\big)\\[3pt]
&=&(\mathrm{id}_{T(C)}\otimes \beta \otimes \mathrm{id}_{T(C)})(\bigtriangleup(c_1)\otimes T^\sigma_{w_{n-1}^{-1}}\circ\bigtriangleup^{\otimes n-1}(c_2\otimes \cdots \otimes c_n)\big)\\[3pt]
&=&(\mathrm{id}_C\otimes T^\sigma_{\chi_{1,n-1}}\otimes
\mathrm{id}_C^{\otimes n-1})(\mathrm{id}_C^{\otimes 2}\otimes
T^\sigma_{w_{n-1}^{-1}})\bigtriangleup^{\otimes n}(c_1\otimes
\cdots \otimes
c_n)\\[3pt]
&=&T^\sigma_{w_n^{-1}}\circ\bigtriangleup^{\otimes n}(c_2\otimes
\cdots \otimes c_n),\end{eqnarray*}where the last equality follows
from the fact that $(1_{\mathfrak{G}_1 }\times\chi_{1,n-1}\times
1_{\mathfrak{G}_{}n-1 })(1_{\mathfrak{G}_2 }\times
w_{n-1}^{-1})=w_{n}^{-1}$ and the expression is
reduced.\end{proof}

Let $\phi=\phi_1+\phi_2: C\rightarrow\mathscr{T} _\beta^2(C)$ and
$\Phi$ be the algebraic map induced by the universal property of
tensor algebras which extends $\phi$.

\begin{proposition}Under the notation above, the triple $(T(C), \Phi, \beta)$ is a braided coalgebra.
\end{proposition}
\begin{proof}We first show that
\[\left\{
\begin{array}{lll}
\beta_1\beta_2 (\Phi\otimes \mathrm{id}_{T(C)})&=& (\mathrm{id}_{T(C)}\otimes \Phi)\beta,\\[3pt]
\beta_2\beta_1(\mathrm{id}_{T(C)}\otimes \Phi)&=&(\Phi\otimes
\mathrm{id}_{T(C)})\beta.
\end{array} \right.
\]
For any $x\in C^{\otimes i}$ and $y\in C^{\otimes j}$ we verify
the first one on $x\underline{\otimes}y$. The second one can be
verified similarly. We use induction on $i$.

When $i=1$,
\begin{eqnarray*}\beta_1\beta_2 (\Phi\otimes \mathrm{id}_{T(C)})(x\underline{\otimes}y)&=&\beta_1\beta_2 (\phi_1\otimes \mathrm{id}+\phi_2\otimes \mathrm{id})(x\underline{\otimes}y)\\[3pt]
&=&(\mathrm{id}\otimes \phi_1+\mathrm{id}\otimes \phi_2)\beta(x\underline{\otimes}y)\\[3pt]
&=& (\mathrm{id}_{T(C)}\otimes
\Phi)\beta(x\underline{\otimes}y).\end{eqnarray*} For any $c\in
C$, we have
\begin{eqnarray*}\beta_1\beta_2 (\Phi\otimes \mathrm{id}_{T(C)})\big((c\otimes x)\underline{\otimes}y\big)&=&\beta_1\beta_2\beta_3\beta_4\beta_2(\Phi\otimes \Phi\otimes\mathrm{id}_{T(C)})(v\underline{\otimes} x\underline{\otimes}y)\\[3pt]
&=&\beta_1\beta_3\beta_2\beta_3\beta_4(\Phi\otimes \Phi\otimes\mathrm{id}_{T(C)})(c\underline{\otimes} x\underline{\otimes}y)\\[3pt]
&=&\beta_1\beta_3\beta_2\big(\Phi\otimes \beta_1\beta_2(\Phi\otimes\mathrm{id}_{T(C)})\big)(c\underline{\otimes} x\underline{\otimes}y)\\[3pt]
&=&\beta_1\beta_3\beta_2(\Phi\otimes \mathrm{id}_{T(C)}\otimes\Phi)\beta_2(v\underline{\otimes} x\underline{\otimes}y)\\[3pt]
&=&\beta_3\big(\beta_1\beta_2(\Phi\otimes \mathrm{id}_{T(C)})\otimes\Phi\big)\beta_2(c\underline{\otimes} x\underline{\otimes}y)\\[3pt]
&=&\beta_3(\mathrm{id}_{T(C)}\otimes \Phi\otimes\Phi)\beta_1\beta_2(c\underline{\otimes} x\underline{\otimes}y)\\[3pt]
&=&(\mathrm{id}_{T(C)}\otimes \Phi)\beta\big((c\otimes
x)\underline{\otimes}y\big).\end{eqnarray*}

The next step is to show $(\Phi\otimes
\mathrm{id}_{T(C)})\Phi=(\mathrm{id}_{T(C)}\otimes \Phi)\Phi$.
Notice that for any $c\in C$,
\begin{eqnarray*}\lefteqn{(\Phi\otimes \mathrm{id}_{T(C)})\Phi(c)}\\[3pt]
&=&(\Phi\otimes \mathrm{id}_{T(C)})(c_{(1)}\underline{\otimes} c_{(2)}+1\underline{\otimes} c+c\underline{\otimes} 1)\\[3pt]
&=&c_{(1)}\underline{\otimes} c_{(2)}\underline{\otimes}
c_{(3)}+1\underline{\otimes} c_{(1)}\underline{\otimes} c_{(2)}+
c_{(1)}\underline{\otimes} c_{(2)}\underline{\otimes}
1\\[3pt]
&&+c_{(1)}\underline{\otimes}1 \underline{\otimes}
c_{(2)}+1\underline{\otimes} 1
\underline{\otimes}c+c_{(1)}\underline{\otimes} c_{(2)}\underline{\otimes} 1+1\underline{\otimes}c \underline{\otimes}1+c\underline{\otimes}1\underline{\otimes}1\\[3pt]
&=&(\mathrm{id}_{T(C)}\otimes \Phi)\Phi(c).\end{eqnarray*} By the
uniqueness of the universal property of $T(C)$, we only need to
show that both $(\Phi\otimes \mathrm{id}_{T(C)})\Phi$ and
$(\mathrm{id}_{T(C)}\otimes \Phi)\Phi$ are algebra morphisms from
$T(C)$ to $\mathscr{T} _\beta^3$. Since $\Phi: T(C)\rightarrow
\mathscr{T} _\beta^2(C)$ is an algebra morphism, we have that
$$\Phi\circ m=(m\otimes m)(\mathrm{id}_{T(C)}\otimes \beta
\otimes \mathrm{id}_{T(C)})(\Phi\otimes \Phi).$$ So
\begin{eqnarray*}
\lefteqn{(\Phi\otimes \mathrm{id}_{T(C)})\circ\Phi\circ
m}\\[3pt]
&=&(\Phi\otimes \mathrm{id}_{T(C)})(m\otimes
m)(\mathrm{id}_{T(C)}\otimes \beta \otimes
\mathrm{id}_{T(C)})(\Phi\otimes \Phi)\\[3pt]
&=&\big((\Phi\circ m)\otimes m\big)(\mathrm{id}_{T(C)}\otimes
\beta \otimes
\mathrm{id}_{T(C)})(\Phi\otimes \Phi)\\[3pt]
&=&(m\otimes m \otimes m)(\mathrm{id}_{T(C)}\otimes \beta \otimes
\mathrm{id}_{T(C)} \otimes \mathrm{id}_{T(C)}\otimes
\mathrm{id}_{T(C)})\\[3pt]
&&\circ(\Phi\otimes \Phi  \otimes \mathrm{id}_{T(C)} \otimes
\mathrm{id}_{T(C)})(\mathrm{id}_{T(C)}\otimes \beta \otimes
\mathrm{id}_{T(C)})(\Phi\otimes \Phi)\\[3pt]
&=&(m\otimes m\otimes m) \beta_2\\[3pt]
&&\circ\big(\mathrm{id}_{T(C)} \otimes \mathrm{id}_{T(C)}\otimes(\Phi\otimes \mathrm{id}_{T(C)}) \beta \otimes \mathrm{id}_{T(C)}\big)\big((\Phi\otimes \mathrm{id}_{T(C)})\Phi\otimes \Phi\big)\\[3pt]
&=&(m\otimes m\otimes
m)\beta_2\beta_4\beta_3\\[3pt]
&&\circ(\mathrm{id}_{T(C)}\otimes \mathrm{id}_{T(C)}\otimes
\mathrm{id}_{T(V)}\otimes \Phi\otimes
\mathrm{id}_{T(C)})\big((\Phi\otimes \mathrm{id}_{T(C)})\Phi\otimes \Phi\big)\\[3pt]
&=&m_{\beta,3}\big((\Phi\otimes \mathrm{id}_{T(C)})\Phi\otimes(
\Phi_2\otimes \mathrm{id}_{T(C)})\Phi\big).
\end{eqnarray*}
It follows that $(\Phi\otimes \mathrm{id}_{T(C)})\Phi$ is an
algebra morphism. Similarly, the map $(\mathrm{id}_{T(C)}\otimes
\Phi)\Phi$ is also an algebra morphism.
\end{proof}

Now we begin to study the relation between the braided coalgebra
$(T(C), \Phi, \beta)$ and the quantum quasi-shuffle algebra. We
show that they are dual to each other in the following sense.

Let $<,>: V\times W\rightarrow \mathbb{K}$ and
$<,>^\prime:V^\prime\times W^\prime\rightarrow \mathbb{K}$ be two
bilinear non-degenerate forms. For any $f\in \mathrm{Hom}(V,
V^\prime)$, the adjoint operator $\mathrm{adj}(f)\in
\mathrm{Hom}(W^\prime, W)$ of $f$ is defined to be the one such
that $<x,\mathrm{adj}(f)(y)>=<f(x),y>^\prime$ for any $x\in V$ and
$y\in W^\prime$. It is clear that $\mathrm{adj}(f\circ
g)=\mathrm{adj}(g)\circ\mathrm{adj}(f)$.

\begin{remark}If there is a non-degenerate bilinear form between two vector spaces $A$ and $C$, and $(A,m,\sigma)$ is a braided algebra, then
$(C,\mathrm{adj}(m),\mathrm{adj}(\sigma))$ is a braided coalgebra.
The converse is also true.\end{remark}

From now on, we assume that there always exists a non-degenerate
bilinear form $<,>$ between two vector spaces $A$ and $C$. It can
be extended to a bilinear form $<,>: A^{\otimes n}\times
C^{\otimes n}\rightarrow \mathbb{K}$ for any $n \geq 1$ in the
usual way: for any $a_1,\ldots, a_n\in A$ and $c_1,\ldots,c_n\in
C$,
$$<a_1\otimes \cdots \otimes a_n, c_1\otimes \cdots \otimes
c_n>=\prod_{i=1}^n <a_i, c_i>.$$ It induces a non-degenerate
bilinear form  $<,>: T(A)\times T(C)\rightarrow \mathbb{K}$ by
setting that $<x,y>=0$ for any $x\in A^{\otimes i}$ , $y\in
C^{\otimes j}$ and $i\neq j$. Then we can define a non-degenerate
bilinear form $<,>: T(A)\underline{\otimes}T(A)\times
T(C)\underline{\otimes}T(C)\rightarrow \mathbb{K}$ by requiring
that $<u\underline{\otimes} v,x\underline{\otimes}y>=<u,x><v,y>$
for any $u,v\in T(A)$ and $x,y\in T(C)$.

If $(C,\bigtriangleup,\sigma)$ is a braided coalgebra, we denote
$\tau=\mathrm{adj}(\sigma)$ and $\alpha=\mathrm{adj}(\beta)$. Then
$\alpha$ is a braiding on $T(A)$ and
$\alpha_{i,j}=T^\tau_{\chi_{ji}^{-1}}=T^\tau_{\chi_{ij}}$.

\begin{theorem}Under the assumptions above, we have that $\mathrm{adj}(\Phi)=\Join_\sigma$.\end{theorem}
\begin{proof}For any $c_1,\ldots,c_n\in
C$, we notice that
\begin{eqnarray*}\Phi(c_1\otimes \cdots\otimes c_n)&=&m_{\beta,2}\big(\Phi(c_1)\underline{\otimes}\Phi(c_2\otimes \cdots\otimes c_n\big)\\[3pt]
&=&m_{\beta,2}\big((1\underline{\otimes}c_1)\underline{\otimes}\Phi(c_2\otimes \cdots\otimes c_n)\big)\\[3pt]
&&+m_{\beta,2}\big((v_1\underline{\otimes}1)\underline{\otimes}\Phi(c_2\otimes \cdots\otimes c_n)\big)\\[3pt]
&&+m_{\beta,2}\big((c_{1(1)}\underline{\otimes}c_{1
(2)})\underline{\otimes}\Phi(c_2\otimes \cdots\otimes
c_n)\big)\\[3pt]
&=&(\beta_{1,?}\otimes \mathrm{id}_C^{\otimes n-1-?})(\mathrm{id}_C\otimes \Phi)(c_1\otimes \cdots\otimes c_n)\\[3pt]
&&+(\mathrm{id}_C\otimes \Phi)(c_1\otimes \cdots\otimes c_n)\\[3pt]
&&+(\mathrm{id}_C\otimes\beta_{1,?}\otimes \mathrm{id}_C^{\otimes
n-1-?})(\bigtriangleup\otimes \Phi)(c_1\otimes \cdots\otimes
c_n).\end{eqnarray*} Here, since $\Phi(C^{\otimes k})\subset
C^{\otimes 0}\underline{\otimes}C^{\otimes k}+C^{\otimes
k}\underline{\otimes}C^{\otimes 0}+\cdots+C^{\otimes
k}\underline{\otimes}C^{\otimes k}$, we denote by $\beta_{1,?}$
the action of $\beta$ on $C\underline{\otimes}C^{\otimes ?}$ where
$C^{\otimes ?}$ is the left factor of some component in
$\Phi(C^{\otimes k})$.

We denote by $\mathrm{adj}(\Phi)_{(k,l)}$ the action of
$\mathrm{adj}(\Phi)$ on $A^{\otimes
k}\underline{\otimes}A^{\otimes l}$. Then on $A^{\otimes
i}\underline{\otimes}A^{\otimes j}$, we have
\begin{eqnarray*}\mathrm{adj}(\Phi)_{(i,j)}&=&\mathrm{adj}\big((\beta_{1,?}\otimes
\mathrm{id}_C^{\otimes i+j-1-?})(\mathrm{id}_C\otimes
\Phi)+(\mathrm{id}_C\otimes
\Phi)\\[3pt]
&&\ \ \ \ \ \ +(\mathrm{id}_C\otimes\beta_{1,?}\otimes
\mathrm{id}_C^{\otimes i+j-1-?})(\bigtriangleup\otimes \Phi)\big)_{(i,j)}\\[3pt]
&=&\big(\mathrm{id}_A\otimes \mathrm{adj}(\Phi)_{(i,j-1)}\big)(\alpha_{i,1}\otimes \mathrm{id}_A^{\otimes j-1})\\[3pt]
&&+\big(\mathrm{id}_A\otimes \mathrm{adj}(\Phi)_{(i-1,j)}\big)\\[3pt]
&&+\big(\mathrm{adj}(\bigtriangleup)\otimes
\mathrm{adj}(\Phi)_{(i-1,j-1)}\big)(\mathrm{id}_A\otimes\beta_{i-1,1}\otimes
\mathrm{id}_A^{\otimes j-1}).\end{eqnarray*}This shows that the
map $\mathrm{adj}(\Phi)$ shares the same inductive formula with
the quantum quasi-shuffle product built on $(A,
\mathrm{adj}(\bigtriangleup),\tau)$. Hence we have the conclusion.
\end{proof}

\section{Applications}
In the past decade, the connection between Rota-Baxter algebras
and quasi-shuffles was discovered and well studied (see \cite{GK},
\cite{EG} and the references therein). Quasi-shuffle algebras
built on commutative algebras provide free objects in the category
of commutative Rota-Baxter algebras. Later, in \cite{Lod}, it is
showed that quasi-shuffle algebras built on polynomial algebras
are free in the category of commutative tridendriform algebras.
Motivated by these works, here, we use quantum quasi-shuffle
algebras to provide examples of Rota-Baxter algebras and
tridendriform algebras.

We first recall the definition of Rota-Baxter algebras. For more
information, one can see \cite{GK}.

\begin{definition}Let $\lambda$ be an element in $\mathbb{K}$. A pair $(R,P)$ is called a \emph{Rota-Baxter algebra of weight $\lambda$} if $R$ is a $\mathbb{K}$-algebra and $P$ is a linear endomorphism of $R$ satisfying that for any $x,y\in R$, $$P(x)P(y)=P(xP(y))+P(P(x)y)+\lambda P(xy).$$\end{definition}

Let $(A, m,1_A,\sigma)$ be a unital braided algebra. Then
apparently $(A, \lambda\cdot m,\sigma)$ is a braided algebra for
any $\lambda\in \mathbb{K}$. We denote by $\Join_{\sigma,\lambda}$
the quantum quasi-shuffle product with respect to $(A,
\lambda\cdot m,\sigma)$. By Lemma 3 in \cite{Ba}, the space
$A\otimes T_{\sigma}^{qsh}(A)$ is a unital associative algebra
with the product
$$\lozenge_{\sigma,\lambda}=(m\otimes
\Join_{\sigma,\lambda})(\mathrm{id}_A\otimes \beta \otimes
\mathrm{id}_{T(A)}).$$ We denote by
$\mathcal{R}_{\sigma,\lambda}(A)$ the pair $(A\otimes
T_{\sigma}^{qsh}(A),\lozenge_{\sigma,\lambda})$. We can view
$\mathcal{R}_{\sigma,\lambda}(A)$ as $T^+(A)=\oplus_{i\geq
1}A^{\otimes i}$ at the level of vector spaces. Recently, we have
two products $\Join_{\sigma,\lambda}$ and
$\lozenge_{\sigma,\lambda}$ on $T^+(A)$. This is an example of
2-braided algebras which produce quantum multi-brace algebras (for
the definitions, see \cite{JR}). We define an endomorphism
$P_A:\mathcal{R}_{\sigma,\lambda}(A)\rightarrow\mathcal{R}_{\sigma,\lambda}(A)$
by
\begin{eqnarray*}P_A(a_0\underline{\otimes}u)&=&1_A\underline{\otimes}a_0\otimes u,\ \mathrm{if}\ u\in T^+(A),\\[3pt]
P_A(a_0\underline{\otimes}\nu)&=&1_A\underline{\otimes}\nu\cdot
a_0,\ \mathrm{if}\ \nu \in \mathbb{K} .\end{eqnarray*}

\begin{theorem}Under the assumptions above, the pair $(\mathcal{R}_{\sigma,\lambda}(A), P_A)$ is a Rota-Baxter algebra of weight $\lambda$.\end{theorem}
\begin{proof}Observe that $\beta(u\underline{\otimes}1_A)=1_A\underline{\otimes}u$ for any $u\in T(A)$. Therefore for any $a, b\in A$ and $x\in A^{\otimes i}$,$ y\in A^{\otimes j}$, we have
\begin{eqnarray*}\lefteqn{P\big((a\underline{\otimes}x)\lozenge_{\sigma,\lambda}P(b\underline{\otimes}y)\big)}\\[3pt]
&=&P\big((a\underline{\otimes}x)\lozenge_{\sigma,\lambda}(1_A\underline{\otimes}b\otimes
y)\big)=P\Big(a\underline{\otimes}\big(x\Join_{\sigma,\lambda}(b\otimes
y)\big)\Big)\\[3pt]
&=&1_A\underline{\otimes}a\otimes
\big(x\Join_{\sigma,\lambda}(b\otimes y)\big),\end{eqnarray*}
\begin{eqnarray*}\lefteqn{P\big(P(a\underline{\otimes}x)\lozenge_{\sigma,\lambda}(b\underline{\otimes}y)\big)}\\[3pt]
&=&P\big((1_A\underline{\otimes}a\otimes  x)\lozenge_{\sigma,\lambda}(b\underline{\otimes}y)\big)\\[3pt]
&=&1_A\underline{\otimes}\big((\mathrm{id}_A\otimes
\Join_{\sigma,\lambda,(i+1,j)})(\beta_{i+1,1}\otimes
\mathrm{id}_A^{\otimes j})(a\otimes  x\otimes b\otimes
y)\big),\end{eqnarray*}and
\begin{eqnarray*}\lefteqn{\lambda P\big((a\underline{\otimes}x)\lozenge_{\sigma,\lambda}(b\underline{\otimes}y)\big)}\\[3pt]
&=& 1_A\underline{\otimes}\big((\lambda\cdot m\otimes
\Join_{\sigma,\lambda})(\mathrm{id}_A\otimes \beta_{i1} \otimes
\mathrm{id}_A^{\otimes j})(a\otimes x\otimes b\otimes y)\big)
.\end{eqnarray*} By taking a summation, we get
\begin{eqnarray*}\lefteqn{P\big((a\underline{\otimes}x)\lozenge_{\sigma,\lambda}P(b\underline{\otimes}y)\big)+P\big(P(a\underline{\otimes}x)\lozenge_{\sigma,\lambda}(b\underline{\otimes}y)\big)+\lambda P\big((a\underline{\otimes}x)\lozenge_{\sigma,\lambda}(b\underline{\otimes}y)\big)}\\[3pt]
&=&1_A\underline{\otimes}\Big(a\otimes
\big(x\Join_{\sigma,\lambda}(b\otimes
y)\big)\\[3pt]
&&\ \ \ \ \ \ \ \ \ +(\mathrm{id}_A\otimes
\Join_{\sigma,\lambda,(i+1,j)})(\beta_{i+1,1}\otimes
\mathrm{id}_A^{\otimes j})(a\otimes  x\otimes b\otimes
y)\\[3pt]
&&\ \ \ \ \ \ \ \ \ +(\lambda\cdot m\otimes
\Join_{\sigma,\lambda})(\mathrm{id}_A\otimes \beta_{i1} \otimes
\mathrm{id}_A^{\otimes j})(a\otimes x\otimes b\otimes y)\Big)\\[3pt]
&=&1_A\underline{\otimes}\big((a\otimes x)\Join_{\sigma,\lambda}(b\otimes y)\big)\\[3pt]
&=&P_A(a\underline{\otimes}x)\lozenge_{\sigma,\lambda}P_A(b\underline{\otimes}y).\end{eqnarray*}\end{proof}

\begin{definition}A triple $(R,P,\sigma)$ is called a \emph{braided Rota-Baxter algebra of weight $\lambda$} if $(R,\sigma)$ is a braided algebra and $P$ is an endomorphism of $R$ such that $(R,P)$ is a Rota-Baxter algebra of weight $\lambda$ and $\sigma(P\otimes P)=(P\otimes P)\sigma$.\end{definition}

\begin{example}Let $(A,m,1_A,\sigma)$ be a unital braided algebra and $(\mathcal{R}_{\sigma,\lambda}(A),P_A)$ be the Rota-Baxter algebra defined before. Then $(\mathcal{R}_{\sigma,\lambda}(A),P_A,\beta)$ is a braided Rota baxter of weight $\lambda$.

Indeed, the only thing we need to verify is that $\beta(P_A\otimes
P_A)=(P_A\otimes P_A)\beta$. For any $a, b\in A$ and $x\in
A^{\otimes i}$,$ y\in A^{\otimes j}$, we have
\begin{eqnarray*}\lefteqn{\beta(P_A\otimes
P_A)\big((a\underline{\otimes}x)\underline{\otimes}(b\underline{\otimes}y)\big)}\\[3pt]
&=&\beta\Big(\big(1_A\underline{\otimes}(a\otimes x)\big)\underline{\otimes}\big(1_A\underline{\otimes}(b\otimes y)\big)\Big)\\[3pt]
&=&1_A\underline{\otimes}(\beta_{1,j+1}\otimes \mathrm{id}_A^{\otimes i+1})\Big(1_A\underline{\otimes}\beta_{i+1,j+1}\big((a\otimes x)\underline{\otimes}(b\otimes y)\big)\Big)\\[3pt]
&=&(P_A\otimes
P_A)\beta\big((a\underline{\otimes}x)\underline{\otimes}(b\underline{\otimes}y)\big).\end{eqnarray*}\end{example}

\begin{proposition}Let $(R,P,\sigma)$ be a braided Rota-Baxter algebra of weight $\lambda$. We define $$x\star_P y=xP(y)+P(x)y+\lambda xy,$$ for any $x,y \in R$. If $\sigma(P\otimes \mathrm{id})=(\mathrm{id}\otimes P)\sigma$ and $\sigma(\mathrm{id}\otimes P)=(P\otimes \mathrm{id})\sigma$, then $(R,\star_P,P,\sigma)$ is again a braided Rota-Baxter algebra of weight $\lambda$.\end{proposition}
\begin{proof}It is well-known that $(R,\star_P,P)$ is a Rota-Baxter algebra of weight $\lambda$.
We denote by $m$ the multiplication of $R$. Then we have
\begin{eqnarray*}\lefteqn{\sigma(\star_P\otimes \mathrm{id})}\\[3pt]
&=&\sigma\big((m\otimes \mathrm{id})( \mathrm{id}\otimes P\otimes \mathrm{id})+(m\otimes \mathrm{id})( P\otimes \mathrm{id}\otimes \mathrm{id})+\lambda m\otimes \mathrm{id}\big)\\[3pt]
&=&(\mathrm{id}\otimes m)\sigma_1\sigma_2( \mathrm{id}\otimes P\otimes \mathrm{id})+(\mathrm{id}\otimes m)\sigma_1\sigma_2( P\otimes \mathrm{id}\otimes \mathrm{id})\\[3pt]
&&+(\mathrm{id}\otimes \lambda m)\sigma_1\sigma_2\\[3pt]
&=&\big((\mathrm{id}\otimes m)( \mathrm{id}\otimes\mathrm{id} \otimes P)+(\mathrm{id}\otimes m)(\mathrm{id} \otimes P\otimes \mathrm{id})+(\mathrm{id}\otimes \lambda m)\big)\sigma_1\sigma_2\\[3pt]
&=&(\mathrm{id}\otimes\star_P
)\sigma_1\sigma_2.\end{eqnarray*}

The another condition $\sigma(\mathrm{id}\otimes\star_P
)=(\star_P\otimes \mathrm{id})\sigma_2\sigma_1$ can be verified
similarly.\end{proof} Given any braided Rota-Baxter algebra
$(R,P,\sigma)$, the above proposition provides another example of
2-braided algebra.

Now we turn to tridendriform algebras which were introduced by
Loday and Ronco (\cite{LR}).

\begin{definition}Let $V$ be a vector space, and $\prec$, $\succ$ and $\cdot$ be three binary operations on $V$. The quadruple $(V,\prec, \succ, \cdot)$ is called a \emph{tridendriform algebra} if the following relations are
satisfied: for any $x,y,z\in V$,
\begin{eqnarray*}(x\prec y)\prec z&=&x\prec(y\ast z),\\[3pt]
(x\succ y)\prec z&=&x\succ (y\prec z),\\[3pt]
(x\ast y)\succ z&=&x\succ(y\succ z),\\[3pt]
(x\succ y)\cdot z&=&x\succ(y \cdot z),\\[3pt]
(x\prec y)\cdot z&=&x\cdot (y\succ z),\\[3pt]
(x\cdot y)\prec z&=&x\cdot (y\prec z),\\[3pt]
(x\cdot y)\cdot z&=&x\cdot (y \cdot z),\end{eqnarray*}where $x\ast
y=x\prec y+x\succ y+ x\cdot y$.\end{definition}

\begin{theorem}Let $(A,m,\sigma)$ be a braided algebra. We define three operations $\cdot$, $\prec$ and $\succ$ on $T(A)$ recursively by: for any $a,b\in A$ and any $x\in A^{\otimes i}$, $x\in A^{\otimes i}$,
\begin{eqnarray*}(a\otimes x)\cdot (b\otimes y)&=&(m\otimes\Join_{\sigma (i,j)} )(\mathrm{id}_A\otimes
\beta_{i,1}\otimes \mathrm{id}_A^{\otimes j-1})(a\otimes x\otimes
b\otimes y),\\[3pt]
(a\otimes x)\prec (b\otimes y)&=&a\otimes \big(x\Join_\sigma (b\otimes y)\big),\\[3pt]
(a\otimes x)\succ (b\otimes y)&=&(\mathrm{id}_A\otimes
\Join_{\sigma (i+1,j)})(\beta_{i+1,1}\otimes
\mathrm{id}_A^{\otimes j})(a\otimes x\otimes b\otimes
y).\end{eqnarray*}Then $(T(A), \prec,\succ,\cdot)$ is a
tridendriform algebra.
\end{theorem}
\begin{proof}All the verifications are direct. We just need that $\Join_\sigma$ is associative and compatible with the braiding $\beta$ in the sense of braided algebras. For instance, we show the third condition. For any $a\in A$ and $x,y,z\in T(A)$,
\begin{eqnarray*}
\lefteqn{(x\Join_\sigma y)\succ (a\otimes
z)}\\[3pt]
&=&(\mathrm{id}_A\otimes \Join_{\sigma})(\beta_{?,1}\otimes
\mathrm{id}_{T(A)})(\Join_\sigma \otimes \mathrm{id}_A\otimes \mathrm{id}_{T(A)})(x \underline{\otimes} y\underline{\otimes} a\otimes z)\\[3pt]
&=&(\mathrm{id}_A\otimes \Join_{\sigma})(\beta_{?,1}(\Join_\sigma
\otimes \mathrm{id}_A)\otimes
\mathrm{id}_{T(A)})(x \underline{\otimes} y\underline{\otimes} a\otimes z)\\[3pt]
&=&(\mathrm{id}_A\otimes
\Join_{\sigma})(\mathrm{id}_A\otimes\Join_\sigma  \otimes
\mathrm{id}_{T(A)})\\[3pt]
&&\circ(\beta_{?,1}\otimes \mathrm{id}_{T(A)}\otimes\mathrm{id}_{T(A)})(\mathrm{id}_{T(A)}\otimes\beta_{?,1}\otimes \mathrm{id}_{T(A)})(x \underline{\otimes} y\underline{\otimes} a\otimes z)\\[3pt]
&=&(\mathrm{id}_A\otimes
\Join_{\sigma})(\mathrm{id}_A\otimes\mathrm{id}_{T(A)}
\otimes\Join_\sigma
)\\[3pt]
&&\circ(\beta_{?,1}\otimes \mathrm{id}_{T(A)}\otimes\mathrm{id}_{T(A)})(\mathrm{id}_{T(A)}\otimes\beta_{?,1}\otimes \mathrm{id}_{T(A)})(x \underline{\otimes} y\underline{\otimes} a\otimes z)\\[3pt]
&=&(\mathrm{id}_A\otimes \Join_{\sigma})(\beta_{?,1}\otimes \mathrm{id}_{T(A)})\\[3pt]
&&\circ(\mathrm{id}_{T(A)}\otimes\mathrm{id}_A\otimes\Join_\sigma )(\mathrm{id}_{T(A)}\otimes\beta_{?,1}\otimes \mathrm{id}_{T(A)})(x \underline{\otimes} y\underline{\otimes} a\otimes z)\\[3pt]
&=&x\succ\big(y\succ  (a\otimes z)\big).
\end{eqnarray*}\end{proof}

\begin{remark}Let $(R,\cdot,P)$ be a Rota-Baxter algebra of weight 1. Define $a\prec b=a\cdot P(b)$ and $a\succ b=P(a)\cdot b$. Then $(R,\prec,\succ,\cdot)$ is a tridendriform algebra (see \cite{Lod}). Using this fact, we can prove the above theorem by a more easier argument: embed $A$ into the unital braided algebra $(\widetilde{A},\widetilde{m},1,\widetilde{\sigma})$, then the tridendriform algebra structure in Theorem 5.8 comes from the Rota-Baxter algebra $(\mathcal{R}_{\widetilde{\sigma},1}(\widetilde{A}),P_{\widetilde{A}})$.\end{remark}

\section*{Acknowledgements}
I learned the original construction of quantum quasi-shuffles from
Professor Marc Rosso in 2008. I am grateful to him for encouraging
me to continue the study of this subject. Finally, I would like to
thank Professor Daniel Sternheimer for his suggestion that
improves the title of this paper. This work was partially
supported by NSF of China (Grant No. U0935003), and the grant for
new teachers from DGUT (Grant No. ZJ100501).

\bibliographystyle{amsplain}

redpajama_set_name: RedPajamaArXiv


text:
President Donald Trump earlier this month did not rule out the possibility of firing Jeff Sessions during an interview with "Fox & Friends." The president said during the interview that Sessions should have told him that he was going to recuse himself from special counsel Robert Mueller's investigation. | Manuel Balce Ceneta/AP Photo
Trump says Sessions will be attorney general at least until the midterms
By REBECCA MORIN
Attorney General Jeff Sessions, who has repeatedly come under fire from President Donald Trump, will stay in his position until at least the midterm elections in November, the president told Bloomberg News on Thursday.
But Trump declined to comment on whether he would keep Sessions on beyond November.
"I just would love to have him do a great job," Trump said in the interview.
The president regularly bashes Sessions for recusing himself from overseeing special counsel Robert Mueller's Russia probe, which has proven to be a major legal headache for the White House. Deputy Attorney General Rod Rosenstein now oversees the investigation.
Trump has assailed the probe as a "witch hunt" and has said he's done nothing illegal.
Trump earlier this month did not rule out the possibility of firing Sessions during an interview with "Fox & Friends." The president said during the interview that Sessions should have told him that he was going to recuse himself from Mueller's investigation.
The president has also personally lobbied Republican senators to turn against Sessions.
POLITICO reported that Trump made a call to Sen. Lindsey Graham about firing the attorney general, and that the South Carolina Republican asked Trump to hold off on any action until the midterm elections.
Earlier this week, Graham told NBC's "Today" show that the relationship between Trump and Sessions is "beyond repair."

redpajama_set_name: RedPajamaCommonCrawl


The Wyevale Garden Centre chain has recently shed a number of stores, which means we can't make use of the vouchers from our loyalty card at the Oxford Garden Centre anymore. Consequently, we took a detour on our way back from another visit to Bletchley Park today and ended up getting a cut price lunch with extra points at the World's End branch (not in deepest, darkest anywhere, unless you have a very low opinion of the area round Aylesbury).
All vouchers duly spent and now I'm minded to close the end by watching The World's End (2013), which was filmed in that vicinity and is one of my favourite comedy / horror films.


text:
\section{Introduction}
\label{sec:intro}

Image denoising is a fundamental image reconstruction problem, but still an active topic for low-level vision researchers. The main objective of image denoising is to recover the clean latent image $f$ from the noise corrupted version $g$, which follows the image degradation model $g= f+ \eta$, where $\eta$ is the additive noise. It has been shown that a efficient denoising algorithm can solve many other image reconstruction problems such as super resolution, deblurring, inpainting, compression etc.\cite{DBLP:journals/corr/RomanoEM16}

Initially, learning the image prior was considered as the effective way of denoising the image. In particular, models like non-local similarity (NSS) models\cite{buades2005non}, Markov random field (MRF) models\cite{li2009markov}, Sparse models\cite{elad2006image}, etc. was used for denoising. \\
However, all the previous models require complex iterative computation, thus becomes time-consuming in the testing stage. On the contrary, for a real-time application, fast algorithms are needed. As a solution , discriminative models\cite{schmidt2014shrinkage} attracted attention because these models eliminated the iterative steps required at the testing stage.
\\In recent times, as a discriminative model, deep convolutional network\cite{mao2016image},\cite{zhang2018ffdnet},\cite{remez2017deep},\cite{8451132} has started gaining considerable attention because of its fast and efficient denoising capability. In this regard Kai Zhang et al.\cite{zhang2017beyond} proposed DnCNN network, they first utilized batch normalization and residual learning for denoising task.
\begin{figure}
       \begin{subfigure}[b]{0.25\textwidth}
                \includegraphics[width=\linewidth]{O05.png}
                \caption{Original}
            \label{fig:or7}
        \end{subfigure
       \begin{subfigure}[b]{0.25\textwidth}
                \includegraphics[width=\linewidth]{N05.png}
                \caption{Noisy}
                \label{fig:n07}
        \end{subfigure}%

        \begin{subfigure}[b]{0.25\textwidth}
                \includegraphics[width=\linewidth]{D05.png}
                \caption{DnCNN(0.55M parameters)}
                \label{fig:m07}
        \end{subfigure
        \begin{subfigure}[b]{0.25\textwidth}
                \includegraphics[width=\linewidth]{M05.png}
                \caption{Proposed(0.13M parameters)}
                \label{fig:d07}
        \end{subfigure}
        \caption{ \footnotesize	Denoising Result on noise level $
        \sigma$ = 25, (a) Original, (b) Noisy (PSNR:24dB/SSIM:0.60) (c)DnCNN (PSNR:30.38dB/SSIM:0.91) (d) Proposed (PSNR:30.29dB/SSIM:0.91). Proposed model with $76\%$ lesser parameter than current state of art\cite{zhang2017beyond} yielded visually similar image. }
        \label{fig:parrot}
\end{figure}
Their model otperformed many of benchmark model such as; BM3D\cite{dabov2006image}, TNRD\cite{chen2017trainable}, etc.

Indeed, Deep CNN based models provide better performance than other models. However, these models comprise of a good deal of matrix multiplication. Therefore, good computing resources are required for implementation. Also, attempt to reduce the complexity of network results in a reduction in performance. In this study, we tried to make a tradeoff between the model complexity and performance.\\
Meanwhile, Mean Square Error (MSE) or L2 norm of the difference image remains the most preferred error measure in the researcher community; it may be because L2 norm is convex, differentiable, easier to optimize. Additionally, it also provides the maximum likelihood (ML) estimate solution. Still, as an error measure, the L2 norm ignores the local characteristic of the image like, contrast, luminance, structural information, etc., which are an integral part of the human visual system. In this study, we also tried to incorporate this measures into loss function to train our network.
\\ \textbf{Contribution: }The main contributions of this study are as follows:
\begin{itemize}
    \item We present a new CNN architecture for image denoising with lesser complexity than the benchmark architectures. We also propose an optional addition to the above mentioned architecture to further reduce the complexity and execution time. Our model with lesser number of parameter produced similar result like more complex deep CNN models.
    \item We forwarded the feature extracted in the first layer to next layer through skip connections. Our experimental result shows that, these low-level features contain complementary information that high level feature lacks for texture generation. Our solution reconstructs better texture than baseline solution.
    \item We also showed the effect of inclusion of local characteristic into the cost function by using MS-SSIM. We also propose specific sequence of multi stage training to generate optimal solution.
\end{itemize}
Remaining part of the paper is organized in the following way. In Section 2 the proposed architecture is introduced, then in Section 3 the training procedure and the cost function is explained, and in Section 4 the evaluation results are represented, and finally, Section 5 provides the conclusion about the findings of this study.
\begin{figure}
    \centering
    \includegraphics[width= 0.5\textwidth, height=1in]{net5.png}
    \caption{Proposed Architecture DB: Dense Block, TL: Transition Layer}
    \label{fig:model}
\end{figure}
\section{Proposed Architecture}
It is known that the initial layer of deep neural network learns low-level feature like edge information, corner points, etc, whereas the deeper layer learns more complex feature like face orientation, larger shapes, etc. Perhaps, these high-level features are useful in classification task, but for image restoration these features might become less significant. Indeed, the low-level features plays an important role in image restoration. With this foundation, we decided to propagate low-level features through the network and let the network infer the importance of these features for reconstruction. Figure \ref{fig:model} illustrates the proposed network. First layer is a convolution layer with 64 filters of $3\times3$ kernel size. Next, a series of dense block (DB)and transition layer (TL) is used to extract features, and then finally a convolution layer with $3\times3$ kernel to reconstruct the image. We used total 6 pairs of dense block and transition layer in our network.
Inside the DB, the feature-maps of all preceding convolution layers are used as the input to current layer, and its own feature-maps are used as inputs into all subsequent layers. The growth rate of every DB was set to 16, and four convolution layer followed by batch normalization and ReLU non-linearity was used in each DB. After every dense block, a transition layer consisting $1\times1$ convolution layer followed by batch normalization and ReLU, had been used to reduce the depth of the feature map and, also to combine the feature extracted at different layer. According to the NIN paper\cite{lin2013network}, $1\times1$ convolution is similar to cross-channel parametric pooling. This cascaded cross channel parametric pooling structure allows complex and learnable interactions of cross channel information. The input of the transition layer is a mixture of the low-level and high-level feature over the volume. The $1\times1$ convolution may drop the low-level feature or can propagate them to the next layer.
\\Adapting dense net\cite{huang2017densely} architecture helped in reusing the feature map. As a result, the total number of parameter reduced significantly. Furthermore, a better receptive field was obtained by using more number of layers with fewer filters. Additionally, the use of skip connection also makes the error surface more smooth and convex\cite{li2018visualizing}. In addition to this, traditional convolution can also be replaced with depthwise separable convolution\cite{howard2017mobilenets} to further reduce the number of parameters drastically.
\begin{table*}[]
\centering
\resizebox{\textwidth}{!}{%
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline
Dataset & Noise Level & \multicolumn{9}{c|}{Method} \\ \hline
 &  & BM3D\cite{dabov2006image}  & WNNM\cite{gu2014weighted} & EPLL\cite{zoran2011learning} & MLP\cite{burger2012image} & CSF\cite{schmidt2014shrinkage} & TNRD\cite{chen2017trainable} & DnCNN\cite{zhang2017beyond}  & FFDNet\cite{zhang2018ffdnet} & Proposed v1 \\ \hline
 \hline
\multirow{3}{*}{BSD68} & 15 & 31.07 & 31.37 & - & 31.24 & 31.42 &31.73 &  \textcolor{red}{31.75} & 31.63 & \textcolor{green}{31.70} \\ \cline{2-11}
 & 25 & 28.57 & 28.83 & 28.68 & 28.96 & 28.74 & 28.92 & \textcolor{red}{29.23} & 29.19 & \textcolor{green}{29.20} \\ \cline{2-11}
 & 50 & 25.62 & 25.87 & 25.67 & 26.03 & - & 25.97 & 26.23 & \textcolor{red}{26.29} & \textcolor{green}{26.25} \\ \hline
 \hline
\multirow{3}{*}{SET12} & 15 & 32.37 & 32.69 & 32.13 & - & 32.31 & 32.50 & \textcolor{red}{32.88} & 32.75 & \textcolor{green}{32.82} \\ \cline{2-11}
 & 25 & 29.96 & 30.25 & 29.69 & 30.02 & 29.83 & 30.05 & \textcolor{red}{30.45} & \textcolor{green}{30.43} & 30.41 \\ \cline{2-11}
 & 50 & 26.72 & 27.05 & 26.47 & 26.78 & - & 26.81 & \textcolor{green}{27.23} & \textcolor{red}{27.32} & 27.17 \\ \hline

\end{tabular}%
}
\caption{The average PSNR(dB) results of different methods, top 2 results are marked in red and green colour respectively.}
\label{psnr}
\end{table*}
\begin{figure*}
        \begin{subfigure}[b]{0.25\textwidth}
                \includegraphics[width=\linewidth]{O7.png}
                \caption{Original}
                \label{fig:or7}
        \end{subfigure}%
        \begin{subfigure}[b]{0.25\textwidth}
                \includegraphics[width=\linewidth]{N7.png}
                \caption{Noisy}
                \label{fig:n07}
        \end{subfigure}%
        \begin{subfigure}[b]{0.25\textwidth}
                \includegraphics[width=\linewidth]{M7.png}
                \caption{Using Low-level feature}
                \label{fig:m07}
        \end{subfigure}%
        \begin{subfigure}[b]{0.25\textwidth}
                \includegraphics[width=\linewidth]{D7.png}
                \caption{Without low-level feature}
                \label{fig:d07}
        \end{subfigure}
        \caption{\footnotesize	Denoising Result of Parrot image on noise level $
        \sigma$ = 25. (The reader is encouraged to zoom in for a better view). Reconstructed image using low-level feature has better texture than the other solution. (a) Original, (b) Noisy, (c) Proposed v1 (d) We removed the skip connection before all the transition layer so that low-level feature does not propagate through the network. } \label{fig:parrot}
\end{figure*}

\begin{figure*}[htb]
    \centering
\begin{subfigure}{0.32\textwidth}
  \includegraphics[width=\linewidth]{08.png}
  \caption{Original}
  \label{fig:1}
\end{subfigure}\hfil
\begin{subfigure}{0.32\textwidth}
  \includegraphics[width=\linewidth]{N8.png}
  \caption{Noisy}
  \label{fig:2}
\end{subfigure}\hfill
\begin{subfigure}{0.32\textwidth}
  \includegraphics[width=\linewidth]{b8.png}
  \caption{BM3D}
  \label{fig:3}
\end{subfigure}
\medskip
\begin{subfigure}{0.32\textwidth}
  \includegraphics[width=\linewidth]{t8.png}
  \caption{TNRD}
  \label{fig:4}
\end{subfigure}\hfill
\begin{subfigure}{0.32\textwidth}
  \includegraphics[width=\linewidth]{d8.png}
  \caption{DnCNN}
  \label{fig:5}
\end{subfigure}\hfil
\begin{subfigure}{0.32\textwidth}
  \includegraphics[width=\linewidth]{m8.png}
  \caption{Proposed v1}
  \label{fig:6}
\end{subfigure}
\caption{\footnotesize	Denoising result on lena image with noise level 25 by different method. Proposed method performed better in preserving fine details while removing noise. The lower inset is the zoomed version of marked area, and the upper inset is the difference image of same region.}
\label{fig:method}
\end{figure*}

\section{Training Details}
Use of residual learning in image reconstruction is already established\cite{zhang2017beyond}. According \cite{he2016deep} to when the original mapping is more similar to identity mapping then it is more easy to learn and optimize the residual mapping. Then clean image can restored by subtracting the residual map(i.e. the noise map) from the noisy image.
\\Now, one obvious choice for error measure is L2 norm of difference image or the Mean Square Error. However, L2 does not consider the local characteristic of the image. To overcame this drawback, we first trained the entire network using MSE. Then, retrained only the last layer of the network with the combination L2 norm and differentiable Multi Scale Structural Similarity Index (MS-SSIM) as a multi objective optimization problem keeping other layer untouched. Why we only trained the last layer only with MS-SSIM is discussed in the Section 4. \\The cost function for training the final layer of the image is given by
{\[\ l(\Theta) =(1- MS-SSIM(x,(y-R(y;\Theta))) \]}
{\[\ + \frac{1}{2N} \sum_{i=1}^{N} || R(y_{i}; \Theta) - (y_{i} - x_{i})||_{F}^{2} \]}
The differentiable implementation of MS-SSIM can be found in the study by Hang Zhao \textit{et al.}\cite{zhao2017loss}.
\\We trained our network for three specific noise level, particularly for $\sigma =$ 15, 25, 50. For validation two benchmark dataset, Berkeley segmentation data set (BSD68) containing 68 natural images, and famous 12 images of SET12, is used.

\section{Result and Discussion}

We evaluated both of our model namely; proposed v1: as described above section, and proposed v2: replacing the convolution layer by depthwise separable convolution. Table \ref{psnr} depicts the comparison of the proposed v1 network with other method applied in this two dataset. It can be seen that proposed method has outperformed many benchmark algorithm like BM3D, TNRD, WNMM, CSF and also yielded comparable PNSR with DnCNN algorithm.
\begin{table}[]
\centering
\resizebox{0.45\textwidth}{!}{%
\begin{tabular}{cccccc}
 &  & \multicolumn{4}{c}{METHOD} \\ \hline
\multicolumn{1}{|c|}{Dataset} & \multicolumn{1}{c|}{Noise Level} & \multicolumn{1}{c|}{BM3D} & \multicolumn{1}{c|}{TNRD} & \multicolumn{1}{c|}{DnCNN} & \multicolumn{1}{c|}{Proposed v1} \\ \hline
\multicolumn{1}{|c|}{\multirow{3}{*}{BSD 68}} & \multicolumn{1}{c|}{$\sigma$=15} & \multicolumn{1}{c|}{0.8741} & \multicolumn{1}{c|}{0.8947} & \multicolumn{1}{c|}{0.8947} & \multicolumn{1}{c|}{\textbf{0.8949}} \\ \cline{2-6}
\multicolumn{1}{|c|}{} & \multicolumn{1}{c|}{$\sigma$=25} & \multicolumn{1}{c|}{0.8025} & \multicolumn{1}{c|}{0.8206} & \multicolumn{1}{c|}{0.8321} & \multicolumn{1}{c|}{\textbf{0.8326}} \\ \cline{2-6}
\multicolumn{1}{|c|}{} & \multicolumn{1}{c|}{$\sigma$=50} & \multicolumn{1}{c|}{0.6744} & \multicolumn{1}{c|}{0.7104} & \multicolumn{1}{c|}{0.7252} & \multicolumn{1}{c|}{\textbf{0.7289}} \\ \hline
\multicolumn{1}{|c|}{\multirow{3}{*}{SET 12}} & \multicolumn{1}{c|}{$\sigma$=15} & \multicolumn{1}{c|}{0.8989} & \multicolumn{1}{c|}{0.9004} & \multicolumn{1}{c|}{0.9073} & \multicolumn{1}{c|}{\textbf{0.9099}} \\ \cline{2-6}
\multicolumn{1}{|c|}{} & \multicolumn{1}{c|}{$\sigma$=25} & \multicolumn{1}{c|}{0.8553} & \multicolumn{1}{c|}{0.8573} & \multicolumn{1}{c|}{0.8675} & \multicolumn{1}{c|}{\textbf{0.8678}} \\ \cline{2-6}
\multicolumn{1}{|c|}{} & \multicolumn{1}{c|}{$\sigma$=50} & \multicolumn{1}{c|}{0.7679} & \multicolumn{1}{c|}{0.7753} & \multicolumn{1}{c|}{\textbf{0.7913}} & \multicolumn{1}{c|}{\textbf{0.7913}} \\ \hline
\end{tabular}%
}
\caption{Comparision of SSIM with other state of art models.}
\label{ssim}
\end{table}
\\Figure \ref{fig:parrot} gives an example to show the effect of adding low-level feature to the input of every transition layer. Reconstructing the texture or the fine details is one of the most challenging tasks for image denoising models. Indeed, most of the algorithm produces a smoothed version of the image. However, it can be seen that adding low-level feature provides better texture. In Figure \ref{fig:parrot} the fine texture near the nose and also in the area near chin is noticable. Table \ref{ssim} reports comparative SSIM of our model with other three benchmark algorithm.
\begin{table}[]
\centering
\resizebox{0.4\textwidth}{!}{%
\begin{tabular}{|c|c|c|c|c|}
\hline
\textbf{Dataset} & \textbf{Noise Level} & \textbf{DnCNN} & \textbf{Proposed v1} & \textbf{Proposed v2}\\ \hline
\hline
\multicolumn{1}{|l|}{\multirow{3}{*}{BSD68}} & 15 & 31.75 & 31.70 & 31.62 \\
\multicolumn{1}{|l|}{} & 25 & 29.23 & 29.20 & 29.08 \\
\multicolumn{1}{|l|}{} & 50 & 26.23 & 26.25 & 26.12 \\ \hline
\hline
\multirow{3}{*}{SET12} & 15 & 32.88 & 32.82 & 32.69 \\
 & 25 & 30.45 & 30.41 & 30.32 \\
 & 50 & 27.23 & 27.17 & 27.09 \\ \hline
 \hline
Parameter &  & 556032 & 382080 & 133248 \\ \hline
\end{tabular}%
}
\caption{Comparison of Parameter and PSNR(dB)}
\label{para}
\end{table}
In most of the times, We had scored best SSIM. Retraining last layer with MS-SSIM helped in achieving this. Furthermore, we also witnessed a small improvement in PSNR after training with MS-SSIM.  We also tried to retrain all the layer with MS-SSIM and, but doing so resulted in similar SSIM but lower PSNR images.
\\ Table \ref{para} shows comparison between the number of trainable parameter and PSNR. Model trained with depthwise separable convolution has nearly $76\%$ less parameter than the DnCNN model, but has achieved similar PSNR, also our model with normal convolution has $30\%$ less parameter than DnCNN.

The visual comparisons of different methods are given in Figure \ref{fig:method}. Proposed model has kept fine details better than the other model. The shape of the reconstructed eye is better our model, it can be also verified by examining the difference image. Overall, proposed solution provides good perceptual quality image.

\section{Conclusion}
 In this paper a new CNN network for image denoising is proposed. This method has less complexity than the state of the art method, but still resultant images are similar with better texture. This light weight model can be utilised in low resources devices, such as smart phones, to perform state of art denoising.
\bibliographystyle{IEEEbib}


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