As a government recognized export house since 1976 ,we specialize in the export of industrial leather safety products to Europe, USA, Canada, Australia, Middle Eastand South East Asia. Our manufacturing units in Kolkata, India are equipped with the latest machinery required for making leather gloves, jackets ,sleeves, shoe protectors, aprons, etc. for industrial safety use. we also make handbags, wallets, briefcase and belts for the consumer markets.


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Athletes For Imagining More by Shannon Rowbury for Imagining More
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Athletes For Imagining More
by Shannon Rowbury for Imagining More
The AthletesFor campaigns have closed. We hope you will join us in supporting our favorite charities and causes next year, but in the meantime, should you wish to support the Imagining More Foundation, please click here.
My idea of Imagining More came from my experiences training in Mexico. I grew up in the city of San Francisco and in a home environment where I thought anything was possible. I encountered strong women in my daily life (my mother worked and my elementary school had a woman principal and vice principal). It was not until I got older that I started to see signs of resistance towards women striving for more.
About Imagining More
Imagining More was established to promote education, arts and athletics. It was born from the limitations that young people face when pursuing extracurricular activities during their academic years. Our goal is to support multi-cultural individuals who wish to pursue education, art projects and/or sport programs, as well as to create awareness about the importance of developing into a multi-dimensional individual.
What Imagining More Has Accomplished
Borderless Sport & Art Scholarships:
- Provided scholarships and resources to several low income / underserved individuals in support of:
- Their college education
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- Their participation in sports programs (athletic gear, program fees, etc.)
Participated in numerous speaking engagements at schools & businesses
- Encouraged and created awareness about the importance of developing into a multi-dimensional individual through sport and/or art
Donated athletic apparel, shoes and equipment to low income / underserved schools and youth running clubs in the USA and Mexico
Provided money & resources to high schools in support of low income / underserved students who do not have the means to participate in extracurricular art projects and/or sports programs
What Imagining More Will Do with Funds Raised through Athletes For
Post Collegiate Scholarships
- $1500 to cover 1 scholarship to be applied towards:
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High School and Youth Coaching Clinic
- $1500 to cover 5 complimentary admissions for coaches at underserved schools
About Shannon Rowbury
Shannon Rowbury is one of the most decorated USA athletes of all time. She is a three-time Olympian, two-time World Championships Bronze medalist, six-time United States champion, and the American Record Holder at 1500m, 2 miles, and 5000m.
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Q: Accessing buttons outside the form C# i'm a newbie in C#. My boss has asked me to get co-ordinates of cursor when it's over some button outside the form (e.g on Desktop or any other random Window, like below).

I will then have to access the button properties for some computations (e.g width and height of button or it's name). I'm not sure if it's possible, my boss says it is as he's done something like that in the past. Gave me a hint that it can be done by knowing if the element is clickable and that there's an API for it. Thanks in advance!
Code Below
public partial class Form1 : Form
    {
        public string label2Y;
        public string label1X;
        public Form1()
        {
            InitializeComponent();
        }

        private void Form1_MouseMove(object sender, MouseEventArgs e)
        {
            label1.Text = Cursor.Position.X.ToString();
            label2.Text = Cursor.Position.Y.ToString();
        }

        private void Form1_Click(object sender, EventArgs e)
        {
            label3.Text = Cursor.Position.X.ToString();
            label4.Text = Cursor.Position.Y.ToString();
        }

    }

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If you are a wire rack table lover and you like to have your small or not so small collection of special wine bottles. Surely you also like to show them off when they come guests or just to contemplate artwork which hangs on a wall. Well today we bring you all a very simple way to make a wire rack table minimalist design. Only we need a wooden plank about 25cm wide by 3 or 4 cm thick by the length we want, sandpaper, a meter and a pencil. The width can also be modified, but the righteous are 25cm to create a sturdy shelf where you can expose several bottles of wine.
The wood can be purchased at any woodworking or if we find a wood in any container that is in good condition, we like its color and see fit to present it on a wall, still better, cheaper we get the wine shelf .First we cut the board to measures that fit our needs, as long or as short as you want. Then we mark the holes where we will hold wine rack table. Once drilled marked with a circular saw the hole at an angle of 30 degrees between hole and hole leaving a space of at least 10cm. (If you do not have the tools or saw, we can go to a carpenter and in 30 minutes we have all the facts holes). Once we have the holes made with the sandpaper to remove any chips soften you may have.
Now that we have prepared the board, we can only hang on the wall like a shelf and insert each bottle in a hole. And we have our shelf ready to show off our beautiful collection of wine rack table. In this example the shelf for wine has been the natural wood color, but each one can give you the look you want. It can be varnished with transparent varnish, or change of tone with tinted varnish, paint it any color with synthetic enamel, etc. The important thing is that each what suits your home decor. You see, now your wine rack table bottles will look beautiful in your new minimalist shelf wood you have done yourselves and your taste for how many bottles you will.


text:
© Grand Bend Strip
Casey Lessard
Crediton, South Huron DHS, VIPs, Vol. 1, #16
Remembering Ryan
Ryan VanValkengoed, 17, of Crediton went missing after leaving a friend's house a short distance from home the evening of January 11. Police divers found his body in the Ausable River Monday afternoon. Ryan was the oldest of three sons of Bob and Lorie VanValkengoed, owners of Advanced Auto Parts and Salvage and Lorie's Advanced Hair Care.
As told to Casey Lessard
Bob VanValkengoed: He was very responsible. With this incident, we knew there was a problem right from the start.
Lorie VanValkengoed: We were hoping this was the first time he did something out of character. But it wasn't.
Lorie: When we first got married, we were probably married about a week and I remember saying to Bob, "Let's have children right away," and him saying, "Yep." As a woman, I thought there were about 15 more sentences that needed to go with that, so I waited a week and asked again, and he said, "Yep."
The hardest time Ryan ever gave us was giving birth. He was 19 days overdue. He was due December 9, and Christmas Eve, the doctor told me to come see him. I said, "You have no idea. If you put me in the hospital over Christmas, I guarantee I will make your life a living hell." He said he just wanted to make sure I would make it through Christmas.
On December 26, we went into London to be induced and on December 28 at 2:32 in the morning, he was born by Caesarean section.
Bob: It's the only time in his life he's been late. He was pretty good otherwise.
He was always a good kid. I went to London three or four times a week, and every morning I had to get up at 5 o'clock and he would get upset if I didn't wake him up. He wasn't even two years old. So I would wake him up and put his work clothes on and we would go to work. He would sleep on my lap. We would take a load of products to London, and I always remember going to the restaurant for breakfast. At that age, he was just so busy fooling around or crawling under the table. We would drop off scrap at Zubick's and every time he would get a chocolate bar. For years, if I didn't wake him up, he would just be so grumpy that I didn't wake him up.
After that, he was always in the shop. He was blonde as snow, but he'd come out of there covered in black from oil and mud.
One day I was working and I had bought an electric car for him. He wasn't two yet, and I looked over; he had hooked a set of chain falls (for pulling engines out of cars) onto the electric car and it was sitting eight feet off the ground.
Lorie: He was interested in possibly taking over the business (Advanced Auto). We just talked about it over the holidays.
Bob: It was in his blood. He liked it. He could print invoices and take care of the business end of it. I always checked and there was never a time when he forgot something. And if someone came after hours, he would tell them, "I'll deal with you this time, but you shouldn't make this a regular habit." He was good at training his customers. He was a real businessman.
Lorie: School was never an interest for him. He wanted to run his own business. One day, I got a call from the school saying Ryan had skipped. I asked him when he got home, and he said, "Yeah, I skipped." I asked what he did, where he went. "To the cafeteria." I said, "You skipped school to go to the cafeteria?" He said he was talking with the principal Jeff Reaburn. They were talking about starting your own business. So I called the school and said, "My son skipped, he was talking with the principal, so please give him detention."
He was a thrill seeker. One day, he fell out of a tree on the property and he came to the shop and said to Bob he fell. He was walking and talking, and Bob said, "You look good to me." We found out later he had fallen 30'. His first cousin said he landed and looked fine but started crying because he couldn't find his shoe. It wasn't the fall that hurt him.
Bob: We always had golf carts or dirt bikes. All the kids rode around on this one golf cart. I told someone that we went through 50 gallons of gas in one summer.
I bought him a new dirt bike in the spring, and he always wore a helmet. I was raised with a bike and I still ride without a helmet. But Ryan always wore a helmet. I always said, "What did I say to him to convince him to wear a helmet?"
Lorie: I cut hair out of my home, and I was with a client one day and Ryan was at the top of the stairs. Ryan asked if I could come up the stairs. I said I was busy – I think I was doing a colour. He called Jacob instead. Ryan had wiped out severely on his dirt bike. Jacob, to help Ryan had wrapped his wounds in toilet paper. It took me about two days to pick the toilet paper out of the wounds as it healed. To this day, he had scars (and he was proud of every one of them).
Bob: He was in bed three days, and Jacob took his meals up to him.
Lorie: All of us were very close with Ryan. He connected with each one of us in different ways.
Bob: Just a great guy to be around. Some of my hired help aren't around this week because they're taking it very hard.
Lorie: Ryan had a good group of friends.
Bob: [He didn't like big parties.] Even on a Friday night, he was never comfortable if there were more than five people.
Lorie: The only time he liked that was for concerts. In fact, I have a $600 bill on my credit cards for Linkin Park.
Bob: He would take matters into his own hands and order tickets. One day he called me and said he needed (continued on p.4) (cont'd from p.3) $900 on my credit card. I said, "What are you doing?" He said, "We're going to see Motley Crue and Aerosmith." He said the tickets were $150 a ticket and if he got six guys to go he could get a limo lined up. He was 15 when he did that. He was organized and knew what he liked.
Bob: He was very thrifty. He came from the movies one night and said, "Dad, I got free popcorn." I asked how he did that. He said, "It was easy; I just went into the garbage bag and grabbed an empty bag and told the guy to fill it." I said, "Don't you find it gross to grab a bag out of the garbage?" He said, "Do you think I'm that stupid? I told the guy I needed a new bag!"
Bob: Lorie had gone to the States for a business meeting for the weekend. I was with the boys myself. I woke up at about 12:15 a.m. I went and checked him out and he wasn't there. So I laid down again and kept getting up again every couple hours. All of a sudden, at 5:45 a.m., he's still not there. I thought maybe this was the first time he slept over at a friend's house. I had to take one of the other boys to hockey practice and on the way, I called the shop but got no answer. He would sleep out in the shop sometimes. I kept thinking he was staying at one of his buddy's places. I had breakfast with a friend of mine that morning before 8 o'clock, and through the morning I kept trying to call a couple of his friends and got no answer.
Finally I got through at 2 o'clock when I called his cousin. His cousin said, "Oh, Ryan should have been home." There were five or six people at the house he was at Friday, and Ryan was one of them. That's when I started getting scared.
In the meantime, one of my nephews drives over here and he's concerned. That's when I called the police and told them it was very out of character. I thought they would just fill out a missing person's report, but he took it very seriously. Within hours they had the dogs out and police officers all over the place. It was on the news Saturday night.
Lorie: I was at a big conference, and one of the ladies called my room Saturday night. She told me Bob had been trying to get in touch with me. I looked at my friends and to
ld them, "It's not good." He wouldn't call me for anything unless it was very serious. I phoned home and when I came downstairs, my one friend said, "You look like you're going to be ill." I kind of half-smiled and said, "Ryan's missing." A couple other friends gathered around and asked, what do you want to do, do you want to sit down, what can we do for you? I said, "Actually, I'd like to breathe for a minute."
I sat and collected my thoughts, and one friend who had disappeared showed up and said, "We're packed and ready to go." It was a surreal ride home. We were about four hours away and I think we did it in two-point-five. The closer we got to home, the more it started to sink in.
Bob: From Saturday afternoon to Monday, we hardly slept. You doze off for a few hours here and there. I sat in the chair in the kitchen, and from that chair, I could see the door to see if anybody was coming in, I could see the road to see if anyone was coming there, and I could look at the clock. For three days, I kept looking at the clock, the door, the window.
Everyone was helping. Even people I knew didn't have their driver's license were driving around trying to find him.
Lorie: Two officers, Ralph Christmas and Jeff Adkin, were in charge. They sat with us, they gave us minute-by-minute updates. We had many people call offering to volunteer. The police had to treat it as a possible crime scene, so volunteers couldn't come in. They tried to get a helicopter from Orillia, but it was grounded due to fog, so local pilots took officers up and did an air search. There were at least a couple of planes. There were numerous neighbours and their kids out looking. From young to old and everybody in between.
Bob: The police found some of his belongings by the tree by the river. They said that's the last trace of where he's been. Monday afternoon, they found him about 25' from the bridge. The autopsy said he had drowned.
My brother-in-law didn't want to be the one to tell us, so he asked a neighbour to do it, and all he said was, "It's not good news. They found him." And we knew.
The house was full of company – friends and family and relatives. And it was devastating. I'll never forget that.
Lorie: The O'Briens had taken the kids for two nights and took their kids out of school to keep them company. Before the news spread, I went to tell my boys. They had some questions and some tears.
I called Const. Adkin and told him I was ready to come and see Ryan, so I got to spend some time with him before he left. That was a really good time.
Lorie: You couldn't sit and write enough thank you cards. The first one would go to the police officers, then the professionals who did the search, the pilots who volunteered, the volunteers, the fire department.
Bob: The phone calls, the food. I've never seen so much food in my life.
Lorie: I'll miss the stupid stuff. Folding laundry and making three piles.
Bob: Four people for supper, not five.
Lorie: It seems like it's a lifetime away. I've told everybody there's part of me that believes I knew this was coming. I don't know if it's just concern for a child normally or something more than that, but I've prepared for this for years because I believe the day he was born, the day of his death was pre-planned.
Bob: I wouldn't have believed it was this bad to go through something like this. You keep blaming yourself and wonder what you did right, and your mind is racing with what went wrong.
Lorie: But if we had to change one thing, there's nothing we could change. For a teenager, he was good.
Bob: We just feel it's a big dream and everything's going to go away and be back to normal.
CreditonSouth Huron District High School
The Author Casey Lessard
Deep trouble: Cyprus, 1970
The trip to Egypt: Christmas 1956
Blues take Dashwood teen to Ottawa
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text:
\section{Introduction}
\subsection{Hall algebras for quivers with potential}
Let $Q=(I,E)$ be a symmetric quiver, let $W$ be a potential of $Q$, and let $d\in\mathbb{N}^I$ be a dimension vector of $Q$.
Denote by $\mathcal{X}(d)$ the stack of representations of $Q$ of dimension $d$ over $\mathbb{C}$ and by $\mathcal{X}(d)_0$ the (derived) zero locus of the regular function \[\text{Tr}\,W:\mathcal{X}(d)\to\mathbb{A}^1_\mathbb{C}.\]
Let $D_{\text{sg}}\left(\mathcal{X}(d)_0\right)$
be the category of singularities. It is equivalent to the category of (coherent) matrix factorizations $\text{MF}(\mathcal{X}(d), W)$ for the regular function $\text{Tr}\,W$.
The categorical Hall algebra is the category $\bigoplus_{d\in\mathbb{N}^I} D_{\text{sg}}\left(\mathcal{X}(d)_0\right)$ with multiplication
\begin{equation}\label{productHA}
m_{d,e}:=p_{d,e*}q_{d,e}^*: D_{\text{sg}}\left(\mathcal{X}(d)_0\right)\otimes D_{\text{sg}}\left(\mathcal{X}(e)_0\right)\to D_{\text{sg}}\left(\mathcal{X}(d+e)_0\right),
\end{equation}
where $q_{d,e}:\mathcal{X}(d,e)\to \mathcal{X}(d)\times \mathcal{X}(e)$ and $p_{d,e}:\mathcal{X}(d,e)\to\mathcal{X}(d+e)$ are the natural maps from the stack of extensions $\mathcal{X}(d,e)$. The K-theoretic Hall algebra (KHA) of $(Q,W)$ is $K_0$ of the categorical Hall algebra.

The above construction is suggested by Kontsevich--Soibelman in \cite{ks} where they studied its cohomological version, see \cite{P} for more details about the categorical and K-theoretic constructions.

\subsection{Preprojective Hall algebras}
For $Q$ a quiver, let $\mathfrak{P}(d)$
be the (derived) stack of representations of the preprojective algebra of $Q$ of dimension $d\in\mathbb{N}^I$. The category $\bigoplus_{d\in\mathbb{N}^I} D^b\left(\mathfrak{P}(d)\right)$
admits a Hall-type product \cite{VV}. The algebra obtained by taking $K_0$ of this category is called the  preprojective K-theoretic Hall algebra (KHA) of $Q$. It was studied by Schiffmann--Vasserot \cite{sv1}, Varagnolo--Vasserot \cite{VV}, Yang--Zhao \cite{yz2} and it was seen to be related to positive parts of quantum affine algebras. Conjecturally, an equivariant version of the preprojective KHA is isomorphic to the positive part of the Okounkov--Smirnov quantum affine algebra \cite{os}. It is thus natural to ask whether one can construct a full quantum group from a KHA.

For $Q$ a quiver,
there is a tripled quiver with potential $\left(\widetilde{Q}, \widetilde{W}\right)$ whose Hall algebra is equivalent, via the Köszul equivalence/ dimensional reduction for categories of singularities \cite{I}, to the preprojective Hall algebra of $Q$.

\subsection{The coproduct}\label{coproduct}

For $d\in\mathbb{N}^I$, write $\mathcal{X}(d)=R(d)/G(d)$.
For $\varepsilon\in E$, let $\mathbb{C}^*$ act on $R(d)$ by scalar multiplication of the linear map corresponding to $\varepsilon$.
Let $T\subset \left(\mathbb{C}^*\right)^E$ be a torus such that $W$ is $T$-invariant.  Denote by $q_\varepsilon$ the $T$-weight corresponding to $\varepsilon\in E$.

Let $\text{MF}_{\text{qcoh}, T}\left(\mathcal{X}(d)\times \mathcal{X}(e), W\right)$ be the category of quasi-coherent matrix factorizations \cite[Section 2.2]{T}, let $\lambda$ be a cocharacter of $G(d)\times G(e)$ which acts with non-negative weights on $R(d,e)$ and has fixed locus $R(d)\times R(e)$, and
let \[\text{MF}_{\text{qcoh}, T}\left(\mathcal{X}(d)\times\mathcal{X}(e), W\right)_{\text{above}}\subset \text{MF}_{\text{qcoh}, T}\left(\mathcal{X}(d)\times\mathcal{X}(e), W\right)\] be the subcategory of quasi-coherent matrix factorizations
with $\lambda$-weights bounded above and with objects whose $\lambda$-weight $w$ components are in $\text{MF}_T(\mathcal{X}(d), W)_w$.
Consider the functor:
\[\Delta'_{d,e}:=q_{d,e*}p_{d,e}^*: \text{MF}_T\left(\mathcal{X}(d+e), W\right)\to
\text{MF}_{\text{qcoh}, T}\left(\mathcal{X}(d)\times\mathcal{X}(e), W\right)_{\text{above}}.\]

Let $K_0^{T\times T(d)}(\text{pt})=K_0^T(\text{pt})\left[z_{ij}^{\pm 1}\Big|\,i\in I, 1\leq j\leq d_i\right]$.
Consider the set $\mathcal{I}_{d,e}$ of $K_0^{T\times T(d)\times T(e)}(\text{pt})$ with functions $1-q^{-1}_\varepsilon z_{ij}^{-1}z_{i'j'}$ for $i,i'\in I$,
$j>d_i$, $j'\leq d_{i'},$ and $q_\varepsilon$ is a weight of $T$ corresponding to an edge $\varepsilon$ from $i$ to $i'$.
The functor $\Delta'_{d,e}$ induces a map
\begin{equation}\label{cp}
\Delta'_{d,e}: K_0^T\left(\text{MF}(\mathcal{X}(d+e), W)\right)\to K_0^T\left(\text{MF}(\mathcal{X}(d)\times \mathcal{X}(e), W)\right)_{\mathcal{I}_{d,e}}.
\end{equation}
We make the following \textbf{Künneth Assumption} on $(Q,W)$ and $T$:
\begin{equation}\label{assumption}
K_0^T\left(D_{\text{sg}}(\mathcal{X}(d)_0)\right)\otimes K_0^T\left(D_{\text{sg}}(\mathcal{X}(e)_0)\right)\cong K_0^{T\times T}\big(D_{\text{sg}}\left((\mathcal{X}(d)\times\mathcal{X}(e))_0\right)\big)
\end{equation}
for any dimension vectors $d$ and $e$;
see in this direction the Thom-Sebastiani theorem \cite{Pr}.
The pair $(Q,0)$ and any torus $T$ satisfy the Künneth Assumption. The tripled quiver $\left(\widetilde{Q}, \widetilde{W}\right)$ and $T\subset (\mathbb{C}^*)^{E}\times\mathbb{C}^*_q$, see Subsection \ref{trqu}, satisfy the Künneth Assumption.

The functor $\Delta'_{d,e}$ (together with the equivalence $\text{MF}\cong D_{\text{sg}}$) induces a map into a completion, see Subsections \ref{comple} and \ref{cocompl} for more details:
\[\Delta'_{d,e}: K_0^T\left(D_{\text{sg}}(\mathcal{X}(d+e)_0)\right)\to K_0^T\left(D_{\text{sg}}(\mathcal{X}(d)_0)\right)\hat{\otimes} K_0^T\left(D_{\text{sg}}(\mathcal{X}(e)_0)\right).\]

\subsection{The extended algebra}
Let $\textbf{A}^{\geq}_T$ be the extended Hall algebra generated by $\text{KHA}_T(Q,W)$ and elements $h^+_{i,n}$ for $i\in I$, $n\geq 0$ with the relations \eqref{rel}.
Denote by $\textbf{A}^{\geq}_T(d)$ the subspace generated by $K_0^T\left(D_{\text{sg}}(\mathcal{X}(d)_0)\right)$ and $h^+_{i,n}$. Consider the formal series $h_i(w):=\sum_{n\geq 0} h^+_{i,n}w^{-n}$.
Define the coproduct:
\begin{align*}
\Delta_{d,e}: \textbf{A}^{\geq}_T(d+e)&\to \textbf{A}^{\geq}_T(d)\hat{\otimes} \textbf{A}^{\geq}_T(e)\\
x&\mapsto \left(\prod_{\substack{i\in I\\ j>d_i}}
h_i(z_{ij})\right)* \Delta'_{d,e}(x).
\end{align*}

Consider also the algebra $\textbf{A}^{\leq}_T$ generated by  $\text{KHA}_T(Q,W)^{\text{op}}$ and generators $h^-_{i,n}$ for $i\in I$, $n\geq 0$, see Subsection \ref{exal}, which also admits a coproduct $\Delta$.

\begin{thm}\label{thm1}
Let $(Q,W)$ be a symmetric quiver with potential. Let $T$ be a torus such that $W$ is $T$-invariant and such that $(Q,W)$ and $T$ satisfy the Künneth Assumption \eqref{assumption}. Then
$\left(\textbf{A}^{\geq}_T, m, \Delta\right)$ and $\left(\textbf{A}^{\leq}_T, m, \Delta\right)$ are bialgebras.
\end{thm}

One can also state a result only involving the $\text{KHA}_T$, but then one needs a twist of the multiplication $m\boxtimes m$ in order to formulate the compatibility between $m$ and $\Delta$, see \cite{d} for the  analogous theorem for CoHA and for a definition of the twist in cohomology.

When $Q$ is the Jordan quiver, $W$ is zero, and $T=\mathbb{C}^*$, then $\textbf{A}^{\geq}_T\cong U_q(L\mathfrak{b})$, where $\mathfrak{b}$ is the Borel subalgebra of $\mathfrak{sl}_2$.
For the tripled quiver $\left(\widetilde{Q},\widetilde{W}\right)$
associated to the Jordan quiver and $T=\left(\mathbb{C}^*\right)^2$, $\textbf{A}_T^{\geq}$ has a subalgebra isomorphic to $U_{q, t}^{\geq}\left(\widehat{\widehat{\mathfrak{gl}_1}}\right)$ with the product and coproduct constructed in \cite{n1}.

\subsection{The Hopf pairing}
We assume now that we are in the case of a tripled quiver $\left( \widetilde{Q}, \widetilde{W}\right)$. The construction also works for other pairs such as $(Q,0)$, but not for all pairs $(Q,W)$ satisfying the Künneth Assumption.

Assume that $T$ is chosen as in \eqref{torusT}. Then $\widetilde{R(d)}^T$ is a point. Let $\mathbf{k}$ be the fraction field of $K_0(BT)$. In Subsection \ref{pair}, we define a pairing
\[(\,,\,): K_0^T\left(D_{\text{sg}}\left(\widetilde{\mathcal{X}(d)}_0\right)\right)\otimes K_0^T\left(D_{\text{sg}}\left(\widetilde{\mathcal{X}(d)}_0\right)\right)
\to
\mathbf{k}.\] The definition is a direct generalization of the pairing used by Negu\c{t} in \cite[Exercise IV.2]{n2} for the cyclic type $A$ quiver.
The pairing $(\,,\,)$ is extended naturally to $\textbf{A}^{\leq}_T\otimes \textbf{A}^{\geq}_T$.

\begin{thm}\label{thm2}
The pairing $(\,,\,)$ induces a non-degenerate pairing
\[(\,,\,): \textbf{A}^{\leq}_T\otimes \textbf{A}^{\geq}_T
\to
\mathbf{k}.\]
Let $x,x'\in\textbf{A}^{\leq}_T$, $y,y'\in\textbf{A}^{\geq}_T$.
Then
\begin{align*}
    (x*x',y)&=(x\otimes x', \Delta(y))\\
    (x, y*y')&=(\Delta^{\text{op}}(x), y\otimes y').
\end{align*}
\end{thm}

Negu\c{t} proved versions of Theorems \ref{thm1} and \ref{thm2}  for the preprojective KHA of the Jordan quiver \cite{n1} and of the cyclic quiver of type $A$ \cite{n2}.
Yang--Zhao \cite{yz1} proved versions of Theorems \ref{thm1} and \ref{thm2} for spherical subalgebra of preprojective Hall algebras.

Analogous to the constructions of Negu\c{t} in \cite[pages 88-89]{n2}, there are antipode maps $S:\textbf{A}^{\geq}_T\to \textbf{A}^{\geq}_T$, $S:\textbf{A}^{\leq}_T\to \textbf{A}^{\leq}_T$ which make $\textbf{A}^{\geq}_T$ and $\textbf{A}^{\leq}_T$ into Hopf algebras.
One can thus construct a Hopf algebra $\textbf{A}_T$ as a
Drinfeld double \cite[Subsection 1.4]{n2}.

It is interesting to see whether the representations constructed in \cite[Section 4]{P} can be extended to representations of $\textbf{A}_T$. In the case of (the tripled quiver associated to) the the Jordan quiver, we expect some of those representations to be related to work of Schiffmann--Vaserot \cite{sv1} and Feigin--Tsymbaliuk \cite{ft} who constructed actions of $U_{q, t}\left(\widehat{\widehat{\mathfrak{gl}_1}}\right)$ on \[\bigoplus_{d\geq 0} K_0^T\left(\text{Hilb}(\mathbb{A}^2_{\mathbb{C}}, d)\right).\] It is also interesting to see whether there are natural categorifications of the double algebras $\textbf{A}_T$.

\subsection{Plan of the paper} In Section \ref{s2}, we discuss definitions and preliminary results. In Section \ref{s3}, we prove shuffle type formulas for $m$ and $\Delta$ and we use them to prove Theorem \ref{thm1}. We next show that the KHA for tripled quivers $\left(\widetilde{Q}, \widetilde{W}\right)$ satisfies the Künneth Assumption. In Section \ref{s4} we define the Hopf pairing $(\,,\,)$ and prove Theorem \ref{thm2} using that $K_0^T\left(D_{\text{sg}}\left(\widetilde{\mathcal{X}(d)}_0\right)\right)$ has no $K_0^{T\times T(d)}(\text{pt})$-torsion. We end Section \ref{s4} with some explicit examples.

\subsection{Acknowledgements.} I thank the Institute of Advanced Studies for support during the preparation of the paper. This material is based upon work supported by the National Science Foundation under Grant No. DMS-1926686.

\section{Preliminaries}\label{s2}

\subsection{Notations}
All stacks considered in the paper are defined over $\mathbb{C}$.

For a stack $\mathcal{X}=X/G$ where $G$ is a reductive group acting on a (possibly quasi-smooth) quasi-projective scheme $X$, denote by $\text{QCoh}(\mathcal{X})$ the unbounded derived category of quasi-coherent sheaves on $\mathcal{X}$, by $D^b\text{Coh}(\mathcal{X})$ the derived category of bounded complexes of coherent sheaves, and by $\text{Perf}(\mathcal{X})$ the subcategory of $D^b\text{Coh}(\mathcal{X})$ of perfect complexes. All the functors used in the paper are derived and we drop $R$ and $L$ from their notations.
Denote by
\begin{align*}
    G_0(\mathcal{X})&:=K_0\left(D^b\text{Coh}(\mathcal{X})\right)_{\mathbb{Q}},\\
    K_0(\mathcal{X})&:=K_0\big(\text{Perf}(\mathcal{X})\big)_{\mathbb{Q}}.
\end{align*}
We denote by $H^{\text{BM}}_{\cdot}(\mathcal{X})$ the Borel-Moore homology of $\mathcal{X}$ with rational coefficients.
For $\mathcal{X}$ a quasi-smooth stack, denote by $\mathcal{X}^{\text{cl}}$ its classical stack. For the stacks considered in this paper, there is an isomorphism $G_0(\mathcal{X})\cong G_0\left(\mathcal{X}^{\text{cl}}\right)$, see \cite[Equation 2.2]{VV}. We will be using the notations
\begin{align*}
    K_0^T\big(D_{\text{sg}}(\mathcal{X}(d)_0)\big)&:=K_0\big(D_{\text{sg},T}(\mathcal{X}(d)_0)\big)_{\mathbb{Q}}.\\
    K_0^T\left(\text{MF}(\mathcal{X}(d), W)\right)&:=K_0\left(\text{MF}_T(\mathcal{X}(d), W)\right)_{\mathbb{Q}}.
\end{align*}
For a regular immersion $\iota:\mathcal{X}\hookrightarrow \mathcal{X}'$, denote by $N_\iota$ the normal bundle of $\mathcal{X}$ in $\mathcal{X}'$.

All the potential appearing in the paper are induced from $\text{Tr}\,W$; we slightly abuse notation and write $W$ without mentioning the dimension vector.
We use the notations $m$ and $*$ for the product of the KHA.
All the tori $T$ which appear as equivariant parameters of the $\text{KHA}$ satisfy
\begin{equation}\label{torus3}
T\subset \left(\mathbb{C}^*\right)^E\text{ and }W\text{ is }T\text{-invariant}.
\end{equation}
We also use in some situations an extra assumption \eqref{torusT} on tori $T$.

\subsection{Categories of singularities.}

A reference for this subsection is \cite[Subsection 2.2]{T}.
The construction of Hall algebras for quivers with potential involves categories of singularities \[D_{\text{sg}}(\mathcal{X}(d)_0):=D^b\text{Coh}\left(\mathcal{X}(d)_0\right)\big/\text{Perf}\left(\mathcal{X}(d)_0\right).\] Here, $\mathcal{X}(d)_0$ is the (derived) zero locus of $\text{Tr}\,W: \mathcal{X}(d)\to\mathbb{A}^1_\mathbb{C}$. We have an exact sequence
\[K_0(\mathcal{X}(d)_0)\to G_0(\mathcal{X}(d)_0)\to K_0\left(D_{\text{sg}}(\mathcal{X}(d)_0)\right)\to 0.\]

$D_{\text{sg}}$ is equivalent to the category of matrix factorizations $\text{MF}\left(\mathcal{X}(d), W\right)$. The objects of $\text{MF}$ are $(\mathbb{Z}/2\mathbb{Z})\times T\times G(d)$-equivariant factorizations $(P, d_P)$, where $P$ is a $T\times G(d)$-equivariant coherent sheaf, $\langle 1\rangle$ is the twist corresponding to a non-trivial $\mathbb{Z}/2\mathbb{Z}$-character on $R(d)$, and \[d_P: P\to P\langle 1\rangle\] satisfies $d_P\circ d_P=f$. There are natural pullback and proper pushforward functors for $\text{MF}$ induced by the natural pullback and proper pushforward functors on the ambient smooth stacks. We will freely switch between $D_{\text{sg}}$ and $\text{MF}$ throughout this paper.

Assume there is an extra $\mathbb{C}^*$-action corresponding to a subgroup of $\left(\mathbb{C}^*\right)^E$ such that $\text{Tr}\,W$ is of weight $2$.
Consider the corresponding category of graded matrix factorizations $\text{MF}^{\text{gr}}(\mathcal{X}(d), W)$.

For graded version of categories of singularities and for the quasi-coherent version of matrix factorizations, see \cite[Section 2.2]{T}; the only use of quasi-coherent matrix factorizations is to make sense of the functor $q_*p^*=\Delta'$ from Subsection \ref{coproduct}.

\subsection{Quivers}

\subsubsection{}
Let $Q=(I,E)$ be a quiver and let $d\in\mathbb{N}^I$. We denote by $\mathcal{X}(d)=R(d)/G(d)$ the stack of representations of $Q$ of dimension $d$. Fix maximal torus and Borel subgroups $T(d)\subset B(d)\subset G(d)$. We use the convention that the weights of the Lie algebra of $B(d)$ are negative; it determines a dominant chamber of weights of $G(d)$. For dimension vectors $a$ and $b$ with sum $d$, let $\lambda:\mathbb{C}^*\to G(d)$ be a fixed antidominant cocharacter corresponding to the partition $(a, b)$ of $d$. Consider the diagram of attracting loci for $\lambda$:
\[\mathcal{X}(a)\times\mathcal{X}(d)\cong\mathcal{X}(d)^\lambda\xleftarrow{q_{a,b}}\mathcal{X}(d)^{\lambda\geq 0}=\mathcal{X}(a,b)\xrightarrow{p_{a,b}}\mathcal{X}(d).\]
We may use the notations $p_\lambda, q_\lambda$ or $p, q$ depending on the context.

Let $W$ be a potential of $Q$ and let $T$ be a torus satisfying \eqref{torus3}. Then the $K$-theoretic Hall algebra of $(Q,W)$ and $T$ is
\[\text{KHA}_T(Q,W):=\bigoplus_{d\in\mathbb{N}^I}K_0^T\big(D_{\text{sg}}(\mathcal{X}(d)_0)\big).\]
The multiplication on Hall algebra \eqref{productHA} is induced by the functor
$p_{a,b*}q^*_{a,b}$ on the corresponding categories of singularities.

\subsubsection{}\label{localtorus}
Let $d\in\mathbb{N}^I$ and let $T$ be a torus satisfying \eqref{torus3}. Let $\mathfrak{S}_d=\times_{i\in I}\mathfrak{S}_{d_i}$ be the Weyl group of $G(d)$. Consider the stacks and maps \[\mathcal{Y}(d):=R(d)/T(d)\xleftarrow{q'}\mathcal{Z}(d):=R(d)/B(d)\xrightarrow{p'}\mathcal{X}(d).\] There is a natural $\mathfrak{S}_d$-action on $K_0^T\big(D_{\text{sg}}(\mathcal{Y}(d)_0)\big)$.
The map $q'$ is a smooth affine bundle map and $p'$ is a smooth proper map with fiber $G(d)/B(d)$. Using the localization theorem in $K$-theory, see for example \cite[Subsection 2.12]{P}, the map \[p'_*q'^*:  K_0^T(D_{\text{sg}}(\mathcal{Y}(d)_0))\to K_0^T(D_{\text{sg}}(\mathcal{X}(d)_0))\] has the formula
\begin{equation}\label{borel}
    p'_*q'^*(y)=\sum_{w\in\mathfrak{S}_d}w\left(\frac{y}{\prod_{\substack{i\in I\\ j<k}}\left(1-z_{ij}^{-1}z_{ik}\right)}\right).\end{equation}
The map $a:\mathcal{Y}(d)\to \mathcal{X}(d)$, base change of $BT(d)\to BG(d)$, induces an injection
\[a^*: K_0^T(D_{\text{sg}}(\mathcal{X}(d)_0))\hookrightarrow K_0^T\left(D_{\text{sg}}\left(\mathcal{Y}(d)_0\right)\right),\]
see \cite[Proposition 2.3]{P}.

\subsubsection{}\label{Phi}

Let $d\in\mathbb{N}^I$ and let $\lambda$ be a cocharacter of $G(d)$. Consider the natural maps
\[\mathcal{Y}(d)^\lambda:=R(d)^\lambda/T(d)\xrightarrow{s_\lambda} \mathcal{Y}(d)^{\lambda\geq 0}:=R(d)^{\lambda\geq 0}/T(d)\xrightarrow{p_\lambda} \mathcal{Y}(d)=R(d)/T(d).\]
We abuse notation and write $p_\lambda$ for the stacks $\mathcal{Y}$.
Define the map
\[\Phi_\lambda:=s_\lambda^*p_\lambda^*:K_0^T\left(D_{\text{sg}}(\mathcal{Y}(d)_0)\right)\to  K_0^T\left(D_{\text{sg}}\left(\mathcal{Y}(d)^\lambda_0\right)\right).\]
There are analogous maps denoted by the same letters
\[\mathcal{X}(d)^\lambda\xrightarrow{s_\lambda}\mathcal{X}(d)^{\lambda\geq 0}\xrightarrow{p_\lambda}\mathcal{X}(d).\]
The following diagram commutes:
\begin{equation}\label{diagram}
    \begin{tikzcd}
    K_0^T\left(D_{\text{sg}}(\mathcal{X}(d)_0)\right)\arrow[d, hook]\arrow[r,"\Phi_\lambda"]&
    K_0^T\left(D_{\text{sg}}(\mathcal{X}(d)^\lambda_0)\right)\arrow[d, hook]\\
    K_0^T\left(D_{\text{sg}}(\mathcal{Y}(d)_0)\right)\arrow[r,"\Phi_\lambda"]&K_0^T\left(D_{\text{sg}}(\mathcal{Y}(d)^\lambda_0)\right).
    \end{tikzcd}
\end{equation}

If $\lambda$ is the antidominant cocharacter corresponding to the partition $(a,b)$ of $d$, we may write $\Phi_{a,b}$ instead of $\Phi_\lambda$. Further, if $\lambda$ acts with weight $w$ on coordinates $z_{ij}$ parametrized by $A=(A_i)_{i\in I}$ for $A_i\subset \{1,\cdots, d_i\}$ and with a different weight $v\neq w$ on coordinates parametrized by its complement $B=(B_i)_{i\in I}$, we use the notation $\Phi_{A,B}$.

\subsection{The Köszul equivalence}\label{isikequi}
Let $X$ be a smooth affine scheme with an action of a reductive group $G$ and let $E$ be a $G$-equivariant vector bundle on $X$. Let $\mathbb{C}^*$ act on the fibers of $E$ with weight $2$ and consider $s\in \Gamma(X, E)$ a section of $E$ of weight $2$.
It induces a map $\partial: E^{\vee}\to \mathcal{O}_X$.
Define the $G$-equivariant regular function \[w:Y:=\text{Tot}_X\left(E^{\vee}\right)\to\mathbb{A}^1_\mathbb{C}\] by the formula
$w(x,v)=\langle s(x), v \rangle$ for $x\in X(\mathbb{C})$ and $v\in E^{\vee}|_x$.
Consider the Köszul stack
\[\mathfrak{K}:=\text{Spec}\left(\mathcal{O}_X\left[E^{\vee}[1];\partial\right]\right)\big/G.\]
Consider the category of graded matrix factorizations $\text{MF}^{\text{gr}}\left(Y/G, w\right)$ with respect to the group $\mathbb{C}^*$ mentioned above. The Köszul equivalence due to Isik \cite{I} is
\begin{equation}\label{koszu}
    \text{MF}^{\text{gr}}\left(Y/G, w\right)\cong D^b(\mathfrak{K}).
    \end{equation}
By \cite[Corollary 3.13]{T5} and \eqref{koszu}, we have that
\begin{equation}\label{grnotgr}
    K_0\left(\text{MF}(Y/G, w)\right)\cong
    K_0\left(\text{MF}^{\text{gr}}(Y/G, w)\right)\cong G_0\left(\mathfrak{K}\right).
\end{equation}

\subsection{The tripled quiver}
\label{trqu}
Let $Q=(I,E)$ be a quiver with vertices set $I$ and edges set $E$. Let $s,t:E\to I$ be the source and target maps. For $e\in E$, denote by $\overline{e}$ the edge with opposite orientation and let $\overline{E}=\{\overline{e}|\,e\in E\}$. For $i\in I$, let $\omega_i$ be a loop at $i$. Let $Q^d=(I,E^d)$ be the doubled quiver, where $E^d=E\sqcup \overline{E}$. Consider the tripled quiver $\widetilde{Q}=\left(I, \widetilde{E}\right)$, where $\widetilde{E}=\{f, \omega_i|\,f\in E^d, i\in I\}$, with potential $\widetilde{W}:=\sum_{e\in E}\omega_{s(e)}\,
[e, \overline{e}]$.
\\

For each $e\in E(Q)$, let $\mathbb{C}^*$ act by multiplication with weight $1$ (by $q_e$) on the linear map corresponding to $e$ and by multiplication with weight $-1$ (by $q_e^{-1}$) on the linear map corresponding to $\overline{e}$.
Let $\mathbb{C}^*_q$ act by $q$ on linear maps corresponding to edges in $\overline{Q}$ and by $q^{-2}$ on the loops $\omega_i$ for $i\in I$. Denote by $\mathbb{C}^*_t\subset \left(\mathbb{C}^*\right)^E$ the group which acts with weight $t$ on the edges on $Q$ and with weight $t^{-1}$ on the edges on $\overline{Q}$.
The KHAs associated to $\left(\widetilde{Q}, \widetilde{W}\right)$ which we are able to extend to Hopf algebras in this paper are equivariant with respect to a torus $T$ such that \begin{equation}\label{torusT}
    \mathbb{C}^*_t\times\mathbb{C}^*_q\subset T\subset (\mathbb{C}^*)^{E}\times\mathbb{C}^*_q.\end{equation}

For $d\in\mathbb{N}^I$, let $\widetilde{\mathcal{X}(d)}$ be the stack of representations of $\widetilde{Q}$ of dimension $d$ and let $\mathfrak{P}(d)$ be the (derived) stack of representations of the preprojective algebra of $Q$ of dimension $d$ \cite{VV}, \cite[Subsection 3.2]{P}. By the Köszul equivalence
from Subsection \ref{isikequi}, see also loc. cit.: \begin{equation}\label{preequi}
\text{MF}^{\text{gr}}_{T}\left(\widetilde{\mathcal{X}(d)}, \widetilde{W}\right)\cong D^b_T\left(\mathfrak{P}(d)\right).\end{equation}
The preprojective categorical Hall algebra of $Q$ is the monoidal category
\[\text{HA}_T(Q):=\bigoplus_{d\in\mathbb{N}^I}D^b_T\left(\mathfrak{P}(d)\right),\]
see \cite{VV} for the definition of the product. Its Grothendieck group is called the preprojective KHA of $Q$.
The categories $\text{MF}^{\text{gr}}_{T}\left(\widetilde{\mathcal{X}(d)}, \widetilde{W}\right)$ and $\text{MF}_{T}\left(\widetilde{\mathcal{X}(d)}, \widetilde{W}\right)$ have the same Grothendieck group, see \cite[Corollary 3.13]{T5}, and thus \[\text{KHA}_T\left(\widetilde{Q}, \widetilde{W}\right)\cong\text{KHA}_T(Q)\]
as vector spaces. The two algebras have the same product up to conjugation by an explicit equivariant element, see \cite[Subsection 3.2.3]{P}, \cite[Subsection 2.3.7]{VV}.

\subsection{The coproduct for KHA}\label{comple}
We explain why $\Delta_{d,e}'$ defined in Subsection \ref{coproduct} induces the map \eqref{cp}. In the setting of Subsection \ref{coproduct},
there is an orthogonal decomposition
\[
\text{MF}_{\text{qcoh}, T}\left(\mathcal{X}(d)\times\mathcal{X}(e), W\right)\cong\bigoplus_{w\in\mathbb{Z}}\text{MF}_{\text{qcoh}, T}\left(\mathcal{X}(d)\times\mathcal{X}(e), W\right)_w,
\]
where the subscript $w$ denotes the category of matrix factorizations on which $\lambda$ acts with weight $w$. The analogous decomposition holds for coherent matrix factorizations.
There are thus functors \[\beta_w:\text{MF}_{\text{qcoh}, T}\left(\mathcal{X}(d)\times\mathcal{X}(e), W\right)_{\text{above}}\to \text{MF}_T\left(\mathcal{X}(d)\times\mathcal{X}(e), W\right)_{w}\] which associate to a matrix factorization $\mathcal{F}$ its $w$-weight component with respect to $\lambda$.
Let $\mathcal{F}\in \text{MF}_T\left(\mathcal{X}(d+e), W\right)$. The formula for $\Delta'_{d,e}$ is:
\[\Delta'_{d,e}(\mathcal{F})=\sum_{w\in \mathbb{Z}} \left[\beta_w q_*p^*(\mathcal{F})\right]\in K_0\Big(\text{MF}_{\text{qcoh}, T}\left(\mathcal{X}(d)\times\mathcal{X}(e), W\right)_{\text{above}}\Big).\]
Then $\Delta'_{d,e}(\mathcal{F})$ is in $K_0\Big(\text{MF}_T\left(\mathcal{X}(d)\times\mathcal{X}(e), W\right)\Big)[\![q_\varepsilon^{-1}z_{i'j'}^{-1}z_{ij}]\!]$, where the monomials in the power series have $j\leq d_i$, $j'>d_{i'}$, and $q_\varepsilon$ is a weight of $T$ associated to an edge $\varepsilon$ from $i'$ to $i$.
Proposition \ref{p2}, part (b) says that this sum is the expansion of a rational function in $K_0\big(\text{MF}_T(\mathcal{X}(d)\times \mathcal{X}(e), W)\big)_{\mathcal{I}_{d,e}}$ in the region $|z_{ij}|\ll |z_{i'j'}|$ for $j\leq d_i$, $j'>d_{i'}$.

\subsection{Cycle maps for quotient stacks}\label{chernstacks}

Let $X$ be a scheme with an action of a reductive group $G$ and assume that $\mathcal{X}=X/G$ is of pure dimension $d$. Denote by $I_G$ the augmentation ideal of $K_0(BG)$ and denote by $\widehat{G_0}(\mathcal{X})$ the $I_G$-adic completion of the $K_0(BG)$-module $G_0(\mathcal{X})$.
Let \[
\widehat{H^{\text{BM}}_\cdot}(\mathcal{X}):=\prod_{a\geq 0}
H^{\text{BM}}_{2d-a}(\mathcal{X}).\]
Then there is a cycle map
\[G_0(\mathcal{X})\to \widehat{G_0}(\mathcal{X})\to \widehat{H^{\text{BM}}_\cdot}(\mathcal{X}),\]
see \cite{EG}, \cite[Appendix C]{VV}. For $X$ a quasi-smooth scheme with an action of $G$, let $\mathcal{X}=X/G$ and let $\mathcal{X}^{\text{cl}}$ be its classical stack. There is a cycle map for $\mathcal{X}$:
\[G_0(\mathcal{X})\cong G_0\left(\mathcal{X}^{\text{cl}}\right)\to \widehat{H^{\text{BM}}_\cdot}\left(\mathcal{X}^{\text{cl}}\right)\cong \widehat{H^{\text{BM}}_\cdot}(\mathcal{X}).\]

\section{The product-coproduct compatibility}\label{s3}

\subsection{Shuffle formulas for product and coproduct}\label{sh}

Let $i,i'$ be vertices and let $\{1,\cdots, \varepsilon(i,i')\}$ be the set of edges from $i$ to $i'$. Consider the action of $\left(\mathbb{C}^*\right)^{\varepsilon(i,i')}$ on $R(d)$ whose $j$th copy acts on $R(d)$ with weight $1$ on the factor $\text{Hom}\left(\mathbb{C}^{d_{s(j)}}, \mathbb{C}^{d_{t(j)}}\right)$ corresponding to the edge $j$. Denote by $q_j$ the weight corresponding to the $j$th copy of $\mathbb{C}^*$.
Define
\begin{equation}\label{zeta}
    \zeta_{ii'}(z):= \frac{\left(1-q_1^{-1}z^{-1}\right)\cdots \left(1-q_{\varepsilon(i,i')}^{-1}z^{-1}\right)}{\left(1-z^{-1}\right)^{\delta_{ii'}}}, \end{equation}
where $\delta_{ii'}$ is $1$ if $i=i'$ and $0$ otherwise.
Let $T$ be a subtorus of $\left(\mathbb{C}^*\right)^E$ which fixes $W$. We also use the notation $q_j$ for the corresponding weight of $T$. Recall the definition of $\Phi_{a,b}$ from Subsection \ref{Phi}.

\begin{prop}\label{p2}
Let $a,b\in\mathbb{N}^I$ with sum $d$.

(a) Let $x\in K_0^T(D_{\text{sg}}(\mathcal{X}(a)_0))$ and $y\in K_0^T(D_{\text{sg}}(\mathcal{X}(b)_0))$. Then
\[\Phi_{a,b}\,m_{a,b}(x,y)=\sum_{\mathfrak{S}_{d}/\mathfrak{S}_a\times\mathfrak{S}_b} w\left(xy\prod_{\substack{i,i'\in I\\ j\leq a_i\\ j'>a_{i'}}}\zeta_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)
\right).\]
(b) Assume that $(Q,W)$ and $T$ satisfy the Künneth Assumption \eqref{assumption}. Let $z$ be in $K_0^T\big(D_{\text{sg}}(\mathcal{X}(d)_0)\big)$. Then
\[\Delta'_{a,b}(z)=\frac{\Phi_{a,b}(z)}{\prod_{\substack{i,i'\in I\\ j>a_i\\ j'\leq a_{i'}}}
\zeta_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)}.\]
\end{prop}

\begin{proof} Consider the diagonal action of $T$ on $\mathcal{X}(a)\times \mathcal{X}(b)$. We are using the notations from Subsection \ref{Phi}.

(a) The map $q_{a,b}:\mathcal{X}(a,b)\to\mathcal{X}(a)\times\mathcal{X}(b)$ is an affine bundle map.  Its restriction to the zero locus is also an affine bundle map, so there is a pullback map $q^*_{a,b}$ in $K$ and $G$-theory and it is an isomorphism.
Further, $q^*_{a,b}\left(\text{Tr}\,W_a+\text{Tr}\,W_b\right)=p^*_{a,b}\left(\text{Tr}\,W_{d}\right)$, so there are maps
\begin{align*}
    q^*_{a,b}&: G_0^T\Big(\left(\mathcal{X}(a)\times\mathcal{X}(b)\right)_0\Big)
    \cong G_0^T(\mathcal{X}(a,b)_0)\\
    q^*_{a,b}&: K_0^T\Big(D_{\text{sg}}\left((\mathcal{X}(a)\times\mathcal{X}(b)\right)_0 \Big)
    \cong K_0^T(D_{\text{sg}}(\mathcal{X}(a,b)_0)).
\end{align*}
Consider the factorizations of $p_{a,b}$:
\begin{align*}
&R(a,b)/G(a,b)\xrightarrow{\iota'_{a,b}} R(d)/G(a,b)\xrightarrow{\pi'_{a,b}} R(d)/G(d)\\
&R(a,b)_0/G(a,b)\xrightarrow{\iota_{a,b}} R(d)_0/G(a,b)\xrightarrow{\pi_{a,b}} R(d)_0/G(d).
\end{align*}
We have that $N_{\iota'_{a,b}}\big|_{\mathcal{X}(a,b)_0}=N_{\iota_{a,b}}$. The weights of the bundle $N_{\iota'_{a,b}}^{\vee}$ are of the form $q^{-1}_{\varepsilon}z_{ij}^{-1}z_{i'j'}$, where $j\leq a_i$, $j'>a_{i'}$, and $\varepsilon\in\{1,\cdots,\varepsilon(i,i')\}$ is an edge between $i$ and $i'$. We have that \begin{align*}
    \iota_{a,b}^*\iota_{a,b*}: \text{MF}_T(\mathcal{X}(a,b), W)&\to \text{MF}_T(\mathcal{X}(a,b), W)\\
    (P, d_P)&\mapsto \left(\iota_{a,b}^*\iota_{a,b*}P, \iota_{a,b}^*\iota_{a,b*}d_P\right).
    \end{align*}
    We have that $\mathcal{H}^{-i}\left(\iota_{a,b}^*\iota_{a,b*}P\right)\cong \Lambda^i\left(N_{\iota_{a,b}}^{\vee}\right)\otimes P$ for $i\in\mathbb{Z}$ and the corresponding factorization is $1\otimes d_P$, so
\[\iota_{a,b}^*\iota_{a,b*}: K_0^T\big(\text{MF}(\mathcal{X}(a,b), W)\big)\to K_0^T\big(\text{MF}(\mathcal{X}(a,b), W)\big)\]
has the formula
\begin{equation}\label{1}
\iota_{a,b}^*\iota_{a,b*}(u)=u\prod_{\substack{i,i'\in I\\ j\leq a_i\\ j'>a_{i'}}}\left(1-q_1^{-1} z_{ij}^{-1}z_{i'j'}\right)\cdots \left(1-q_{\varepsilon(i,i')}^{-1} z_{ij}^{-1}z_{i'j'}\right)
\end{equation}
for $u$ in $K_0^T(D_{\text{sg}}(\mathcal{X}(a,b)_0))$.
The map $\pi_{a,b}$ is smooth, so there is a pullback in $G$-theory.
Using the localization theorem, the maps
\begin{align*}
    \pi_{a,b}^*\pi_{a,b*}: G_0^T\left(R(d)_0/G(a,b)\right)&\to G_0^T\left(R(d)_0/G(a,b)\right)\\
    \pi_{a,b}^*\pi_{a,b*}: K_0^T\Big(D_{\text{sg}}(R(d)_0/G(a,b))\Big)&\to K_0^T\Big(D_{\text{sg}}(R(d)_0/G(a,b))\Big)
\end{align*}
have the following formulas, for example for $v$ in $K_0^T\Big(D_{\text{sg}}(R(d)_0/G(a,b))\Big)$:
\begin{equation}\label{2}
\pi_{a,b}^*\pi_{a,b*}(v)=\sum_{\mathfrak{S}_{d}/\mathfrak{S}_a\times\mathfrak{S}_b} w\left(\frac{v}{\prod_{\substack{i\in I\\ j\leq a_i<k}}\left(1-z_{ij}^{-1}z_{ik}\right)}\right).
\end{equation}
Further, $\iota_{a,b}^*$ and the action of $\mathfrak{S}_d$ commute.
The formula for $m_{a,b}$ follows from \eqref{1} and \eqref{2}.

(b) Consider the factorization of $q_{a,b}$:
\[ R(a,b)_0/G(a,b)\xrightarrow{j_{a,b}}R(a,b)_0/G(a)\times G(b)\xrightarrow{r_{a,b}}\left(R(a)\times R(b)\right)_0/G(a)\times G(b).\]
The map $j_{a,b}$ is a section of a natural affine bundle map $\omega_{a,b}$; the map $r_{a,b}$ has a natural section $\tau_{a,b}$.
There are induced maps
\begin{align*}
    j_{a,b*}: G_0^T(\mathcal{X}(a,b)_0)&\to G_0^T\left(R(a,b)_0/G(a)\times G(b)\right)\\
    j_{a,b*}: K_0^T(D_{\text{sg}}(\mathcal{X}(a,b)_0))&\to K_0^T\Big(D_{\text{sg}}(R(a,b)_0/G(a)\times G(b))\Big).
\end{align*}
The map $\omega_{a,b}$ is smooth, so there is a pullback map in $G$-theory.
The map $j_{a,b*}$ has the following formula, for example for $u$ in $G_0^T(\mathcal{X}(a,b)_0)$:
\begin{equation}\label{3}
j_{a,b*}(u)=\omega_{a,b}^*(u)\prod_{\substack{i\in I\\ k\leq a_i<j}}\left(1-z_{ij}^{-1}z_{ik}\right).
\end{equation}
Finally, $r_{a,b}$ is the restriction to the zero locus of the affine bundle map \[r'_{a,b}:R(a,b)\big/G(a)\times G(b)\to R(a)\times R(b)\big/G(a)\times G(b).\]
The pullback $r^*_{a,b}$ exists and is an isomorphism in $K$ and $G$-theory.
Let $\lambda$ be a cocharacter of $G(a)\times G(b)$ which acts with non-negative weights on $R(a,b)$ and has fixed locus $R(a)\times R(b)$. Similarly to the discussion in Subsection \ref{comple},   $r_{a,b*}$ has image in
\begin{multline*}
r_{a,b*}:
K_0^T\Big(D_{\text{sg}}(R(a,b)_0/G(a)\times G(b))\Big)
\to\\
K_0^T\Big(D_{\text{sg}}(\left(R(a)\times R(b)\right)_0/G(a)\times G(b))\Big)[\![q_\varepsilon^{-1}z_{i'j'}^{-1}z_{ij}]\!],
\end{multline*} where the monomials in the power series have $j\leq a_i$, $j'>a_{i'}$.
Let
$N:=R(a,b)^{\lambda>0}$ be a representation of $G(a)\times G(b)$.
For a pair $(\mathcal{F},d)$ in $\text{MF}_T\big(\mathcal{X}(a)\times \mathcal{X}(b), W\big)$, we have that \[r_{a,b*}r_{a,b}^*\left(\mathcal{F}\right)=\mathcal{F}\otimes \mathbb{C}[N^{\vee}]\] and $r_{a,b*}r_{a,b}^*(d)=d\otimes \text{id}$.
The map $\tau_{a,b}^*$ is an inverse to $r_{a,b}^*$.
Consider $v$ in $K_0^T\big(\text{MF}\big(R(a,b)\big/G(a)\times G(b), W\big)\big)$. Then
\begin{equation}\label{4}
r_{a,b*}(v)=\frac{\tau_{a,b}^*(v)}{\prod_{\substack{i,i'\in I\\ j>a_i\\ j'\leq a_{i'}}}\left(1-q_1^{-1}z_{ij}^{-1}z_{i'j'}\right)\cdots\left(1-q_{\varepsilon(i,i')}^{-1}z_{ij}^{-1}z_{i'j'}\right)}.
\end{equation}
Consider the action of $T\times T$ on $\mathcal{X}(a)\times \mathcal{X}(b)$ induced by the actions of $T$ on both factors. By the Künneth Assumption \eqref{assumption},
we have that
\[
K_0^{T\times T}\Big(D_{\text{sg}}((\mathcal{X}(a)\times\mathcal{X}(b))_0)\Big)\cong K_0^T(D_{\text{sg}}(\mathcal{X}(a)_0))\otimes
K_0^T(D_{\text{sg}}(\mathcal{X}(b)_0)).\]
The map $r_{a,b*}$ has image in
$K_0^T\left(D_{\text{sg}}(\mathcal{X}(a)_0)\right)\hat{\otimes} K_0^T\left(D_{\text{sg}}(\mathcal{X}(b)_0)\right),$
where the completion on the right hand side is defined in Subsection \ref{comple}, see also Subsection \ref{cocompl}.
The formula for $\Delta'_{a,b}$ follows by combining \eqref{3}, \eqref{4}, and $\Phi_{a,b}=p^*_{a,b}\omega^*_{a,b}\tau^*_{a,b}$.
\end{proof}

\subsection{The extensions of the Hall algebra}
\label{exal}
Let $\textbf{A}^{\geq}_T$ be the extended Hall algebra generated by $\text{KHA}_T(Q,W)$ and generators $h^+_{i,n}$ for $i\in I$, $n\geq 0$. Consider the formal power series $h^+_i(w):=\sum_{n\geq 0} h_{i,n}^+w^{-n}$. Define
\[\tau_{ii'}(z):=\frac{\zeta_{i'i}(z)}{\zeta_{ii'}(z^{-1})}.\]

We impose the relations
\begin{equation}\label{rel}
    y*h^+_{i}(w)=h^+_{i}(w)*\left(y\prod_{\substack{i'\in I\\ 1\leq j'\leq d_{i'}}}\tau_{ii'}\left(\frac{z_{i'j'}}{w}\right)\right)
\end{equation}
for all $y$ in $K_0^T(D_{\text{sg}}(\mathcal{X}(d)_0))$ and $i\in I$.

Similarly, define $\textbf{A}^{\leq}_T$ as the extended Hall algebra generated by $\text{KHA}_T(Q,W)^{\text{op}}$ and generators $h^-_{i,n}$ for $i\in I$, $n\geq 0$. Let $h^-_i(w):=\sum_{n\geq 0} h^-_{i,n}w^{n}$. Impose the relations
\begin{equation}\label{rel2}
    y*h^-_{i}(w)=h^-_{i}(w)*\left(y\prod_{\substack{i'\in I\\ 1\leq j'\leq d_{i'}}}\tau_{ii'}\left(\frac{w}{z_{i'j'}}\right)\right)
\end{equation}
for all $y$ in $K_0^T(D_{\text{sg}}(\mathcal{X}(d)_0))$ and $i\in I$.

\subsection{The coproduct}

Let $a, b\in\mathbb{N}^I$ have sum $d$. Let $y$ be in $K_0^T(D_{\text{sg}}(\mathcal{X}(d)_0))$.

\subsubsection{}
Define the coproduct for $\textbf{A}^{\geq}_T$:
\begin{align*}
\Delta\left(h^+_i(w)\right)&=h^+_i(w)\otimes h^+_i(w)\\
    \Delta_{a,b}(y)&=\prod_{\substack{i\in I\\ j>a_i}}h^+_i(z_{ij})*\Delta'_{a,b}(y)\\
    \Delta(y)&=\sum_{a,b\in\mathbb{N}^I}\Delta_{a,b}(y).
\end{align*}

\subsubsection{}
We next define the coproduct for $\textbf{A}^{\leq}_T$.
Define
\[\Delta'_{a,b}(y)=\text{sw}_{b,a} q_{b,a*}p_{b,a}^*(y),\]
where $\text{sw}_{b,a}$ is the map that interchanges the factors:
\[\text{sw}_{b,a}: K_0^T\left(D_{\text{sg}}(\mathcal{X}(b)_0)\right)\otimes K_0^T\left(D_{\text{sg}}(\mathcal{X}(a)_0)\right)\cong K_0^T\left(D_{\text{sg}}(\mathcal{X}(a)_0)\right)\otimes K_0^T\left(D_{\text{sg}}(\mathcal{X}(b)_0)\right).\]
$y$ is $\mathfrak{S}_{d}$-symmetric. A similar computation to Proposition \ref{p2}, part (b) shows that:
\begin{equation}\label{coprneg}
\Delta'_{a,b}(y)=\frac{\Phi_{a,b}(y)}{\prod_{\substack{i,i'\in I\\ j\leq a_i\\ j'>a_{i'}}}
\zeta_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)}.\end{equation}
We define the coproduct for $\textbf{A}^{\leq}_T$:
\begin{align*}
 \Delta\left(h^-_i(w)\right)&=h^-_i(w)\otimes h^-_i(w)\\
    \Delta_{a,b}(y)&=\Delta'_{a,b}(y)*\prod_{\substack{i\in I\\ j\leq a_i}}h^-_i(z_{ij})\\
    \Delta(y)&=\sum_{a,b\in\mathbb{N}^I}\Delta_{a,b}(y).
\end{align*}

\subsubsection{}\label{cocompl}
To make sense of the coproduct formulas, expand in the region $|z_{ij}|\ll |z_{i'j'}|$ for $j\leq a_i$, $j'>a_{i'}$, place all functions in $z_{ij}$ for $j\leq a_i$ to the left of $\otimes$ and all functions in $z_{ij}$ for $j>a_i$ to the right of $\otimes$. The coproduct has infinitely many terms, but each homogeneous $d$ part is in $\textbf{A}^{\geq}_T(d)$ or $\textbf{A}^{\leq}_T(d)$.

\subsection{The counit maps}
The counit maps $\epsilon: \textbf{A}^{\geq}\to \textbf{k}$, $\epsilon: \textbf{A}^{\leq}\to \textbf{k}$ are defined by $\epsilon(y)=0$ for $y$ in $\text{KHA}_T(Q,W)$ and $\epsilon\left(h^{\pm}_i(z)\right)=1$.

\subsection{Proof of Theorem \ref{thm1}}
$\text{KHA}_T$ is an algebra by \cite[Theorem 3.3]{P}. $\textbf{A}^{\geq}_T$ and $\textbf{A}^{\leq}_T$ are also algebras with unit $1$ in degree $0$. By a direct computation using Proposition \ref{p2} or by an argument similar to the one in \cite[Theorem 3.3]{P}, $\text{KHA}_T$, $\textbf{A}^{\geq}_T$, and $\textbf{A}^{\leq}_T$ are also coalgebras with counit $\epsilon$. We drop the suprascript from the notation of the formal series $h_i$, $i\in I$.

Let $c$ and $f$ be in $\mathbb{N}^I$. Let $\textbf{S}$ be the set of quadruples $(e_1, e_2, e_3, e_4)$ such that
\[e_1+e_2=a,\,\, e_3+e_3=b,\,\, e_1+e_3=c,\,\, e_2+e_4=f.\]
We need to show that
\begin{equation*}
    \Delta_{c,f}(x*y)=\sum_{\textbf{S}}\Delta_{e_1,e_2}(x)*\Delta_{e_3,e_4}(y).
\end{equation*}
Let $d\in\mathbb{N}^I$. Let $\lambda$ be a generic cocharacter of $G(d)$. Then \[\mathcal{X}(d)^\lambda\cong \mathcal{Y}(d)^\lambda= \times_{i\in I} \mathcal{Y}(\delta_i)^{\times d_i}\cong \times_{i\in I} \mathcal{X}(\delta_i)^{\times d_i} ,\] where $\delta_i$ is the dimension vector with $1$ in position $i$ and zero otherwise. Consider the inclusion
\[\iota_d:\mathcal{Y}(d)^\lambda\hookrightarrow \mathcal{Y}(d).\]
Let $\mathcal{I}$ be the set in $K_0^{T\times T(d)}(\text{pt})$ of functions $1-q^{-1}_\varepsilon z_{ij}^{-1}z_{i'j'}$ for $i,i'\in I$,
$j\leq d_i$, $j'\leq d_{i'}$ with $(ij)\neq (i'j')$, and $q_\varepsilon$ is a weight of $T$ corresponding to an edge from $i$ to $i'$. Then, by the localization theorem \cite[Theorem 2.5]{P},
\[\iota^*_d: K_0^T\big(D_{\text{sg}}(\mathcal{Y}(d)_0)\big)_\mathcal{I}\cong K_0^T\left(D_{\text{sg}}\left(\mathcal{Y}(d)^\lambda_0\right)\right)_\mathcal{I}.\]
Let $\mathcal{N}$ be the normal bundle of $\mathcal{Y}(d)^\lambda$ in $\mathcal{Y}(d)$. There is a natural projection map \[u:\mathcal{Y}(d)\cong\text{Tot}\left(\mathcal{N}\right)\to \mathcal{Y}(d)^\lambda.\]
Consider the action of $\mathbb{C}^*$ on $\text{Tot}\left(u^*\mathcal{N}\right)$ which is the pullback of the action of $\mathbb{C}^*$ via $\lambda$ on $\text{Tot}\left(\mathcal{N}\right)$. We have that $\text{Tot}\left(u^*\mathcal{N}\right)^\lambda=\mathcal{Y}(d)$.
By \cite[Proposition 2.4]{P} for the vector bundle $u^*\mathcal{N}$ on $\mathcal{Y}(d)$ and the above action of $\mathbb{C}^*$, the following map is injective \[K_0^T\left(D_{\text{sg}}(\mathcal{Y}(d)_0)\right)\hookrightarrow K_0^T\left(D_{\text{sg}}(\mathcal{Y}(d)_0)\right)_\mathcal{I}.\]
We write $\iota_d$ for maps from $\mathcal{Y}$ and $\mathcal{X}$; we write $\iota_A$ for maps from stacks $\mathcal{Y}_A$ with coordinates in the set $A=(A_i)_{i\in I}$.
It thus suffices to show that
\begin{equation}\label{prcopr}
    \left(\iota^*_c\otimes \iota_f^*\right)\Delta_{c,f}(x*y)=\left(\iota^*_c\otimes \iota_f^*\right)\sum_{\textbf{S}}\Delta_{e_1,e_2}(x)*\Delta_{e_3,e_4}(y).
\end{equation}

Let $A=(A_i)_{i\in I}$ and $B=(B_i)_{i\in I}$ be sets such that $A_i\sqcup B_i=\{1,\cdots, a_i+b_i\}$, $|A_i|=a_i$, $|B_i|=b_i$. By Proposition \ref{p2}, part (a), we have that
\[\Phi_{a,b}(x*y)=\sum_{A,B}\left(x_Ay_B\prod_{\substack{i,i'\in I\\ j\in A_i\\ j'\in B_{i'}}}\zeta_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)\right),\]
where the sum is after all sets $A$ and $B$ as above.
Define the $I$-tuples of sets $A^1=\left(A^1_i\right)_{i\in I}$, $A^2=\left(A^2_i\right)_{i\in I}$, $B^1=\left(B^1_i\right)_{i\in I}$, $B^2=\left(B^2_i\right)_{i\in I}$:
\begin{align*}
  A^1_i&=A_i\cap \{1,\cdots, c_i\},\\
  A^2_i&=A_i\cap \{c_i+1,\cdots, c_i+f_i\},\\
  B^1_i&=B_i\cap \{1,\cdots, c_i\},\\
  B^2_i&=B_i\cap \{c_i+1,\cdots, c_i+f_i\}.
\end{align*}
We introduce the following sets of indices:
\begin{align*}
  \textbf{E}&=\{(i,i',j,j')|\,i,i'\in I,  (j,j')\in (A^1_i,B^1_{i'}), (A^1_i,B^2_{i'}), (A^2_i,B^2_{i'})\}\\
  \textbf{F}&=\{(i,i',j,j')|\,i,i'\in I,  (j,j')\in (A^2_i,A^1_{i'}), (B^2_i,A^1_{i'}), (B^2_i,B^1_{i'})\}\\
  \textbf{G}&=\{(i,i',j,j')|\,i,i'\in I,  (j,j')\in (A^1_i,B^1_{i'}),  (A^2_i,B^2_{i'})\}\\
  \textbf{H}&=\{(i,i',j,j')|\, i,i'\in I, (j,j')\in (A^1_i, B^2_{i'})\}\\
  \textbf{K}&=\{(i,i',j,j')|\, i,i'\in I, (j, j')\in (B^2_i, A^1_{i'})\}.
\end{align*}
Write $\Phi_{A^1,A^2}(x)=\sum x_{A^1}\otimes x_{A^2}$; we will drop the sum sign in the formula of $\Phi$ in what follows.
We have that:
\begin{align*}
    \left(\iota^*_c\otimes \iota_f^*\right)
    \Delta_{c,f}(x*y)
    &=\prod_{\substack{i\in I\\j>c_i}}
    h_i(z_{ij})*
    \frac{\sum_{A,B}
    \iota^*_c(x_{A^1}y_{B^1})
    \otimes
    \iota_f^*(x_{A^2}y_{B^2})
    \prod_{\substack{i,i'\in I\\ j\in A_i\\ j'\in B_{i'}}}\zeta_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)
    }
    {\prod_{\substack{i,i'\in I\\j>c_i\\j'\leq c_{i'}}}\zeta_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)}\\
    &=\prod_{\substack{i\in I\\j\in A^2_i\sqcup B^2_i}}h_i(z_{ij})*\sum_{\substack{j\in A^1_i\sqcup A^2_i\\j'\in B^1_{i'}\sqcup B^2_{i'}}}
    \iota^*_c(x_{A^1}y_{B^1})\otimes \iota_f^*(x_{A^2}y_{B^2})
    \frac{\prod_{\textbf{E}} \zeta_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)}{\prod_{\textbf{F}} \zeta_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)}.
\end{align*}
Using Proposition \ref{p2}, part (b), we have that:
\begin{multline*}
    \Delta_{e_1,e_2}(x)*\Delta_{e_3,e_4}(y)=
    \left(
    \frac{ \prod_{\substack{i\in I\\j>e_{1i}}}h_i(z_{ij})*\Phi_{e_1,e_2}(x)}{
    \prod_{\substack{i,i'\in I\\j>e_{1i}\\j'\leq e_{1i'}}}\zeta_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)
    }
    \right)*
    \left(
    \frac{\prod_{\substack{i\in I\\j>a_i+e_{3i}}}h_i(z_{ij})*\Phi_{e_3,e_4}(y)}{
    \prod_{\substack{i,i'\in I\\j>a_i+e_{3i}\\j'\leq a_{i'}+e_{3i'}}}\zeta_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)
    }
    \right)
    \end{multline*}
Using that $\iota^*$ and $\Phi$ are compatible and using the formula for the (restriction of the) product from Proposition \ref{p2}, part (a), we obtain:
\begin{multline*}
    \left(\iota^*_c\otimes \iota_f^*\right)\big(\Delta_{e_1,e_2}(x)*\Delta_{e_3,e_4}(y)\big)=
    \sum_{A^1,\cdots, B^2}
    \left(
    \frac{\prod_{\substack{i\in I\\j\in A^2_i}}h_i(z_{ij})*\left(\iota^*_{A^1}\otimes \iota^*_{A^2}\right)\left(\Phi_{A^1,A^2}(x)\right)}{
    \prod_{\substack{i,i'\in I\\j\in A^2_i\\j'\in A^1_{i'}}}\zeta_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)
    }
    \right)\\
    \left(
    \frac{\prod_{\substack{i\in I\\j\in B^2}}h_i(z_{ij})*\left(\iota^*_{B^1}\otimes \iota^*_{B^2}\right)\left(\Phi_{B^1,B^2}(y)\right)
    }{
    \prod_{\substack{i,i'\in I\\j\in B^2_i\\j'\in B^1_{i'}}}\zeta_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)
    }
    \right)
    \prod_{\textbf{G}}\zeta_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right),
\end{multline*}
where the sum after all sets $A^1,\cdots, B^2$ as above.
We commute $x_{A^1}$ past the $h_i$s using the relations \eqref{rel}:
\begin{equation}\label{a}
x_{A^1}*\prod_{\substack{i'\in I\\j'\in B^2_{i'}}} h_{i'}(z_{i'j'})=
\prod_{\substack{i'\in I\\j'\in B^2_{i'}}} h_{i'}(z_{i'j'})*\left(x_{A^1}\prod_{\textbf{H}}
\tau_{i'i}\left(\frac{z_{ij}}{z_{i'j'}}\right)\right).
\end{equation}
We have that:
\begin{equation}\label{b}
\prod_{\textbf{H}}
\tau_{i'i}\left(\frac{z_{ij}}{z_{i'j'}}\right)=
\prod_{\textbf{H}}\frac{\zeta_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)}{\zeta_{i'i}\left(\frac{z_{i'j'}}{z_{ij}}\right)}=
\frac{
\prod_{\textbf{H}} \zeta_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)}
{\prod_{\textbf{K}} \zeta_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)}.
\end{equation}
Combining the formula in \eqref{b} with the formula for $\Delta_{e_1,e_2}(x)*\Delta_{e_3,e_4}(y)$, we obtain that:
\begin{multline*}
    (\iota^*_c\otimes\iota^*_f)\left(\Delta_{e_1,e_2}(x)*\Delta_{e_3,e_4}(y)\right)=
    \sum_{A^1,\cdots, B^2}\left(
    \prod_{\substack{i\in I\\j\in A^2_i}}h_i(z_{ij})
    \prod_{\substack{i\in I\\j\in  B^2_i}}h_i(z_{ij})\right)*\\
    \sum_{\substack{j\in A^1_i\sqcup A^2_i\\j'\in B^1_{i'}\sqcup B^2_{i'}}} \iota^*_c\left(x_{A^1}y_{B^1}\right)\otimes \iota_f^*\left(x_{A^2}y_{B^2}\right)
    \frac{\prod_{\textbf{E}} \zeta_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)}{\prod_{\textbf{F}} \zeta_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)}.
\end{multline*}
We thus obtain the equality claimed in \eqref{prcopr}.

\subsection{Properties of the tripled quiver}

For $Q$ a quiver, recall the definition of the tripled quiver from Subsection \ref{trqu}. Consider a torus $T\subset (\mathbb{C}^*)^{E}\times\mathbb{C}^*_q$, see Subsection \ref{trqu}.
We prove the following:

\begin{prop}\label{propo}
The tripled quiver $\left(\widetilde{Q}, \widetilde{W}\right)$ and the torus $T$ satisfy the Künneth Assumption \eqref{assumption}.
\end{prop}

We say that a stack $\mathcal{X}$ has \textbf{Property I} if the cycle map, see Subsection \ref{chernstacks}, is injective
\[G_0(\mathcal{X})\hookrightarrow \widehat{H^{\text{BM}}_\cdot}(\mathcal{X}).\]

\begin{prop}\label{I}
Let $\mathcal{X}$ be a stack, let $\mathcal{Z}$ be a closed substack, and let $\mathcal{U}=\mathcal{X}\setminus \mathcal{Z}$.
If $\mathcal{U}$ and $\mathcal{Z}$ have Property I, then $\mathcal{X}$ has Property I.

\end{prop}

\begin{proof}
The statement follows from the commutative diagram:
\begin{equation*}
    \begin{tikzcd}
    G_0(\mathcal{Z})\arrow[d, hook]\arrow[r]& G_0(\mathcal{X})\arrow[d]\arrow[r, two heads]& G_0(\mathcal{U})\arrow[d, hook]\\
    \widehat{H^{\text{BM}}_{\cdot}}(\mathcal{Z})\arrow[r]& \widehat{H^{\text{BM}}_{\cdot}}(\mathcal{X})\arrow[r]& \widehat{H^{\text{BM}}_{\cdot}}(\mathcal{U}).
    \end{tikzcd}
\end{equation*}
\end{proof}

\begin{prop}\label{II}
Let $\mathcal{X}_1,\cdots, \mathcal{X}_s$ be stacks with Property $I$. Then
\[\otimes_{i=1}^s G_0(\mathcal{X}_i)\hookrightarrow G_0\left(\times_{i=1}^s\mathcal{X}_i\right).\]
\end{prop}

\begin{proof}
This follows from the diagram:
\begin{equation*}
    \begin{tikzcd}
    \otimes_{i=1}^s G_0(\mathcal{X}_i)\arrow[d, hook]\arrow[r]& G_0\left(\times_{i=1}^s\mathcal{X}_i\right)\arrow[d]\\
    \otimes_{i=1}^s \widehat{H^{\text{BM}}_{\cdot}}(\mathcal{X}_i)\arrow[r, hook]&\widehat{H^{\text{BM}}_{\cdot}}\left(\times_{i=1}^s\mathcal{X}_i\right).
    \end{tikzcd}
\end{equation*}
\end{proof}

We say that a stack $\mathcal{X}$ has \textbf{Property K} if for any quotient stack $\mathcal{Y}$, we have a surjective map:
\[G_0(\mathcal{X})\otimes G_0(\mathcal{Y})\twoheadrightarrow G_0\left(\mathcal{X}\times\mathcal{Y}\right).\]

\begin{prop}\label{propK}
Let $\mathcal{X}$ be a stack, let $\mathcal{Z}$ be a closed substack, and let $\mathcal{U}$ be its complement.

(a) If $\mathcal{X}$ has Property K, then $\mathcal{U}$ has Property K.

(b) If $\mathcal{U}$ and $\mathcal{Z}$ have Property K, then $\mathcal{X}$ has Property K.

\end{prop}

\begin{proof}
The statement follows by diagram chasing through the maps:
\begin{equation*}
    \begin{tikzcd}
    G_0(\mathcal{Z})\otimes G_0(\mathcal{Y})\arrow[d, two heads]\arrow[r]& G_0(\mathcal{X})\otimes G_0(\mathcal{Y})\arrow[d]\arrow[r, two heads]&
    G_0(\mathcal{U})\otimes G_0(\mathcal{X})\arrow[d]\\
    G_0(\mathcal{Z}\times\mathcal{Y})\arrow[r]& G_0(\mathcal{X}\times\mathcal{Y})\arrow[r, two heads]& G_0(\mathcal{U}\times\mathcal{Y}).
    \end{tikzcd}
\end{equation*}
\end{proof}

Let $\mathfrak{W}(d)$ be the (quasi-smooth) stack of pairs $(x,u)$, where $x$ is a representation of $Q$ of dimension $d$ and $u$ is a $\mathbb{C}[Q]$-endomorphism of $x$, see \cite[Proof of Lemma 4.3, page 22]{Da}, \cite[Proof of Proposition 4.6, page 42]{SV2}. We define it as follows. Let \begin{align*}
    \mu: R(d)\times\mathfrak{g}(d)&\to R(d)\\
    (x,u)&\mapsto xu-ux.
    \end{align*}
Then $\mu$ induces a map $\partial:\mathcal{R}(d)^{\vee}\to \mathcal{O}_{R(d)\times \mathfrak{g}(d)},$ where $\mathcal{R}(d)$ is the corresponding vector bundle on $R(d)\times \mathfrak{g}(d)$ with fiber $R(d)$. Consider the Köszul stack
\[\mathfrak{W}(d):=\left(\text{Spec}\,\mathcal{O}_{R(d)\times \mathfrak{g}(d)}\left[\mathcal{R}(d)^{\vee}[1];\partial\right] \right)\big/ G(d).\]
Let $\mathfrak{N}(d)$ be the substack of $\mathfrak{W}(d)^{\text{cl}}$ of pairs $(x,u)$ where $u$ is nilpotent.

\begin{prop}\label{de}
Let $d, e \in\mathbb{N}^I$.
The stacks $\mathfrak{N}(d)/T$ and $\mathfrak{W}(d)/T$ satisfy Properties I and K.
In particular, we have that
\[G_0^T\left(\mathfrak{W}(d)\right)\otimes G_0^T\left(\mathfrak{W}(e)\right)\cong G_0^{T\times T}\left(\mathfrak{W}(d)\times\mathfrak{W}(e)\right).\]
\end{prop}

\begin{proof}
The stack $\mathfrak{W}(d)$ has Property I or K if and only if the stack $\mathfrak{W}(d)^{\text{cl}}$ has Property I or K, respectively.

Let $\mathfrak{N}(d)=N(d)/G(d)$.
As explained in \cite[Proof of Proposition 4.6, pages 42-43]{SV2}, the variety $N(d)$ has a stratification by locally closed subvarieties $S_J$ (depending on the Jordan type of $u$) which come with natural vector bundle maps
\begin{equation}\label{Jstrata}
S_J\big/T\times G(d)\to \left(\mathcal{X}(e_1)/T\right)\times_{BT}\cdots\times_{BT}\left(\mathcal{X}(e_s)/T\right),
\end{equation}
where $e_i$ are dimension vectors of $Q$.
The stacks $\mathcal{X}(d)/T$ have Properties I and K. Thus the stacks $S_J\big/T\times G(d)$ have Properties I and K. By Propositions \ref{I} and \ref{propK}, $\mathfrak{N}(d)/T$ has Properties I and K.

Consider different dimension vectors $d_i$ and positive integers $l_i$ for $1\leq i\leq k$ such that $\sum_{i=1}^k l_id_i=d$. Call this data $A$. Define $\mathfrak{W}_A$ as the substack of $\mathfrak{W}(d)^{\text{cl}}$ of representations $(x,u)$ such that
\[x\cong R_{11}\oplus\cdots R_{1l_1}\oplus\cdots\oplus R_{k1}\oplus\cdots R_{kl_k},\]
where $R_{ij}$ are representations of $Q$ with $\dim R_{ij}=d_{i}$ for $1\leq i\leq k$, $1\leq j\leq l_i$ and such that $u$ has eigenvalues $\varepsilon_{ij}$ on $R_{ij}$ for any $i$ and $j$ as above and $\varepsilon_{ij}\neq \varepsilon_{ik}$ for $1\leq j\neq k\leq l_i$.
The stack $\mathfrak{W}(d)^{\text{cl}}$ has a stratification by the (locally closed) substacks $\mathfrak{W}_A$.
Let $U_i\subset\mathbb{C}^{i}$ be the complement of the diagonals.
Then
\[\mathfrak{W}_A\cong \times_{i=1}^k\left(\mathfrak{N}(d_i)^{\times l_i}\times U_{l_i}\right).\]
Write $T=T_1\times T_2$, where $T_2=\mathbb{C}^*_q\cap T$. Then $T$ acts on $\mathfrak{N}(d_i)^{\times l_i}\times U_{l_i}$ via $T_1$ on the first factor and via $T_2$ on the second factor. We thus have that:
\[\mathfrak{W}_A/T\cong \prod^{1\leq i\leq k}_{BT}\left(\mathfrak{N}(d_i)^{\times l_i}/T_1\times U_{l_i}/T_2\right),\] where the right hand side is the product of the $k$ factors over the stack $BT$.
By Proposition \ref{propK}, the stack $U_i/T_2$ has Property K. By Proposition \ref{propK} and using that $\mathfrak{N}(d)/T$ has Property K, the stack $\mathfrak{W}(d)^{\text{cl}}/T$ has Property K. In particular,
\begin{equation}\label{sur}
G_0^T(\mathfrak{W}(d))\otimes G_0^T(\mathfrak{W}(e))\twoheadrightarrow G_0^{T\times T}\left(\mathfrak{W}(d)\times\mathfrak{W}(e)\right).
\end{equation}

To show that $\mathfrak{W}(d)^{\text{cl}}$ has Property I, it suffices to show that $\mathfrak{W}_A$ has Property I. Consider the diagram:
\begin{equation*}
    \begin{tikzcd}
    G_0^T\left( \times_{i=1}^k\left(\mathfrak{N}(d_i)^{\times l_i}\times \mathbb{C}^{l_i}\right)\right)\arrow[d,"\sim"]\arrow[r,"a"]& \widehat{H^{\text{BM}}_T}\left( \times_{i=1}^k\left(\mathfrak{N}(d_i)^{\times l_i}\times \mathbb{C}^{l_i}\right)\right)\arrow[d, hook]\\
    G_0^T\left( \times_{i=1}^k\left(\mathfrak{N}(d_i)^{\times l_i}\times U_{l_i}\right)\right)\arrow[r,"b"]& \widehat{H^{\text{BM}}_T}\left( \times_{i=1}^k\left(\mathfrak{N}(d_i)^{\times l_i}\times U_{l_i}\right)\right).
    \end{tikzcd}
\end{equation*}
In order to show that $b$ is injective, it is enough to prove that $a$ is injective, or that the following map is injective:
\begin{equation}\label{injN}
G_0^T\left(\times_{i=1}^k\mathfrak{N}(d_i)^{\times l_i}\right)\hookrightarrow \widehat{H^{\text{BM}}_T}\left( \times_{i=1}^k\mathfrak{N}(d_i)^{\times l_i}\right).\end{equation}
The stacks $\left( \times_{i=1}^k\mathfrak{N}(d_i)^{\times l_i}\right)/T$ have stratification by strata which are vector bundles over $\left(\mathcal{X}(f_1)/T\right)\times_{BT}\cdots\times_{BT}\left(\mathcal{X}(f_t)/T\right)$, where $f_i$ are some dimension vectors of $Q$, see \eqref{Jstrata}. The following map is injective
\[\bigotimes_{K_0^T(\text{pt})}G_0^T\left(\mathcal{X}(f_i)\right)\hookrightarrow \bigotimes_{\widehat{H_T}(\text{pt})}
\widehat{H^{\text{BM}}_T}\left(\mathcal{X}(f_i)\right),\]
and thus the map in \eqref{injN} is injective. Thus the stack $\mathfrak{W}(d)^{\text{cl}}$ has Property I. Using Proposition \ref{II}, we obtain that \[G_0^T(\mathfrak{W}(d))\otimes G_0^T(\mathfrak{W}(e))\hookrightarrow G_0^{T\times T}\left(\mathfrak{W}(d)\times\mathfrak{W}(e)\right),\]
and thus the map in \eqref{sur} is an isomorphism.

\end{proof}

Recall that $\mathfrak{P}(d)$ is the stack of dimension $d$ representations of the preprojective algebra of $Q$.

\begin{prop}\label{PpropI}
The stack $\mathfrak{P}(d)/T$ has Property I.
\end{prop}

\begin{proof}
The statement follows by Totaro's approximation trick for quotient stacks and using that Nakajima quiver varieties have property I \cite[Proposition 4.3]{SV2}.
Let $v$ be a dimension vector of $Q$, let $\mathfrak{M}(d,v)$ be the corresponding Nakajima quiver variety, and let $\mathcal{E}(d,v)\to \mathfrak{P}(d)$ be the equivariant vector bundle of dimension $\sum_{i\in I}d_iv_i$ such that $\mathfrak{M}(d,v)\subset \mathcal{E}(d,v)$ is the GIT semistable locus with respect to a stability condition. Let $\mathcal{U}(d,v):=\mathcal{E}(d,v)\setminus \mathfrak{M}(d,v)$.

Let $c$ be an integer.
By a standard dimension bound, see \cite[Proof of Proposition 3.12]{Da}, \cite[Proof of Lemma 4.1]{DM}, for $v_i$ large enough for all $i\in I$ we have that \[\text{codim}\left(\mathcal{U}(d,v)\text{ in }\mathcal{E}(d,v)\right)>c.\]
In particular, for any non-zero class $y$ in $G_0^T\left(\mathfrak{P}(d)\right)$, there exists $v$ such that  $y$ is nonzero under the composition
\[G_0^T\left(\mathfrak{P}(d)\right)\cong G_0^T\left(\mathcal{E}(d,v)\right)\to G_0^T\left(\mathfrak{M}(d,v)\right).\]
The conclusion follows from chasing through the diagram:
\begin{equation*}
    \begin{tikzcd}
    G_0^T\left(\mathfrak{P}(d)\right)\arrow[d]\arrow[r]& G_0^T\left(\mathfrak{M}(d,v)\right)\arrow[d, hook]\\
    \widehat{H^{\text{BM}}_T}\left(\mathfrak{P}(d)\right)\arrow[r]& H^{\text{BM}}_T\left(\mathfrak{M}(d,v)\right)
    \end{tikzcd}
\end{equation*}
where the right vertical arrow is an injection by \cite[Proposition 4.3]{SV2}.
\end{proof}

\begin{prop}\label{knot}
The categories $D^b_T\left(\mathfrak{P}(d)\right)$, $D^b_T\left(\mathfrak{W}(d)\right)$, and $D_{\text{sg}, T}(\mathcal{X}(d)_0)$ have the same $K_0$.
\end{prop}

\begin{proof}
Recall the Köszul equivalence from Subsection \ref{trqu}:
    \begin{equation}\label{equi10}
        \text{MF}^{\text{gr}}_{T}\left(\widetilde{\mathcal{X}(d)}, \widetilde{W}\right)\cong D^b_T\left(\mathfrak{P}(d)\right),\end{equation}
    where the grading is with respect to the group $\mathbb{C}^*$ which acts with weight $2$ on the maps corresponding to the loops $\omega_i$ for $i\in I$.

Next, use the Köszul equivalence/ dimensional reduction for the cut $\overline{E}\subset \widetilde{E}$, that is, define graded categories with respect to the group $\mathbb{C}^*$ acting with weight $2$ on the linear maps corresponding to $\overline{e}$ for $e\in E$ and
use dimensional reduction with respect to the coordinates of those maps, see \cite[Subsection 3.2.1]{P}. We obtain the equivalence
\begin{equation}\label{equi11}
\text{MF}^{\text{gr}}_{T}\left(\widetilde{\mathcal{X}(d)}, \widetilde{W}\right)\cong D^b_T\left(\mathfrak{W}(d)\right).
\end{equation}
The conclusion now follows from \cite[Corollary 3.13]{T5}.
\end{proof}

\begin{proof}[Proof of Proposition \ref{propo}]
By Proposition \ref{knot},
the Künneth Assumption is thus equivalent to
\[G_0^T(\mathfrak{W}(d))\otimes G_0^T(\mathfrak{W}(e))\cong
G_0^{T\times T}(\mathfrak{W}(d)\times \mathfrak{W}(d)),\]  and this follows from Proposition \ref{de}.
\end{proof}

Assume that $T$ satisfies \eqref{torusT}. Let $R=K_0^{T\times T(d)}(\text{pt})$ and let $K=\text{Frac}\,R$. The following result follows as \cite[Proposition 4.6]{SV2}, \cite[Lemma 2.4.2]{VV}.

\begin{prop}\label{Rtors}
Let $d\in\mathbb{N}^I$. Then $G_0^T\left(\mathfrak{P}(d)\right)$ is $R$-torsion free. Thus the following map is injective
\[K_0^T\left(D_{\text{sg}}\left(\widetilde{\mathcal{X}(d)}_0\right)\right)\hookrightarrow K_0^T\left(D_{\text{sg}}\left(\widetilde{\mathcal{X}(d)}_0\right)\right)\otimes_R K.\]
\end{prop}

\begin{proof}
By \eqref{preequi} and Proposition \ref{PpropI}, it suffices to show that the following map is injective
\[G_0^T\left(\mathfrak{P}(d)\right)\hookrightarrow G_0^T\left(\mathfrak{P}(d)\right)\otimes_{R}K.\]
Consider the diagram
\begin{equation*}
    \begin{tikzcd}
    G_0^T\left(\mathfrak{P}(d)\right)\arrow[d,"a"]\arrow[r,"b"]& G_0^T\left(\mathfrak{P}(d)\right)\otimes_{R}K\arrow[d,"a"]\\
    \widehat{H^{\text{BM}}_T}\left(\mathfrak{P}(d)\right)\arrow[r,"c"]& \widehat{H^{\text{BM}}_T}\left(\mathfrak{P}(d)\right)\otimes_{R}K.
    \end{tikzcd}
\end{equation*}
The maps $a$ are injective by Proposition \ref{PpropI}. It suffices to show that $ H^{\text{BM}}_T\left(\mathfrak{P}(d)\right)$ is $R$-torsion free.
This follows as in \cite[Proposition 4.6]{SV2} where the result is stated for nilpotent versions of $\mathfrak{P}(d)$. In loc. cit., $\mathfrak{B}(d)^{\text{cl}}$ is denoted by $\text{M}(d)/G(d)$.
Let $\theta:\mathbb{C}^*\to T$ be a cocharacter which acts with non-zero weight on the linear maps corresponding to $e\in E$ and on $\omega_i$ for $i\in I$, and with zero weight on the linear maps corresponding to $\overline{e}$ for $e\in E$.
Such a cocharacter always exists for tori $T$ satisfying \eqref{torusT}. Let $\overline{\mathcal{X}(d)}$ be the moduli stack of representations of $\overline{Q}$ of dimension $d$. Then $\mathfrak{P}(d)^\theta\cong\overline{\mathcal{X}(d)}$. By the localization theorem, it suffices to show that $H^{\text{BM}}_T\left(\mathfrak{P}(d)\right)$ is $S_\theta$-torsion free, where $S_\theta$ is defined in loc. cit. This statement is proved in loc. cit.
\end{proof}

\section{The Hopf pairing}\label{s4}

\subsection{The antipode map}
Let $(Q,W)$ be an arbitrary symmetric quiver with potential and let $T$ be a torus satisfying \eqref{torus3}.
Let $y$ be in $K_0^T(D_{\text{sg}}(\mathcal{X}(d)_0))$. The antipode map $S$ for $\textbf{A}^{\geq}_T$ is defined by:
\begin{align*}
    S\left(h^+(z)\right)&=\left(h^+(z)\right)^{-1}\\
    S(y)&=\left[\prod_{\substack{i\in I\\j\leq d_i}}\left(-h_i^+(z_{ij})\right)^{-1}\right]*y.
\end{align*}
The antipode map $S$ for $\textbf{A}^{\leq}_T$ is defined by:
\begin{align*}
    S\left(h^-(z)\right)&=\left(h^-(z)\right)^{-1}\\
    S(y)&=y*\left[\prod_{\substack{i\in I\\j\leq d_i}}\left(-h_i^-(z_{ij})\right)^{-1}\right].
\end{align*}
Using these antipode maps, $\textbf{A}^{\geq}_T$ and $\textbf{A}^{\leq}_T$ are Hopf algebras. The proof is the same as the one for the tripled quiver of the cyclic type $A$ quiver in \cite[Exercise IV.1]{n2}.

\subsection{The bialgebra pairing}
\label{pair}

\subsubsection{}
Let $(Q,W)$ be an arbitrary symmetric quiver with potential and let $T$ be a torus satisfying \eqref{torus3} and such that $R(d)^T$ is a point for any dimension vector $d\in\mathbb{N}^I$.
Consider the set $\mathcal{J}$ in $K_0^{T\times T(d)}(\text{pt})$ with functions $1-z_{ij}^{-1}z_{ik}$ for $i\in I$, $j\neq k$, and by $1-q^{-1}_\varepsilon z_{ij}^{-1}z_{i'j'}$ for $1\leq j\leq d_i$, $1\leq j'\leq d_{i'}$ and where $q_\varepsilon$ is a weight of $T$ corresponding to an edge $\varepsilon$ from $i$ to $i'$ of $Q$. None of the weights $q^{-1}_\varepsilon z_{ij}^{-1}z_{i'j'}$ are $1$ for $(i,j)=(i',j')$ by the assumption that $R(d)^T$ is a point.

Let $\iota_d: BT(d)\hookrightarrow \mathcal{Y}(d)$. We abuse notation and write
\begin{equation*}\label{iota}
    \iota_d^*: K_0^T\left( D_{\text{sg}}(\mathcal{X}(d)_0)\right)\hookrightarrow K_0^T\left( D_{\text{sg}}(\mathcal{Y}(d)_0)\right)\to K_0^{T\times T(d)}(\text{pt}).
    \end{equation*}
Recall the definition of $\zeta_{ii'}$ from \eqref{zeta}. Define
\[
\widetilde{\zeta}_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)=
\begin{cases}
\zeta_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)\text{ if }(i,j)\neq (i',j')\\
\left(1-q_1^{-1}\right)\cdots \left(1-q_{\varepsilon(i,i)}^{-1}\right)\text{ otherwise}.
\end{cases}
\]
For $y$ in $K_0^T(D_{\text{sg}}(\mathcal{X}(d)_0))$, define
\[\Psi(y):=\frac{\iota_d^*(y)}{
\prod_{\substack{i,i'\in I\\j,j'}}\widetilde{\zeta}_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)}.\]
Define the pairing
\begin{multline*}
    \langle\,,\,\rangle: K_0^T(D_{\text{sg}}(\mathcal{X}(d)_0))\otimes K_0^T(D_{\text{sg}}(\mathcal{X}(d)_0))\xrightarrow{\otimes} K_0^T(D_{\text{sg}}(\mathcal{X}(d)_0))\to\\ K_0^T(D_{\text{sg}}(\mathcal{X}(d)_0))_\mathcal{J}\xrightarrow{\Psi}K_0^T(BT(d))_\mathcal{J}.
\end{multline*}

\subsubsection{}\label{tripled}
Let $Q$ be an arbitrary quiver, let $\left(\widetilde{Q}, \widetilde{W}\right)$ be its tripled quiver, and let $T$ be a torus satisfying \eqref{torusT}. Then $\widetilde{R(d)}^T$ is a point. Let $\textbf{k}=\text{Frac}\,K_0^{T}(\text{pt})$.
Consider a new variable $p$. For $i\in I$, define
\[\mu_i(z)=\frac{1-p^2z^{-1}}{1-q^2z^{-1}}.\]
Define the map
\[\Phi: \text{image}\,(\Psi)\to \textbf{k}
\] by the formula:
\[\Phi\left(\frac{y}{
\prod_{\substack{i,i'\in I\\j,j'}}\widetilde{\zeta}_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)}
\right)=\frac{1}{d!}
\int^{|q|<1<|p|}_{|z_{ij}|=1}
\frac{\iota_d^*(y)}{\prod_{\substack{i,i'\in I\\j,j'}}
\widetilde{\zeta}_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)
\prod_{\substack{i\in I\\j,k}}\mu_i\left(\frac{z_{ij}}{z_{ik}}\right)
}
\prod_{\substack{i\in I\\ j\leq d_i}} Dz_{ij}
\bigg|_{p\to q}
.\]
In the above, $Dz=\frac{dz}{2\pi iz}$. To evaluate $\Phi$, we first compute the integral in the region indicated
also assuming that $|q_e|=1$ for $e\in E$, and then we take the limit $p\to q$. The integral is equal to the sum of residues of the integrand in the region $|q|<1<|p|$, $|z_{ij}|=|q_e|=1$ for $i\in I$, $j\leq d_i$, $e\in E$.

 Define the pairing
 \begin{align*}
     (\,,\,):K_0^T\left(D_{\text{sg}}\left(\widetilde{\mathcal{X}(d)}_0\right)\right)\otimes K_0^T\left(D_{\text{sg}}\left(\widetilde{\mathcal{X}(d)}_0\right)\right)&\to\textbf{k}\\
     (x,y)&=\Phi\,\langle x,y\rangle.
 \end{align*}
Extend the pairing $(\,,\,)$ by the formula:
\begin{align*}
    (\,,\,):\textbf{A}^{\leq}_T\otimes \textbf{A}^{\geq}_T&\to \textbf{k}\\
    (h_{i'}^-(z), h_i^+(w))&=\tau_{i'i}\left(\frac{w}{z}\right).
\end{align*}

\textbf{Remark.}
By the localization theorem \cite[Theorem 2.5]{P}, there is an isomorphism
\[K_0^T\left( D_{\text{sg}}(\mathcal{Y}(d)_0)\right)_{\mathcal{I}}
\xrightarrow{\sim} K_0^{T\times T(d)}(\text{pt})_{\mathcal{I}}.\]
Using a shuffle description of the various maps involved, $\Psi$ can be thought as an inverse of the map in \eqref{iota}. It would be interesting to prove Theorem \ref{thm2} using such a geometric description of $\Phi$ and adjunction of the various functors used in the definitions of $m$ and $\Delta$.

\subsection{Proof of Theorem \ref{thm2}}

We check that $(x, y*y')=(\Delta^{\text{op}}(x), y\otimes y')$. Assume that $\deg(y)=a$, $\deg(y')=b$, $\deg(x)=a+b=d$.
For elements in $\text{KHA}_T(Q,W)$, the statement follows from the argument in \cite[Exercise IV.2]{n2}, which we now explain. Let $A_i=\{1,\cdots, a_i\}$ and $B_i=\{a_i+1,\cdots, d_i\}$.
Let $\textbf{E}$ be the set of indices
\[\textbf{E}=\{(i,i',j,j')|\,i,i'\in I, (j,j')\in (A_i,A_{i'}), (B_i,B_{i'}), (B_i,A_{i'})\}.\]
We will use the shorthand notations $z_A$ for $z_{ij}$ with $i\in I$ and $j\in A_i$ and $z_B$ for $z_{ij}$ with $i\in I$ and $j\in B_i$.

Using Proposition \ref{p2}, part (a), and the fact that the region we integrate in is symmetric in the all variables $z_{ij}$, we obtain that:
\[(x,y*y')=\frac{1}{a!b!}
\int^{|q|<1<|p|}_{|z_A|=|z_B|=1}
\frac{\iota^*_d(x)\iota_a^*(y)\iota_b^*(y')}{\prod_{\textbf{E}}\widetilde{\zeta}_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)\prod_{\substack{i\in I\\j,k}}\mu_i\left(\frac{z_{ij}}{z_{ik}}\right)}
\prod_{\substack{i\in I\\ j\in A_i}} Dz_{ij} \prod_{\substack{i\in I\\ j\in B_i}} Dz_{ij}.\]
Call the integrand in the above formula $u$. By the formula in \eqref{coprneg}, we have that:
\begin{align*}
    (\Delta^{\text{op}}(x), y\otimes y')&=
    (\Delta(x), y'\otimes y)\\
&=\frac{1}{a!b!}
\int^{|q|<1<|p|}_{|z_B|\ll|z_A|}
\frac{\iota^*_d(x)\iota_b^*(y')\iota_a^*(y)}{\prod_{\textbf{E}}\widetilde{\zeta}_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)\prod_{\substack{i\in I\\j,k}}\mu_i\left(\frac{z_{ij}}{z_{ik}}\right)}
\prod_{\substack{i\in I\\ j\in A_i}} Dz_{ij} \prod_{\substack{i\in I\\ j\in B_i}} Dz_{ij}\\
&=\frac{1}{a!b!}
\int^{|q|<1<|p|}_{|z_A|\gg|z_B|}
\frac{\iota^*_d(x)\iota_a^*(y)\iota_b^*(y')}{\prod_{\textbf{E}}\widetilde{\zeta}_{ii'}\left(\frac{z_{ij}}{z_{i'j'}}\right)\prod_{\substack{i\in I\\j,k}}\mu_i\left(\frac{z_{ij}}{z_{ik}}\right)}
\prod_{\substack{i\in I\\ j\in A_i}} Dz_{ij} \prod_{\substack{i\in I\\ j\in B_i}} Dz_{ij}.
\end{align*}
The last equality holds because $\iota_d^*(x)$ is $\mathfrak{S}_d$-symmetric.
It suffices to show that
\begin{equation}\label{int2}
\int^{|q|<1<|p|}_{|z_A|=|z_B|=1}
u
\prod_{\substack{i\in I\\ j\in A_i}} Dz_{ij} \prod_{\substack{i\in I\\ j\in B_i}} Dz_{ij}
\bigg|_{p\to q}=
\int^{|q|<1<|p|}_{|z_A|\gg|z_B|}
u
\prod_{\substack{i\in I\\ j\in A_i}} Dz_{ij} \prod_{\substack{i\in I\\ j\in B_i}} Dz_{ij}
\bigg|_{p\to q}.
\end{equation}
The integrand $u$ has poles involving $z_A$ and $z_B$ in the left hand side region at
\[z_A=qq_\varepsilon^{\pm 1}z_B,\quad z_{ij}=p^{-2}z_{ik},\quad z_{ij}=p^2z_{ik},\]
where $\varepsilon$ in an edge of $Q$, $j\in A_i$, $k\in B_i$.
There is a factor $z_{ij}=q^2z_{ik}$ in the numerator for $i\in A_i$, $j\in B_i$, and thus any residue obtained at $z_{ij}=p^2z_{ik}$ vanishes after setting $p\to q$. There are no other poles of $u$ involving $z_A$ and $z_B$ as we move to $|z_A|\gg |z_B|$. Similarly, we can also replace the factors $\mu_i\left(\frac{z_A}{z_B}\right)$ by $1$ as the two corresponding integrals coincide after passing to $p\to q$. The equality in \eqref{int2} thus follows. The statement $(x*x',y)=(x\otimes x', \Delta(y))$ follows similarly.
\\

The pairing $(\,,\,)$ extends to $\textbf{A}^{\leq}_T\otimes \textbf{A}^{\geq}_T$. We show that
\begin{equation}\label{op}
\left(
\Delta^{\text{op}}(x), h^+_{i}(w)\otimes y\prod_{\substack{i'\in I\\j'\leq d_{i'}}}\tau_{ii'}\left(\frac{z_{i'j'}}{w}\right)
\right)
=
\left(\Delta^{\text{op}}(x), y\otimes h_{i}^+(w)\right)
.\end{equation}
Indeed, $\Delta^{\text{op}}(x)=x\otimes 1+\cdots+\prod_{\substack{i'\in I\\j'\leq d_{i'}}}h^-_{i'}(z_{i'j'})\otimes x$, where the middle terms do not contribute to the pairing, and thus the left hand side is equal to
\begin{align*}
\prod_{\substack{i'\in I\\j'\leq d_{i'}}}
\left(h^-_{i'}(z_{i'j'}), h^+_{i}(w)\right)
(x,y)
\prod_{\substack{i'\in I\\j'\leq d_{i'}}}\tau_{ii'}\left(\frac{z_{i'j'}}{w}\right)&=
\prod_{\substack{i'\in I\\j'\leq d_{i'}}}\tau_{i'i}\left(\frac{w}{z_{i'j'}}\right)(x,y)\prod_{\substack{i'\in I\\j'\leq d_{i'}}}\tau_{ii'}\left(\frac{z_{i'j'}}{w}\right)\\
&=(x,y).
\end{align*}
The right hand side is $(x,y)\left(1, h_{i'}(w)\right)=(x,y)$, so the equality in \eqref{op} holds. Similarly, we have that
\begin{equation*}
\left(
h^-_{i}(w)\otimes y\prod_{\substack{i'\in I\\j'\leq d_{i'}}}\tau_{ii'}\left(\frac{w}{z_{i'j'}}\right), \Delta(x)
\right)
=
\left(y\otimes h_{i}^-(w), \Delta(x)\right)
.\end{equation*}

Finally, let $x$ be a non-zero element in the image of $\Psi$. There exists a monomial $u$ such that $\Phi(xu)\neq 0$. Let $y$ be non-zero in $K_0^T\left(D_{\text{sg}}\left(\widetilde{\mathcal{X}(d)}_0\right)\right)$.
Using Proposition \ref{Rtors}, $\langle y,u\rangle\neq 0$ for any non-zero monomial $u$. Thus there is a monomial $u$ such that $(y,u)=\Phi\langle y,u\rangle\neq 0$.

\subsection{Examples.}
\subsubsection{}
Let $J$ be a Jordan quiver and let $\mathbb{C}^*$ act on representations of $J$ by scaling the linear map. By \cite[Subsection 9.2]{P}, \cite{Ts1}, \cite{VV}, $\text{KHA}_{\mathbb{C}^*}(J,0)\cong U^{>}_q\left(L\mathfrak{sl}_2\right)$ and thus $\textbf{A}^{\geq}_{\mathbb{C}^*}\cong U_q\left(L\mathfrak{b}\right)\cong U^{\geq}_q\left(L\mathfrak{sl}_2\right)$, where $\mathfrak{b}$ is a Borel subalgebra of $\mathfrak{sl}_2$.
For $a\in\mathbb{Z}$,
denote by $e_a:=z^a$ generators of $\text{KHA}_{\mathbb{C}^*}(J,0)\subset \textbf{A}^{\geq}_{\mathbb{C}^*}$ and by $f_a:=z^a$ generators of $\text{KHA}_{\mathbb{C}^*}(J,0)^{\text{op}}\subset \textbf{A}^{\leq}_{\mathbb{C}^*}$.
The formulas for the coproduct are:
\begin{align*}
    \Delta(e_a)&=e_a\otimes 1+\sum_{n\geq 0}h^+_n\otimes e_{a-n}\\
    \Delta(f_a)&=1\otimes f_a+\sum_{n\geq 0}f_{a-n}\otimes h^-_n.
\end{align*}
We next compute the bialgebra pairing on $\textbf{A}^{\leq}_{\mathbb{C}^*}\otimes \textbf{A}^{\geq}_{\mathbb{C}^*}$.
By definition, we have that
\begin{align*}
\Big(h^-(z), h^+(w)\Big)&=\tau\left(\frac{w}{z}\right)=\frac{qw-z}{w-qz}\\
(h^-_b, e_a)&=0\\
(f_a, h^+_b)&=0.
\end{align*}
Finally, we compute
\[(f_m, e_n)=\int^{|p|<1<|q|}_{|z|=1}\frac{z^{n+m}}{1-q^2}\frac{1-q^2}{1-p^2}Dz\bigg|_{p\to q}=\frac{1}{1-q^2}\int _{|z|=1}z^{n+m}Dz=\frac{\delta_{0,n+m}}{1-q^2}.\]
Replacing $q$ with $q^{-1}$ and up to a factor in $(f_m, e_n)$, we obtain the bialgebra pairing on $U^{\leq}_q\left(L\mathfrak{sl}_2\right)\otimes U^{\geq}_q\left(L\mathfrak{sl}_2\right)$ in \cite[Proposition 2.34]{Ts2}. Thus $\textbf{A}_{\mathbb{C}^*}\cong U_q\left(L\mathfrak{sl}_2\right)$.

\subsubsection{} Consider the quiver
$Q_3$ with one vertex and three loops $\ell_1,\ell_2,\ell_3$ with potential $W_3=\ell_1\ell_2\ell_3-\ell_1\ell_3\ell_2$. It is the tripled quiver of the Jordan quiver $J$. Let $T$ be the two dimensional torus acting with weights $t$ on the edge $\ell_1$, $t^{-1}$ on $\ell_2$, and with weights $q$ on $\ell_1$ and $\ell_2$ and $q^{-2}$ on $\ell_3$.  The subalgebra generated by
$K_0^T(BGL(d))\to K_0^T\left(D_{\text{sg}}\left(\widetilde{\mathcal{X}(d)}_0\right)\right)$ has been studied by Negu\c{t} \cite{n1}, \cite{n2}, and is isomorphic to $U_{q, t}^{>}\left(\widehat{\widehat{\mathfrak{gl}_1}}\right)$. Consider the generators $e_{n}:=z^n$ of $\text{KHA}_T(Q_3,W_3)$ and $f_n:=z^n$ of $\text{KHA}_T(Q_3,W_3)^{\text{op}}$. Then
\[\Delta(e_{n})=e_{n}\otimes 1+\sum_{i\geq 0}h^+_i\otimes e_{n-i}.\]
The coproduct is the same as the one in \cite[Subsection 4.6]{n1}. The pairing on generators of $\text{KHA}_T(Q_3,W_3)$ is:
\begin{align*}
(f_{m}, e_n)&=\int^{|p|<1<|q|}_{|z|=1}\frac{z^{n+m}}{(1-q^{-1}t)(1-q^{-1}t^{-1})(1-q^2)}Dz\bigg|_{p\to q}
\\&=\frac{\delta_{0,n+m}}{(1-q^{-1}t)(1-q^{-1}t^{-1})(1-q^2)}.\end{align*}
This is the same as the pairing in \cite[Subsection 4.6]{n1}.

redpajama_set_name: RedPajamaArXiv


text:
Dersu Uzala – il cacciatore che guidò Vladimir Klavdievič Arsen'ev
 Dersu Uzala – romanzo di viaggio del 1923 di Vladimir Klavdievič Arsen'ev che racconta le gesta del cacciatore
 Dersu Uzala - Il piccolo uomo delle grandi pianure – film del 1975 di Akira Kurosawa, trasposizione del romanzo di Arsen'ev
 Dersu Uzala – film del 1961 diretto da Agasi Arutjunovič Babajan

redpajama_set_name: RedPajamaWikipedia


WHAT IS FORMULA ONE - Many people see Formula One (F1) racing as the very best in the whole of motorsport. It's actually a set of rules for single-seater racing cars. The rules apply to how much the cars can weigh and how they're raced.
The F1 in Schools Challenge is a competition, open to all UK based secondary schools, colleges and (organised) youth groups, to design and manufacture CO2 powered model racing cars (dragsters).
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Brisk in Pink are a 16-18 team from Cheltenham Girls High School in Australia. The girls competed in the 2005/06 Season F1 in Schools World Championships which was held on January 15th 2006 at Autosport International, NEC Birmingham.
Felis Onca are from Woodlands Community School in Southampton and this is their website.
Flaming Fury are from Ridgewood School in Doncaster. The team competed in 2005/06 Season National Finals and won 3rd place in our website competition.
These guys are from Arden School in Solihull. Why not take a look their car - 'The Force' - on their website.
This team is from Brumby Engineering College in Scunthope and came 2nd place in the key stage 3 catergory at their Regional Final.
The Rapid Stags are a post 16 team from Bassaleg School in Wales and have been competing in the Challenge for 3 years. Why not take a look at their website.
Rothers Revvers are from Glenrothers High School in Scotland. The team competed in the 2005/06 National Finals, here's their website.
Team Air are from Ynysawdre Comprehensive School in South Wales and won the award for Best Engineered Car at the 2005 Regional Finals.
Team Burnout are from Bryntirion Comprehensive in South Wales, this website won them 3rd Palce in our Website competition at the National Final 2004-05 in January 2005.
These guys are a post 16 team from Carnoustie High School in Scotland. Here's their website.
Team Cobra are from Bournville School in Birmingham and heres their website.
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Team CXTR-2 are from Cotham School in Bristol.
Team Dynamic are from Lutterworth Grammar School - these guys are even doing their own newsletter you can download as a pdf.
Team Extreme are from St. Patricks College in Dungiven, Northern Ireland. Here's their website..
Team Falcon claimed the overall winning team in the 16-18 Category at the 2004-05 National final in January 2005.
Team Firestarter are from Neale-wade Community College in Cambridgeshire and this is their website.
Team Infurno are Yorkshire and Humberside winners in the 16-18 Year Age Group.
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Lecon 2 competed in the 2004-05 Championship Season and here is their website.
Team Ministry of Dynamics are from Lutterworth Grammer School in Leicestershire.
Team Sidewinder won the National Final Overall winners title in 2003-04 Championship Seaon and represented England at our International Final, this is their website.
St Patricks, Dugiven Team Sonic came second place in their age group at the National Final 2004-05 and this is their website.
Team Turbo are a Year 8 Team from Lutterworth High School, this is the first Championship season (2005-06) they have entered a team.
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