text:
\section{Introduction}
The Hall algebra of a finite dimensional algebra $A$ over a finite field was introduced by Ringel \cite{R90} in 1990.
Ringel \cite{R90,R90a} proved that if $A$ is a representation-finite hereditary algebra, the Ringel--Hall algebra of $A$ provides a realization of the positive part of the corresponding quantum group.
Ringel's approach establishes a relation between the representation theory of algebras and Lie theory, and provides an algebraic framework for studying the Lie theory resulting from Hall algebras associated to various abelian categories. To\"en \cite{Toen2006} generalized Ringel's construction to define the derived Hall algebra for a DG-enhanced triangulated category satisfying certain finiteness conditions. Later on, for a triangulated category satisfying the left homological finiteness condition, Xiao and Xu \cite{XiaoXu} showed that To\"en's construction still provides an associative unital algebra. It was expected but so far not successful to realize an entire quantum group through derived Hall algebra over triangulated category. In 2013,
Bridgeland \cite{Bri13} provided a realization of the
whole quantum group via the Hall algebra of 2-cyclic complexes of projective modules over a hereditary algebra.

Lusztig \cite{Lu90,Lu91} invented the geometric version of Ringel--Hall algebra constructions and obtained the canonical basis of the positive part of a quantum group as the direct summands of some semisimple constructible complexes over module varieties of a quiver. Kashiwara \cite{Ka91} applied an algebraic approach to define the crystal basis of the positive part of a quantum group. It is noteworthy that the canonical basis of a quantum group coincides with its crystal basis. In \cite{L3}, Lusztig has also introduced the semicanonical basis of the positive part of the enveloping algebra associated to a quiver $Q$  as certain constructible functions over the varieties of nilpotent representations of the preprojective algebra of $Q$, and the basis is indexed by the irreducible components of the varieties.

Cluster algebras were introduced by Fomin and Zelevinsky in \cite{FZ} and later the quantum cluster algebras were introduced by Berenstein and Zelevinsky in \cite{BZ05}. Inventions of cluster algebras and quantum cluster algebras are aimed to develop a combinatorial approach to the dual semicanonical bases in coordinate rings and dual canonical bases in quantum deformations of varieties related to algebraic groups. Geiss, Leclerc, and Schr\"oer proved in
\cite{GLS} that the cluster monomials of certain cluster algebras are elements of the dual of Lusztig's semicanonical basis. Kimura and Qin \cite{KQ} categorified the quantum cluster algebra of an acyclic quiver via the generalized graded quiver variety, and as a byproduct it is proved that the quantum cluster monomials belong to the corresponding dual canonical basis. Moreover, Qin \cite{Qin2} extended these results to quantum cluster algebras which are injective-reachable.

A natural idea  is to construct a framework to explicitly relate Hall algebras with (quantum) cluster algebras. In \cite{CK2005}, Caldero and Keller suggested the similarity between the multiplication in a cluster algebra and that in a dual Hall algebra. In \cite[Theorem 3.3]{DX}, the similarity was confirmed for the quantum cluster algebra of an acyclic quiver (see \cite{BR} for the generalization). Given a finite acyclic quiver $Q$, let $\mathcal{AH}_q(Q)$ be the subalgebra of a certain skew-field of fractions generated by quantum cluster characters (see Section \ref{quantum_cluster_char} for more details). Then, there exists an algebra homomorphism from the dual Hall algebra associated to the representation category of $Q$ to $\mathcal{AH}_q(Q)$ (cf. \cite{CDX}). It is pitiful that this homorphism may not be surjective, in particular, there might be no preimages of  the initial quantum cluster variables.

The aim of this paper is to construct a surjective algebra homomorphism from a certain Hall algebra to  $\mathcal{AH}_q(Q)$, and then realize the quantum cluster algebra as a sub-quotient algebra of this Hall algebra. In order to achieve this aim, we construct the localized Hall algebra $\mathcal{MH}(Q)$ associated to the morphism category $\mathcal{C}_2(\mathcal{P})$ (see Section \ref{morphism} for the definition), which has indecomposable objects indexed by the isomorphism classes of indecomposable objects in the cluster category of $Q$. The algebra $\mathcal{MH}(Q)$ contains the dual Hall algebra of $Q$ as a subalgebra and is isomorphic to the subalgebra of the extended dual derived Hall algebra of $Q$ generated by all objects ${M\oplus P[1]}$ corresponding to $kQ$-modules $M$ and projective $kQ$-modules $P$ (see Theorem \ref{mor_derived}).

The main result of this paper is Theorem \ref{alg-homo} which gives a surjective homorphism from a twisted version of $\mathcal{MH}(Q)$ to $\mathcal{AH}_q(Q)$.
This surjective algebra homomorphism will motivate the interactions between Hall algebras and quantum cluster algebras. For example, one can compare the categorifications of quantum groups and quantum cluster algebras, and then compare their dual canonical bases.

The paper is organized as follows: In Section 2 we mainly summarize some homological properties of a morphism category, and then define the associated Hall algebra via Bridgeland's approach in Section 3. The characterizations of the multiplicative structure of thus defined Hall algebra are given in Sections 4 and 5. Section 6 is devoted to establishing the relation between this Hall algebra and the extended dual derived Hall algebra. We reformulate the definitions of quantum cluster characters and give the corresponding cluster multiplication formulas in Section 7. Finally, we prove the main theorem of this paper in Section 8.

Let us fix some notations used throughout the paper. Let $k=\mathbb{F}_q$ be always a finite field with
$q$ elements, and $\ZZ[{q}^{\pm\frac{1}{2}}]$ be the ring of integral Laurent polynomials. Let $\A$ be an (essentially small) finitary hereditary abelian $k$-category with enough projectives, where the term ``finitary" means that for any objects $M,N\in\A$, $\Hom_{\A}(M,N)$ and $\Ext^1_{\A}(M,N)$ are both finite dimensional. Let $\mathscr{P}\subset\A$ be the subcategory consisting of projective objects. Denote by $C^b(\A)$ and $D^b(\A)$ the category of bounded complexes over $\A$ and its bounded derived category, respectively. For each $M_\bullet\in C^b(\A)$, its $i$-th homology is denoted by $H_i(M_\bullet)$.  The Grothendieck group of $\A$ and the set of isomorphism classes $[X]$ of objects in $\A$ are denoted by $K(\A)$ and $\Iso(\A)$, respectively. For any object $M\in\A$ we denote by $\hat{M}$ or $\Dim M$ the image of $M$ in $K(\A)$. For a finite set $S$, we denote by $|S|$ its cardinality. For an object $M$ in an additive category, we denote by $\Aut(M)$ the automorphism group of $M$, and set $a_M:=|\Aut(M)|$. We always assume that all the vectors are column vectors.

\section{Preliminaries}

\subsection{Hall algebras}
Given objects $L,M,N \in \mathcal{A}$, let $\Ext_\mathcal{A}^1(M,N)_L \subset \Ext_\mathcal{A}^1(M,N)$ be the subset consisting of those equivalence classes of short exact sequences with middle term isomorphic to $L$.
\begin{definition}\label{Hall algebra of abelian category}
The \emph{Hall algebra} $\mathcal {H}(\mathcal{A})$ of $\mathcal{A}$ is the free $\ZZ[{q}^{\pm\frac{1}{2}}]$-module with basis elements $[M] \in \Iso(\mathcal{A})$, and with the multiplication defined by
\[[M] \diamond [N] = \sum\limits_{[L] \in \Iso(\mathcal{A})} {\frac{{|\Ext_\mathcal{A}^1{{(M,N)}_L}|}}{{|\Hom_\mathcal{A}(M,N)|}}} [L].\]
\end{definition}
\begin{remark}
Given objects $L,M,N\in \A$, set
$$F_{MN}^L:=|\{N'\subset L~|~N'\cong N, L/N'\cong M\}|.$$
By the Riedtmann--Peng formula \cite{Riedtmann,Peng},
$$F_{MN}^{L}=\frac{|\Ext^1_{\A}(M,N)_{L}|}{|\Hom_{\A}(M,N)|}\frac{a_{L}}{a_{M}a_{N}}.$$ Thus in terms of alternative generators $[[M]]=\frac{[M]}{a_M}$, the product takes the form
$$[[M]]\diamond [[N]]= \sum\limits_{[L] \in \Iso(\mathcal{A})}F_{MN}^L[[L]],$$
which is the definition used, for example, in \cite{R90a,Sc}. The associativity of Hall algebras amounts to the following identity
\begin{equation}\label{jiehe}\sum\limits_{[M]}F_{XY}^MF_{MZ}^L=\sum\limits_{[N] }F_{XN}^LF_{YZ}^N,\end{equation} for any objects $L,X,Y,Z\in\A$.
\end{remark}

Given objects $M,N \in \mathcal{A}$, one defines \begin{equation}\label{Euler form}\lr{M,N}:=\dim_k\Hom_{\A}(M,N)-\dim_k\Ext^1_{\A}(M,N),\end{equation}
which descends to give a bilinear form
$$\lr{\cdot ,\cdot }: K(\mathcal{A})\times K(\mathcal{A})\longrightarrow \mathbb{Z},$$
called the \emph{Euler form} of $\mathcal{A}$.

The \emph{twisted Hall algebra} $\H_{\tw}(\A)$ is the same module as $\H(\A)$ but with the twisted multiplication defined by
$$[M]*[N]=q^{\lr{M,N}}[M]\diamond[N].$$

\subsection{Morphism categories}\label{morphism}
Let $C_2(\A)$ be the category whose objects are morphisms $\xymatrix{M_{-1}\ar[r]^f&M_0}$ in $\A$, and each morphism from $\xymatrix{M_{-1}\ar[r]^f&M_0}$ to $\xymatrix{N_{-1}\ar[r]^g&N_0}$ is a pair $(u,v)$ of morphisms in $\A$ such that the following diagram
$$\xymatrix{M_{-1}\ar[r]^f\ar[d]_u&M_0\ar[d]^v\\
N_{-1}\ar[r]^g&N_0}$$ is commutative. Clearly, we may consider $C_2(\A)$ as an extension-closed subcategory of $C^b(\A)$ by identifying each morphism $\xymatrix{M_{-1}\ar[r]^f&M_0}$ with the complex $M_\bullet$ whose components of degrees $-1$ and $0$ are $M_{-1}$ and $M_0$, respectively, and other components are zero. In what follows, we also write $M_\bullet$ as a morphism $\xymatrix{M_{-1}\ar[r]^f&M_0.}$ Let $C_2(\P)\subset C_2(\A)$ be the extension-closed subcategory consisting of morphisms in $\P$. Denote by $C^b(\P)\subset C^b(\A)$ the subcategory consisting of bounded complexes over projectives, and denote by $K^b(\P)$ its homotopy category. Since $\A$ is hereditary, it is well known that $K^b(\P)$ is equivalent to $D^b(\A)$ as triangulated categories. For any objects $M_\bullet$ and $N_\bullet$ in $C_2(\P)$, by \cite[Lemma 3.1]{Gor2}, we know that
\begin{equation}
\begin{split}
\Ext^1_{C_2(\P)}(M_\bullet,N_\bullet)&\cong\Ext^1_{C^b(\P)}(M_\bullet,N_\bullet)\\
&\cong\Hom_{K^b(\P)}(M_\bullet,N_\bullet[1])\\
&\cong\Hom_{D^b(\A)}(M_\bullet,N_\bullet[1]).
\end{split}
\end{equation}

For each object $P\in\P$, define two objects in $C_2(\P)$
\begin{equation}K_P:=\xymatrix{P\ar[r]^1&P}~~\text{and}~~Z_P:=\xymatrix{P\ar[r]&0.}\end{equation}

We have the following well known result:
\begin{lemma}\cite[Lemma 4.1]{Bri13}\label{jx}
Given $M\in\A$, each projective resolution of $M$ is isomorphic to a resolution of the form
\begin{equation*}
\xymatrix{0\ar[r]&\Omega_M\oplus R\ar[r]^{\delta_M\oplus1}&P_M\oplus R\ar[r]&M\ar[r]&0,}
\end{equation*}
for some $R\in\P$ and some minimal projective resolution\footnote{The notations $P_M$ and $\Omega_M$ will be used throughout the paper.}
\begin{equation}\label{mpr}
\xymatrix{0\ar[r]&\Omega_M\ar[r]^{\delta_M}&P_M\ar[r]&M\ar[r]&0.}
\end{equation}
\end{lemma}

Given an object $M\in\A$, take a minimal projective resolution (\ref{mpr}). Then we define an object in $C_2(\P)$
\begin{equation}\label{cm}
C_M:=\xymatrix{\Omega_M\ar[r]^{\delta_M}&P_M.}
\end{equation}
By Lemma \ref{jx}, we know that any two minimal projective resolutions of $M$ are isomorphic, so $C_M$ is well defined up to isomorphism.

Now we describe the indecomposable objects in $C_2(\P)$ in the following
\begin{proposition}\label{fenjie}
Each object $M_\bullet$ in $C_2(\P)$ has a direct sum decomposition
$$M_\bullet=K_P\oplus Z_Q\oplus C_M$$
for some $P, Q\in\P$ and $M\in\A$. Moreover, the objects $P, Q$ and $M$ are uniquely determined up to isomorphism.
\end{proposition}
\begin{proof}
Clearly, $C_2(\A)$ is an abelian category, and thus idempotents split in $C_2(\A)$. For any $Z_\bullet\in C_2(\P)$ and idempotent $e:Z_\bullet\to Z_\bullet$, we obtain in $C_2(\A)$ that $Z_\bullet=X_\bullet\oplus Y_\bullet$. It is clear that
$X_\bullet$ and $Y_\bullet$ are in $C_2(\P)$. That is, idempotents split in $C_2(\P)$.
Moreover, since $C_2(\P)$ is Hom-finite, we conclude that $C_2(\P)$ is a Krull--Schmidt category. Now,
we assume that $M_\bullet=\xymatrix{M_{-1}\ar[r]^f&M_0}$ is a nonzero indecomposable object.

If $\Ker(f)=0$, then we have a short exact sequence
$$\xymatrix{0\ar[r]&M_{-1}\ar[r]^f&M_0\ar[r]&M\ar[r]&0,}$$ where $M=\Coker(f)$. By Lemma \ref{jx}, $M_\bullet\cong C_M\oplus K_P$ for some $P\in\P$. Since $M_\bullet$ is indecomposable, $M_\bullet\cong C_M$ if $M\neq0$; otherwise, $M_\bullet\cong K_P$.

If $\Ker(f)\neq0$, consider the short exact sequence
\begin{equation}\label{KI}\xymatrix{0\ar[r]&\Ker(f)\ar[r]^i&M_{-1}\ar[r]^{\pi}&\im(f)\ar[r]&0.}\end{equation}
Since $\A$ is hereditary, we know that $\im(f)$ is projective, and thus the sequence (\ref{KI}) is splitting. That is, there exists a morphism $r:M_{-1}\to\Ker(f)$ such that $r\circ i=\id.$ Then we have morphisms in $C_2(\P)$
\begin{equation*}\xymatrix{\Ker(f)\ar[r]\ar[d]^i&0\ar[d]\\
M_{-1}\ar[r]^f&M_0,}\quad\quad \xymatrix{M_{-1}\ar[r]^f\ar[d]^r&M_0\ar[d]\\
\Ker(f)\ar[r]&0.}\end{equation*}
Hence $Z_Q=\xymatrix{Q\ar[r]&0,}$ where $Q=\Ker(f)$, is a nonzero direct summand of $M_\bullet$. Since $M_\bullet$ is indecomposable, we obtain that $M_\bullet\cong Z_{Q}$.
\end{proof}

\begin{lemma}\label{hom}
For any $M_\bullet\in C_2(\P)$, $M,N\in\A$ and $P\in\P$, we have that
\begin{flalign}
&|\Hom_{C_2(\P)}(K_P,M_\bullet)|=|\Hom_{\A}(P,M_{-1})|;\label{v_1}\\
&|\Hom_{C_2(\P)}(M_\bullet,K_P)|=|\Hom_{\A}(M_0,P)|;\\
&|\Hom_{C_2(\P)}(C_P,M_\bullet)|=|\Hom_{\A}(P,M_0)|;\\
&|\Hom_{C_2(\P)}(M_\bullet,Z_P)|=|\Hom_{\A}(M_{-1},P)|;\label{v_4}\\
&|\Hom_{C_2(\P)}(C_M,C_N)|=|\Hom_{\A}(M,N)|\cdot|\Hom_{\A}(P_M,\Omega_N)|.\label{v_5}
\end{flalign}
\end{lemma}
\begin{proof}
The identities in $(\ref{v_1}$-$\ref{v_4})$ are taken from \cite[Proposition 3.1]{Bau}, and they can also be easily obtained by direct calculations. We only prove the identity in (\ref{v_5}).

By the comparison lemma in homological algebra, it is easy to get a surjective map
$$\xymatrix{\varphi:\Hom_{C_2(\P)}(C_M,C_N)\ar@{->>}[r]& \Hom_{\A}(M,N).}$$
Then by the universal property of kernels, we can obtain $\Ker(\varphi)\cong\Hom_{\A}(P_M,\Omega_N)$.
\end{proof}

\begin{lemma}\cite[Corollaries 3.1 and 3.2]{Bau}\label{proinj}
The objects $C_P$ and $K_P$, where $P\in\P$ is indecomposable, provide a complete set of indecomposable projective objects in $C_2(\mathscr{P})$; and the objects $Z_P$ and $K_P$ provide a complete set of indecomposable injective objects. Moreover, all $K_P$ are exactly the whole indecomposable projective-injective objects.
\end{lemma}

The existence of almost split sequences in $C_2(\P)$ was studied in \cite{Bau}, see also \cite{Bau2,Cha}. Let us give an example of the Auslander--Reiten quiver of $C_2(\P)$ as follows:
\begin{example}\cite[Example 6.7]{Cha}
Let $\A$ be the category of finite dimensional representations of the quiver
$$1\longrightarrow2\longrightarrow3.$$
Then the Auslander--Reiten quiver of $C_2(\P)$ is the following
$$\xymatrix{&&C_{P_1}\ar[rd]\ar@{.}[rr]&&Z_{P_3}\ar[rd]&&\\&C_{P_2}\ar@{.}[rr]\ar[ru]\ar[rd]&&C_{I_2}\ar@{.}[rr]\ar[ru]\ar[rd]&&Z_{P_2}\ar[rd]&\\C_{P_3}\ar[ru]\ar[rd]\ar@{.}[rr]&&C_{S_2}\ar[ru]\ar[rd]\ar@{.}[rr]&&C_{S_1}\ar[ru]\ar[rd]\ar@{.}[rr]&&Z_{P_1}.\\&K_{P_3}\ar[ru]&&K_{P_2}\ar[ru]&&K_{P_1}\ar[ru]&}$$
\end{example}

For each $M_\bullet\in C_2(\P)$, we have a projective resolution and an injective resolution of $M_\bullet$ in the following
\begin{lemma}\cite[Proposition 3.2]{Bau}\label{resol}
For each $M_\bullet\in C_2(\P)$, we have the following short exact sequences
\begin{flalign}
&0\longrightarrow C_{M_{-1}}\longrightarrow C_{M_0}\oplus K_{M_{-1}}\longrightarrow M_{\bullet}\longrightarrow0;\label{tfj}\\
&0\longrightarrow M_{\bullet}\longrightarrow Z_{M_{-1}}\oplus K_{M_0}\longrightarrow Z_{M_0}\longrightarrow0.
\end{flalign}
\end{lemma}

By Lemma \ref{resol}, we know that the global dimension of $C_2(\P)$ is equal to one. Similarly to the Euler form of $\A$ defined in Section 2.1, for any objects $M_\bullet,N_\bullet\in C_2(\P)$, we define
\begin{equation}
\lr{M_\bullet,N_\bullet}:=\dim_k\Hom_{C_2(\P)}(M_\bullet,N_\bullet)-\dim_k\Ext_{C_2(\P)}^1(M_\bullet,N_\bullet).
\end{equation}
It also induces a bilinear form on the Grothendieck group of $C_2(\P)$. Here we use the same notation as for the Euler form on $\A$, since this should not cause confusion by the context.

\begin{lemma}
For any $M_\bullet,N_\bullet\in C_2(\P)$,
\begin{equation*}\lr{M_\bullet,N_\bullet}=\dim_k\Hom_{\A}(M_0,N_0)+\dim_k\Hom_{\A}(M_{-1},N_{-1})-\dim_k\Hom_{\A}(M_{-1},N_0).\end{equation*}
\end{lemma}
\begin{proof}
By Lemma \ref{resol}, we have a projective resolution (\ref{tfj}) of $M_{\bullet}$. Applying the functor $\Hom_{C_2(\P)}(-,N_\bullet)$ to the short exact sequence (\ref{tfj}), we obtain a long exact sequence
$$0\to\Hom(M_\bullet,N_\bullet)\to\Hom(C_{M_0}\oplus K_{M_{-1}},N_\bullet)\to\Hom(C_{M_{-1}},N_\bullet)\to\Ext^1(M_\bullet,N_\bullet)\to0,$$ since $C_{M_0}\oplus K_{M_{-1}}$ is projective.
Hence,
\begin{flalign*}
\lr{M_\bullet,N_\bullet}&=\dim_k\Hom_{C_2(\P)}(M_\bullet,N_\bullet)-\dim_k\Ext_{C_2(\P)}^1(M_\bullet,N_\bullet)\\
&=\dim_k\Hom_{C_2(\P)}(C_{M_0}\oplus K_{M_{-1}},N_{\bullet})-\dim_k\Hom_{C_2(\P)}(C_{M_{-1}},N_{\bullet}).
\end{flalign*}
Thus, by Lemma \ref{hom}, we complete the proof.
\end{proof}

\begin{lemma}\label{ext}
For any $M,N\in\A$ and $P\in\P$, there are the following
isomorphisms of vector spaces
\begin{flalign}
&\Ext^1_{C_2(\P)}(C_M,C_N)\cong\Ext^1_{\A}(M,N);\\
&\Ext^1_{C_2(\P)}(Z_P,C_M)\cong\Hom_{\A}(P,M)\label{e2}.
\end{flalign}
\end{lemma}
\begin{proof}
Firstly,
\begin{flalign*}\Ext^1_{C_2(\P)}(C_M,C_N)&\cong\Ext^1_{C^b(\P)}(C_M,C_N)\\
&\cong\Hom_{K^b(\P)}(C_M,C_N[1])\\
&\cong\Hom_{D^b(\A)}(M,N[1])\\
&\cong\Ext^1_{\A}(M,N).\end{flalign*}

Secondly,
\begin{flalign*}\Ext^1_{C_2(\P)}(Z_P,C_M)&\cong\Ext^1_{C^b(\P)}(Z_P,C_M)\\
&\cong\Hom_{K^b(\P)}(Z_P,C_M[1])\\
&\cong\Hom_{D^b(\A)}(P[1],M[1])\\
&\cong\Hom_{\A}(P,M).\end{flalign*}
\end{proof}
\begin{proposition}\label{kuozhang0}
Let $M_\bullet, N_\bullet\in C_2(\P)$. Then
\begin{equation}
\Ext^1_{C_2(\P)}(M_\bullet, N_\bullet)\cong\Hom_{\A}(H_{-1}(M_\bullet),H_{0}(N_\bullet))
\oplus\Ext^1_{\A}(H_{0}(M_\bullet),H_{0}(N_\bullet)).
\end{equation}
\end{proposition}
\begin{proof}
By Proposition \ref{fenjie}, we write $M_\bullet=K_{P}\oplus Z_Q\oplus C_M$ and $N_\bullet=K_{T}\oplus Z_W\oplus C_N$ for some $P,Q,T,W\in\P$ and $M,N\in\A$. Then
\begin{flalign*}
\Ext^1_{C_2(\P)}(M_\bullet, N_\bullet)&=\Ext^1_{C_2(\P)}(K_{P}\oplus Z_Q\oplus C_M,K_{T}\oplus Z_W\oplus C_N)\\
&\cong\Ext^1_{C_2(\P)}(Z_Q\oplus C_M,C_N)\\
&\cong\Hom_{\A}(Q,N)\oplus\Ext_{\A}^1(M,N)\\
&\cong\Hom_{\A}(H_{-1}(M_\bullet),H_{0}(N_\bullet))
\oplus\Ext^1_{\A}(H_{0}(M_\bullet),H_{0}(N_\bullet)).
\end{flalign*}
\end{proof}

\section{Hall algebras of morphisms}
Let $\H(C_2(\A))$ be the Hall algebra of the abelian category $C_2(\A)$ as defined in Definition \ref{Hall algebra of abelian category}. Let $\H(C_2(\P))$ be the submodule of $\H(C_2(\A))$ spanned by the isomorphism classes of objects in $C_2(\P)$. Since $C_2(\P)$ is closed under extensions, $\H(C_2(\P))$ is a subalgebra of the Hall algebra $\H(C_2(\A))$.
Define $\H_{\tw}(C_2(\P))$ to be the same module as $\H(C_2(\P))$, but with the twisted multiplication
$$[M_\bullet]\ast[N_\bullet]=q^{\lr{M_\bullet,N_\bullet}}[M_\bullet]\diamond[N_\bullet].$$

\begin{lemma}\label{KM}
For any $M_\bullet\in C_2(\P)$ and $P\in\P$, we have that in $\H_{\tw}(C_2(\P))$
\begin{equation}\label{jh}
[K_P]\ast[M_\bullet]=[M_\bullet]\ast[K_P]=[K_P\oplus M_\bullet].
\end{equation}
\end{lemma}
\begin{proof}
Since $K_P$ is projective-injective,
\begin{flalign*}
[K_P]\ast[M_\bullet]&=q^{\lr{K_P,M_\bullet}}\frac{1}{|\Hom_{C_2(\P)}(K_P,M_\bullet)|}[K_P\oplus M_\bullet]\\
&=[K_P\oplus M_\bullet].
\end{flalign*}
Similarly, $[M_\bullet]\ast[K_P]=[K_P\oplus M_\bullet].$ Thus, we complete the proof.
\end{proof}
\begin{lemma}\label{ybasis}
For any $P,Q\in\P$ and $M\in\A$, we have that in $\H_{\tw}(C_2(\P))$
$$[K_P\oplus Z_Q\oplus C_M]=[K_P]\ast[C_M]\ast[Z_Q].$$
\end{lemma}
\begin{proof}
Since $\Ext_{C_2(\P)}^1(C_M,Z_Q)=0$, we obtain that
\begin{flalign*}[C_M]\ast[Z_Q]&=q^{\lr{C_M,Z_Q}}\frac{1}{|\Hom_{C_2(\P)}(C_M,Z_Q)|}[C_M\oplus Z_Q]\\&=[C_M\oplus Z_Q].\end{flalign*}
Thus, by Lemma \ref{KM}, $$[K_P\oplus Z_Q\oplus C_M]=[K_P]\ast[C_M\oplus Z_Q]=[K_P]\ast[C_M]\ast[Z_Q].$$
\end{proof}

Define the {\em localized Hall algebra} $\M\H(\A)$ to be the localization of $\H_{\tw}(C_2(\P))$ with respect to all elements $[K_P]$. In symbols,
$$\M\H(\A):=\H_{\tw}(C_2(\P))[[K_P]^{-1}: P\in\P].$$

For each $\alpha\in K(\A)$, by writing $\alpha=\hat{P}-\hat{Q}$ for some objects $P,Q\in\P$, we define
$$K_{\alpha}:=[K_P]\ast[K_Q]^{-1}.$$ Then for any $\alpha,\beta\in K(\A)$ and $M_\bullet\in C_2(\P)$, \begin{flalign}&K_\alpha\ast K_\beta=K_{\alpha+\beta}=K_\beta\ast K_\alpha;\label{KK}\\
&K_\alpha\ast[M_\bullet]=[M_\bullet]\ast K_\alpha.\label{KX}\end{flalign}

For each $M\in\A$, we define
$$\X_M:=K_{-\hat{P}_M}\ast[C_M]\in\M\H(\A).$$
Suppose we take a different, not necessarily minimal, projective resolution (\ref{mpr}), and consider the corresponding object, denoted by $C'_M$, in $C_2(\P)$. By Lemma \ref{jx}, $[C'_M]=[K_R\oplus C_M]$ for some $R\in\P$. Thus
\begin{flalign*}K_{-(\hat{P}_M+\hat{R})}\ast[C'_M]&=K_{-(\hat{P}_M+\hat{R})}\ast K_{\hat{R}}\ast [C_M]\\&=
K_{-\hat{P}_M}\ast [C_M]\\&=\X_M.
\end{flalign*}
In other word, $\X_M$ does not depend on the minimality of projective resolutions of $M$.

Given $M\in\A$ and $P\in\P$, we define
\begin{flalign*}\X_{M\oplus P[1]}:=\X_M\ast[Z_P]=K_{-\hat{P}_M}\ast[C_M]\ast[Z_P]=K_{-\hat{P}_M}\ast[C_M\oplus Z_P].
\end{flalign*}
In particular, \begin{equation}\label{XMP}\X_{P[1]}=[Z_P]~~\text{and}~~\X_{M\oplus P[1]}=\X_M\ast \X_{P[1]}.\end{equation}
\begin{proposition}\label{basis}
The algebra $\M\H(\A)$ has a basis consisting of elements
\begin{equation*}K_{\alpha}\ast \X_M\ast \X_{P[1]},\end{equation*}where $\alpha\in K(\A)$, $M\in\A$ and $P\in\P$.
\end{proposition}
\begin{proof}
It can be easily proved by Proposition \ref{fenjie} and Lemma \ref{ybasis}.
\end{proof}
\begin{lemma}\label{XP}
For any $P,Q\in\P$, we have that in $\M\H(\A)$
\begin{equation}\X_{P[1]}\ast \X_{Q[1]}=\X_{Q[1]}\ast \X_{P[1]}=\X_{(P\oplus Q)[1]}.\end{equation}
\end{lemma}
\begin{proof}
By definition,
\begin{flalign*}
\X_{P[1]}\ast \X_{Q[1]}&=q^{\lr{Z_P,Z_Q}}\frac{1}{|\Hom_{C_2(\P)}(Z_P,Z_Q)|}\X_{(P\oplus Q)[1]}\\
&=\X_{(P\oplus Q)[1]}.
\end{flalign*}
\end{proof}
\begin{theorem}\label{main1}
There exists an embedding of algebras
$$\xymatrix{\Psi:\H_{\tw}(\A)\ar@{^{(}->}[r]&\M\H(\A),} \xymatrix{[M]\ar@{|->}[r]&\X_M.}$$
\end{theorem}
\begin{proof}
By Proposition \ref{basis}, clearly, $\Psi$ is injective. We only need to prove that $\Psi$ is a homomorphism of algebras.

By definition,
\begin{flalign*}
\X_M\ast \X_N&=K_{-\hat{P}_M}\ast[C_M]\ast K_{-\hat{P}_N}\ast[C_N]\\
&=K_{-(\hat{P}_M+\hat{P}_N)}\ast[C_M]\ast[C_N].
\end{flalign*}
By Lemma \ref{ext}, $$\Ext^1_{C_2(\P)}(C_M,C_N)\cong\Ext^1_{\A}(M,N).$$ Moreover, by the Horse-Shoe Lemma (cf. \cite{Weibel}), any extension of $C_M$ by $C_N$ is the object $C'_L$ defined by the corresponding extension $L$ of $M$ by $N$. Hence,
$$|\Ext^1_{C_2(\P)}(C_M,C_N)_{C'_L}|=|\Ext^1_{\A}(M,N)_L|.$$

Thus,
\begin{flalign*}
\X_M\ast \X_N&=q^{\lr{C_M,C_N}}\sum\limits_{[L]}\frac{|\Ext^1_{C_2(\P)}(C_M,C_N)_{C'_L}|}{|\Hom_{C_2(\P)}(C_M,C_N)|} K_{-(\hat{P}_M+\hat{P}_N)}\ast[C'_L]\\
&=\frac{|\Hom_{C_2(\P)}(C_M,C_N)|}{|\Ext^1_{C_2(\P)}(C_M,C_N)|}\sum\limits_{[L]}\frac{|\Ext^1_{\A}(M,N)_L|}{|\Hom_{C_2(\P)}(C_M,C_N)|} \X_L\\
&=\sum\limits_{[L]}\frac{|\Ext^1_{\A}(M,N)_L|}{|\Ext^1_{C_2(\P)}(C_M,C_N)|} \X_L\\
&=\sum\limits_{[L]}\frac{|\Ext^1_{\A}(M,N)_L|}{|\Ext^1_{\A}(M,N)|} \X_L\\
&=\frac{|\Hom_{\A}(M,N)|}{|\Ext^1_{\A}(M,N)|}\sum\limits_{[L]}\frac{|\Ext^1_{\A}(M,N)_L|}{|\Hom_{\A}(M,N)|} \X_L\\
&=q^{\lr{M,N}}\sum\limits_{[L]}\frac{|\Ext^1_{\A}(M,N)_L|}{|\Hom_{\A}(M,N|} \X_L.
\end{flalign*}
That is, $\Psi([M])\ast\Psi([N])=\Psi([M]\ast[N])$, and we complete the proof.
\end{proof}

\section{A multiplication formula in $\M\H(\A)$}
Given $M\in\A$ and $P\in\P$, consider an extension of $C_M$ by $Z_P$
\begin{equation}\label{eta}\eta: 0\longrightarrow C_M\longrightarrow X_{\bullet}\longrightarrow Z_P\longrightarrow0.\end{equation}
It induces a long exact sequence in homology
$$H_{-1}(C_M)\longrightarrow H_{-1}(X_\bullet)\longrightarrow H_{-1}(Z_P)\longrightarrow H_0(C_M)\longrightarrow H_0(X_{\bullet})\longrightarrow H_0(Z_P).$$
Since $H_{-1}(C_M)=H_0(Z_P)=0$, $H_{-1}(Z_P)\cong P$ and $H_0(C_M)\cong M$, we obtain the exact sequence
$$0\longrightarrow H_{-1}(X_\bullet)\longrightarrow P\longrightarrow M\longrightarrow H_0(X_{\bullet})\longrightarrow0.$$
Writing $X_\bullet=K_{T}\oplus Z_{Q}\oplus C_B$ for some $T,Q\in\P$ and $B\in\A$, we get the following exact sequence
$$\xymatrix{0\ar[r]& Q\ar[r]& P\ar[r]^-{\delta}& M\ar[r]& B\ar[r]& 0,}$$
where $\delta$ is determined by the equivalence class of $\eta$ via the isomorphism
\begin{equation}\label{tg}
\Ext^1_{C_2(\P)}(Z_P,C_M)\cong\Hom_{\A}(P,M).\end{equation}
By the short exact sequence (\ref{eta}), we obtain that
\begin{equation}P_B\oplus T\cong P_M~~\text{and}~~\Omega_B\oplus Q\oplus T\cong P\oplus\Omega_M.\end{equation}
It is easy to see that the isomorphism (\ref{tg}) induces that
\begin{equation}\label{kuozhang}\Ext^1_{C_2(\P)}(Z_P,C_M)_{K_T\oplus Z_Q\oplus C_B}\cong {}_Q\Hom_{\A}(P,M)_B,\end{equation}
where ${}_Q\Hom_{\A}(P,M)_B:=\{f:P\to M~|~\Ker(f)\cong Q~~\text{and}~~\Coker(f)\cong B\}$. By \cite{Vanden,ZHC},
\begin{equation}\label{xjs}|{}_Q\Hom_{\A}(P,M)_B|=\sum\limits_{[L]}a_LF^P_{LQ}F^M_{BL}.\end{equation}

\begin{theorem}\label{main2}
Given $M\in\A$ and $P\in\P$, we have that in $\M\H(\A)$
\begin{flalign*}\X_{P[1]}\ast \X_M&=q^{-\lr{P,M}}\sum\limits_{[B],[Q]}|{}_Q\Hom_{\A}(P,M)_B| \X_B\ast \X_{Q[1]}\\
&=q^{-\lr{P,M}}\sum\limits_{[B],[Q]}|{}_Q\Hom_{\A}(P,M)_B| \X_{B\oplus Q[1]}.\end{flalign*}
\end{theorem}
\begin{proof}
By definition,
\begin{flalign*}
\X_{P[1]}\ast \X_M&=[Z_P]\ast K_{-\hat{P}_M}\ast[C_M]\\
&=K_{-\hat{P}_M}\ast[Z_P]\ast[C_M]\\
&=q^{\lr{Z_P,C_M}}\sum\limits_{[B],[T],[Q]}\frac{|\Ext^1_{C_2(\P)}(Z_P,C_M)_{K_T\oplus Z_Q\oplus C_B}|}{|\Hom_{C_2(\P)}(Z_P,C_M)|} K_{-\hat{P}_M}\ast[K_T\oplus Z_Q\oplus C_B].
\end{flalign*}
Since $\Hom_{C_2(\P)}(Z_P,C_M)=0$ and then $\lr{Z_P,C_M}=-\lr{P,M}$, by (\ref{kuozhang}), we obtain that
\begin{flalign*}
\X_{P[1]}\ast \X_M
&=q^{-\lr{P,M}}\sum\limits_{[B],[T],[Q]}|{}_Q\Hom_{\A}(P,M)_B| K_{-\hat{P}_M}\ast[K_T\oplus Z_Q\oplus C_B]\\
&=q^{-\lr{P,M}}\sum\limits_{[B],[T],[Q]}|{}_Q\Hom_{\A}(P,M)_B| K_{-\hat{P}_M}\ast K_{\hat{T}}\ast[C_B]\ast[Z_Q]\\
&=q^{-\lr{P,M}}\sum\limits_{[B],[Q]}|{}_Q\Hom_{\A}(P,M)_B| K_{-\hat{P}_B}\ast[C_B]\ast[Z_Q]\\
&=q^{-\lr{P,M}}\sum\limits_{[B],[Q]}|{}_Q\Hom_{\A}(P,M)_B| \X_B\ast \X_{Q[1]}\\
&=q^{-\lr{P,M}}\sum\limits_{[B],[Q]}|{}_Q\Hom_{\A}(P,M)_B| \X_{B\oplus Q[1]}.
\end{flalign*}
\end{proof}

\begin{corollary}
Given $M\in\A$ and $P\in\P$, if $\Hom_{\A}(P,M)=0$, then in $\M\H(\A)$
$$\X_{P[1]}\ast \X_M=\X_M\ast \X_{P[1]}.$$
\end{corollary}

\section{Generators in $\M\H(\A)$}
Let $\Ind(\A)$ be a complete set of indecomposable objects in $\A$,
and let $\Ind(\P)\subset\Ind(\A)$ be a subset consisting of indecomposable projective objects. Define $K_{\geq0}(\A)$ to be the positive cone in $K(\A)$, which consists of classes of objects in $\A$, rather than formal differences of such. We assume that for each nonzero object $M\in\A$, $\hat{M}\neq0$. This condition is equivalent to the statement that the rule
$$\alpha\leq\beta~~\Leftrightarrow~~\beta-\alpha\in K_{\geq0}(\A)$$ defines a partial order on $K(\A)$. It holds, for example, if $\A$ has finite length.

\begin{proposition}\label{generators}
The set $\{K_{{P}}^{\pm},\X_M,\X_{P[1]}~|~M\in\Ind(\A)~~\text{and}~~P\in\Ind(\P)\}$ is a generating set of $\M\H(\A)$.
\end{proposition}
\begin{proof}
For any short exact sequence in $C_2(\P)$
$$0\longrightarrow N_\bullet\longrightarrow L_\bullet \longrightarrow M_\bullet\longrightarrow0,$$
it induces a long exact sequence in homology
{\footnotesize\begin{equation}\label{f}\xymatrix{0\to H_{-1}(N_\bullet)\ar[r]& H_{-1}(L_\bullet)\ar[r]& H_{-1}(M_\bullet)\ar[r]^-f& H_{0}(N_\bullet)\ar[r]& H_{0}(L_\bullet)\ar[r]& H_{0}(M_\bullet)\to0.}\end{equation}}
Setting $Y:=\im(f)$, we obtain two exact sequences
$$\xymatrix{0\to H_{-1}(N_\bullet)\ar[r]& H_{-1}(L_\bullet)\ar[r]& H_{-1}(M_\bullet)\ar[r]^-f&Y\to0}$$
and
$$\xymatrix{0\to Y\ar[r]&H_{0}(N_\bullet)\ar[r]& H_{0}(L_\bullet)\ar[r]& H_{0}(M_\bullet)\to0.}$$
Thus, \begin{equation}\begin{split}\label{filter}&\Dim H_{-1}(L_\bullet)+\Dim H_{0}(L_\bullet)=\\
&\Dim H_{-1}(M_\bullet)+\Dim H_{0}(M_\bullet)+\Dim H_{-1}(N_\bullet)+\Dim H_{0}(N_\bullet)-2\Dim Y.\end{split}\end{equation}

For any object $X_\bullet\in C_2(\P)$, define $\Dim X_\bullet:=\Dim H_{-1}(X_\bullet)+\Dim H_{0}(X_\bullet)$. Then (\ref{filter}) can be rewritten as \begin{equation}\label{filter2}\Dim L_\bullet=\Dim M_\bullet+\Dim N_\bullet-2\Dim Y.\end{equation} For each $\alpha\in K_{\geq0}(\A)$, let $\H_{tw}^{\leq\alpha}(C_2(\P))$ be the submodule of $\H_{tw}(C_2(\P))$ spanned by all $[X_\bullet]$ with $\Dim X_\bullet\leq\alpha$. By (\ref{filter2}),
$$\H_{tw}^{\leq\alpha}(C_2(\P))\ast \H_{tw}^{\leq\beta}(C_2(\P))\subseteq\H_{tw}^{\leq\alpha+\beta}(C_2(\P)).$$
That is, $\H_{tw}(C_2(\P))$ is a $K_{\geq0}(\A)$-filtered algebra.

By Proposition \ref{kuozhang0}, $$\Ext^1_{C_2(\P)}(M_\bullet, N_\bullet)\cong\Hom_{\A}(H_{-1}(M_\bullet),H_{0}(N_\bullet))
\oplus\Ext^1_{\A}(H_{0}(M_\bullet),H_{0}(N_\bullet)).$$
If $f$ in (\ref{f}) is nonzero, then $\Dim L_\bullet<\Dim M_\bullet+\Dim N_\bullet$;
otherwise, $\Dim L_\bullet=\Dim M_\bullet+\Dim N_\bullet$ and we have the short exact sequence $$\xi: 0\longrightarrow H_0(N_\bullet)\longrightarrow H_0(L_\bullet)\longrightarrow H_0(M_\bullet)\longrightarrow0.$$ Furthermore, $$\dim_k\End_{\A}(H_0(L_\bullet))\leq\dim_k\End_{\A}(H_0(M_\bullet)\oplus H_0(N_\bullet)),$$ and the equality holds if and only if $\xi$ is splitting.

Let us give an order on $K_{\geq0}(\A)\times\mathbb{N}$:
$$(\alpha_1,d_1)\leq(\alpha_2,d_2)\Longleftrightarrow``\alpha_1<\alpha_2"~ \text{or}~``\alpha_1=\alpha_2,d_1\leq d_2".$$

For any $X_\bullet\in C_2(\mathscr{P})$, set $\deg X_\bullet:=(\Dim X_\bullet,\dim_k\End_{\A}(H_0(X_\bullet)))$.
Then for any $M_\bullet,N_\bullet\in C_2(\mathscr{P})$,
$$[M_\bullet]\ast[N_\bullet]=a_{M_\bullet\oplus N_\bullet}[M_\bullet\oplus N_\bullet]+\sum\limits_{[L_\bullet]:\deg L_\bullet<\deg M_\bullet\oplus N_\bullet}a_{L_\bullet}[L_\bullet],$$
where $a_{M_\bullet\oplus N_\bullet}$ and all $a_{L_\bullet}$ belong to $\ZZ[{q}^{\pm\frac{1}{2}}]$.

By induction on the number of indecomposable direct summands, we prove that the elements in $\{[K_P],[C_M],[Z_P]~|~M\in\Ind(\A)~~\text{and}~~P\in\Ind(\P)\}$ generate the Hall algebra $\H_{\tw}(C_2(\P))$, and then we easily complete the proof.
\end{proof}

For use below we write down the defining relations of $\mathcal{MH}(\A)$ in the following
\begin{theorem}\label{ydygx}
The algebra $\mathcal{MH}(\A)$ is generated by all $K_{\alpha}$ and $\mathbb{X}_{M\oplus P[1]}$ $($with $\alpha\in K(\A)$, $M\in\A$ and $P\in\P$$)$,
which are subject to the following relations
\begin{equation}\label{h1}
\begin{split}K_{\alpha}\ast K_{\beta}=K_{\alpha+\beta}=K_{\beta}\ast K_{\alpha};\end{split}\end{equation}
\begin{equation}\label{h2}
\begin{split}K_{\alpha}\ast\mathbb{X}_{M\oplus P[1]}
    =\mathbb{X}_{M\oplus P[1]}\ast K_{\alpha};\end{split}\end{equation}
\begin{equation}\label{h3}
\begin{split}
\mathbb{X}_{P[1]}\ast \mathbb{X}_{Q[1]}=
\mathbb{X}_{(P\oplus Q)[1]}
=\mathbb{X}_{Q[1]}\ast\mathbb{X}_{P[1]};
\end{split}\end{equation}
\begin{equation}\label{h4}\mathbb{X}_{M}\ast\mathbb{X}_{N}=q^{\lr{M,N}}\sum_{[L]}\frac{|\mathrm{Ext}_{\A}^{1}(M,N)_{L}|}{|\mathrm{Hom}_{\A}(M,N)|}\mathbb{X}_L;\end{equation}
\begin{equation}\label{h5}\mathbb{X}_{M}\ast\mathbb{X}_{P[1]}=\mathbb{X}_{M\oplus
P[1]};\end{equation}
\begin{equation}\label{h6}\mathbb{X}_{P[1]}\ast\mathbb{X}_{M}=q^{-\lr{P,M}}
\sum\limits_{[B],[Q]}|{}_Q\Hom_{\A}(P,M)_B|\mathbb{X}_{B\oplus Q[1]};\end{equation}
for any $\alpha,\beta\in K(\A)$, $M,N\in\A$ and $P,Q\in\P$.
\end{theorem}
\begin{proof}
By Proposition \ref{basis},
$\mathcal{MH}(\A)$ is spanned by all elements $K_{\alpha}\ast\mathbb{X}_{M\oplus P[1]}=K_{\alpha}\ast\mathbb{X}_{M}\ast\mathbb{X}_{P[1]}$. That is, the multiplication map induces a triangular decomposition of $\mathcal{MH}(\A)$ as a module.
Using the triangular decomposition of $\mathcal{MH}(\A)$, we can easily prove that the relations $(\ref{h1}$-$\ref{h6})$ are the defining relations, which have been obtained in the previous sections.
\end{proof}

\section{Derived Hall algebras}
The derived Hall algebra of the bounded derived category $D^b(\A)$ of $\A$ was introduced in \cite{Toen2006} (see also \cite{XiaoXu}). By definition, the (Drinfeld dual) {\em derived Hall algebra} $\mathcal {D}\mathcal {H}(\A)$ is the free $\ZZ[{q}^{\pm\frac{1}{2}}]$-module with the basis $\{u_{X_\bullet}~|~X_\bullet\in \Iso(D^b(\A))\}$ and the multiplication defined by
\begin{equation}
u_{X_\bullet}\diamond u_{Y_\bullet}=\sum\limits_{[{Z_\bullet}]}\frac{|\Ext^1_{D^b(\A)}({X_\bullet},{Y_\bullet})_{Z_\bullet}|}{\prod\limits_{i\geq0}|\Hom_{D^b(\A)}({X_\bullet}[i],{Y_\bullet})|^{(-1)^i}} u_{Z_\bullet},
\end{equation}
where $\Ext^1_{D^b(\A)}({X_\bullet},{Y_\bullet})_{Z_\bullet}$ is defined to be $\Hom_{D^b(\A)}({X_\bullet},{Y_\bullet}[1])_{{Z_\bullet}[1]}$, which denotes the subset of $\Hom_{D^b(\A)}({X_\bullet},{Y_\bullet}[1])$ consisting of morphisms $f:{X_\bullet}\rightarrow {Y_\bullet}[1]$ whose cone is isomorphic to ${Z_\bullet}[1]$.

Let us  reformulate \cite[Proposition 7.1]{Toen2006} in the form of the Drinfeld dual derived Hall algebras
as the following
\begin{proposition}{\rm(\cite{Toen2006})}
$\mathcal {D}\mathcal {H}(\A)$ is an associative unital algebra generated by the elements in $\{u_{M[i]}~|~M\in\Iso(\A),~i\in \mathbb{Z}\}$ and the following relations
\begin{flalign}
&u_{M[i]}\diamond u_{N[i]}=\sum\limits_{[L]}{\frac{{|\Ext_\mathcal{A}^1{{(M,N)}_L}|}}{{|\Hom_\mathcal{A}(M,N)|}}} u_{L[i]};\\
&u_{M[i+1]}\diamond u_{N[i]}=\sum\limits_{[X],[Y]}q^{-\lr{{Y},{X}}}|{}_X\Hom_{\A}(M,N)_Y| u_{Y[i]}\diamond u_{X[i+1]};\\
&u_{M[i]}\diamond u_{N[j]}=q^{(-1)^{i-j}\lr{N,M}} u_{N[j]}\diamond u_{M[i]}, \quad i-j>1.
\end{flalign}
\end{proposition}

For any ${X_\bullet},{Y_\bullet}\in D^b(\A)$, define
\begin{equation*}
\lr{{X_\bullet},{Y_\bullet}}:=\sum\limits_{i\in\mathbb{Z}}(-1)^i\dim_k\Hom_{D^b(\A)}({X_\bullet},{Y_\bullet}[i]),
\end{equation*}
it also descends to give a bilinear form on the Grothendieck group of $D^b(\A)$. Moreover, it coincides with the Euler form of $K(\A)$ over the objects in $\A$. In particular, for any $M,N\in\A$ and $i,j\in\mathbb{Z}$, we have that $\lr{M[i],N[j]}=(-1)^{i-j}\lr{M,N}$.

Let us twist the multiplication in $\mathcal {D}\mathcal {H}(\A)$ as follows (cf. \cite{SX}):
\begin{equation}u_{X_\bullet}\ast u_{Y_\bullet}=q^{\lr{{X_\bullet},{Y_\bullet}}} u_{X_\bullet}\diamond u_{Y_\bullet}\end{equation}
for any ${X_\bullet},{Y_\bullet}\in D^b(\A)$.
The \emph{twisted derived Hall algebra} $\mathcal {D}\mathcal {H}_{q}(\A)$ is the same module as $\mathcal {D}\mathcal {H}(\A)$, but with the twisted multiplication. Then we have the following
\begin{proposition}\label{twistderived}
$\mathcal {D}\mathcal {H}_q(\A)$ is an associative unital algebra generated by the elements in $\{u_{M[i]}~|~M\in\Iso(\A),~i\in \mathbb{Z}\}$ and the following relations
\begin{flalign}
&u_{M[i]}\ast u_{N[i]}=q^{\lr{M,N}}\sum\limits_{[L]}{\frac{{|\Ext_\mathcal{A}^1{{(M,N)}_L}|}}{{|\Hom_\mathcal{A}(M,N)|}}}u_{L[i]};\\
&u_{M[i+1]}\ast u_{N[i]}=q^{-\lr{M,N}}\sum\limits_{[X],[Y]}|{}_X\Hom_{\A}(M,N)_Y| u_{Y[i]}\ast u_{X[i+1]};\\
&u_{M[i]}\ast u_{N[j]}=q^{(-1)^{i-j}\lr{M,N}} u_{N[j]}\ast u_{M[i]}, \quad i-j>1.
\end{flalign}
\end{proposition}
\begin{remark}
For any $M,N\in\A$ and $i\in\mathbb{Z}$, we have that
$$u_{M[i]}\ast u_{N[i+1]}=u_{M[i]\oplus N[i+1]}.$$
In fact, since $\A$ is hereditary, we obtain that \begin{flalign*}\Hom_{D^b(\A)}(M[i],N[i+1][1])\cong\Ext^2_{\A}(M,N)=0.\end{flalign*}
Thus, \begin{flalign*}u_{M[i]}\diamond u_{N[i+1]}&=\frac{1}{\prod\limits_{j\geq0}|\Hom_{D^b(\A)}(M[i+j],N[i+1])|^{(-1)^j}}u_{M[i]\oplus N[i+1]}\\&=q^{\lr{M,N}} u_{M[i]\oplus N[i+1]}.\end{flalign*}
Hence, $$u_{M[i]}\ast u_{N[i+1]}=q^{-\lr{M,N}} u_{M[i]}\diamond u_{N[i+1]}=u_{M[i]\oplus N[i+1]}.$$
\end{remark}
In order to compare with the subsequent cluster multiplication formulas, we give the following
\begin{corollary}
Given objects $M\in\A$, $P\in\P$ and an injective object $I\in\A$, we have that
\vspace{0.2cm}
\begin{itemize}
\item[(1)] $u_{M}\ast u_{I[-1]}=q^{-\lr{M,I}}\sum\limits_{[B],[I']}|{}_B\Hom_{\A}(M,I)_{I'}| u_{B\oplus I'[-1]};$
\vspace{0.4cm}
\item[(2)] $u_{P[1]}\ast u_M=q^{-\lr{P,M}}\sum\limits_{[B],[Q]}|{}_Q\Hom_{\A}(P,M)_B| u_{B\oplus Q[1]}$.
\end{itemize}
\end{corollary}

Let $\mathbb{T}(\A)$ be the group algebra of the Grothendieck group $K(\A)$ over $\ZZ[{q}^{\pm\frac{1}{2}}]$. For each $\alpha\in K(\A)$, we denote by $\T_\alpha$ the element in $\mathbb{T}(\A)$ corresponding to $\alpha$, thus $\T_\alpha\ast\T_\beta=\T_{\alpha+\beta}$. We equip the module $\mathcal {D}\mathcal {H}_q(\A)\otimes_{\ZZ[{q}^{\pm\frac{1}{2}}]}\mathbb{T}(\A)$ with the structure of an algebra (containing $\mathcal {D}\mathcal {H}_q(\A)$ and $\mathbb{T}(\A)$ as subalgebras) by imposing the relations
$\T_\alpha\ast x=x\ast\T_\alpha$ for any $\alpha\in K(\A)$ and $x\in \mathcal {D}\mathcal {H}_q(\A)$, and denote this algebra by $\mathcal {D}\mathcal {H}^{\e}_q(\A)$. We remark that $\mathcal {D}\mathcal {H}^{\e}_q(\A)$ is just the tensor algebra of $\mathcal {D}\mathcal {H}_q(\A)$ and $\mathbb{T}(\A)$.
\begin{theorem}\label{mor_derived}There exists an embedding of algebras $$\xymatrix{\Phi:\M\H(\A)\ar@{^{(}->}[r]&
\mathcal{D}\mathcal{H}_q^{\e}(\A)}$$
defined on generators by
$$K_\alpha\mapsto\T_\alpha,\quad\X_M\mapsto u_M\quad\text{and}\quad \X_{P[1]}\mapsto u_{P[1]}$$
for $\alpha\in K(\A)$, $M\in\A$ and $P\in\P$.
\end{theorem}
\begin{proof}
By Proposition \ref{twistderived} and the definition of $\mathcal{D}\mathcal{H}_q^{\e}(\A)$, we know that
all the relations in Theorem \ref{ydygx} are preserved under $\Phi$, so $\Phi$ is a homomorphism of algebras.
The injectivity of $\Phi$ follows from the fact that $\Phi$ sends the basis
$$\{K_\alpha\ast\X_{M\oplus P[1]}~|~\alpha\in K(\A),M\in\A,P\in\P\}$$
of $\M\H(\A)$ to a linearly independent set in $\mathcal{D}\mathcal{H}_q^{\e}(\A)$.
\end{proof}

\section{Cluster multiplication formulas for acyclic quivers}
In this section, we recall the definitions of the quantum cluster algebra and quantum Caldero--Chapoton map, and give the multiplication formulas of quantum cluster characters.

\subsection{Quantum cluster algebras}
Let $\Lambda$ be an $m\times m$ skew-symmetric integral matrix, and denote by $\{{\bf e}_1,\cdots,{\bf e}_m\}$ the standard basis of $\mathbb{Z}^{m}$. Let $\mathfrak{q}$ be a formal variable
and $\ZZ[\mathfrak{q}^{\pm\frac{1}{2}}]$ be the ring of integral Laurent polynomials.
Define the  quantum torus  associated to the pair
$(\mathbb{Z}^{m},\Lambda)$ to be the $\ZZ[\mathfrak{q}^{\pm\frac{1}{2}}]$-algebra $\mathcal{T}_{\mathfrak{q}}$ with
a distinguished basis $\{X^{\bf e}: {\bf e}\in \mathbb{Z}^{m}\}$ and the
multiplication given by
\[X^{\bf e}X^{\bf f}=\mathfrak{q}^{\Lambda({\bf e},{\bf f})/2}X^{{\bf e+f}},\]
where we still denote by $\Lambda$ the skew-symmetric bilinear form on $\mathbb{Z}^{m}$ associated to the skew-symmetric matrix $\Lambda$. It is well-known that $\mathcal{T}_{\mathfrak{q}}$ is an Ore domain, and thus is contained in its
skew-field of fractions $\mathcal{F}_{\mathfrak{q}}$.

Let $\tilde{B}=(b_{ij})$ be an $m\times n$ integral matrix with $n\le m$.  We call the pair
$(\Lambda, \tilde{B})$ {\em compatible} if $\tilde{B}^{tr}\Lambda=(D|0)$ for some
$D=diag(d_1,\cdots,d_n)$, where each $d_i$ is a positive integer and  $\tilde{B}^{tr}$ denotes the transpose of  $\tilde{B}$. An {\em initial  quantum seed} for $\mathcal{F}_{\mathfrak{q}}$ is a triple
$(\Lambda, \tilde{B}, X)$  consisting of a compatible pair $(\Lambda,\tilde{B})$ and the set  $X=\{X_1,\cdots,X_m\}$, where each $X_i$ denotes $X^{{\bf e_i}}$. For any $1\leq k\leq n$, we  define the mutation of $(\Lambda, \tilde{B}, X)$ in direction $k$ to obtain the new quantum seed $(\Lambda',\tilde{B}',X')$ as follows:

(1)\ $\Lambda'=E^{tr}\Lambda E$, where the
$m\times m$ matrix $E=(e_{ij})$ is given by
\[e_{ij}=\begin{cases}
\delta_{ij} & \text{if $j\ne k$;}\\
-1 & \text{if $i=j=k$;}\\
max(0,-b_{ik}) & \text{if $i\ne j = k$.}
\end{cases}
\]

(2)\ $\tilde{B}'=(b'_{ij})$ is given by
\[b'_{ij}=\begin{cases}
-b_{ij} & \text{if $i=k$ or $j=k$;}\\
b_{ij}+\frac{|b_{ik}|b_{kj}+b_{ik}|b_{kj}|}{2} & \text{otherwise.}
\end{cases}
\]

(3)\  $X'=\{X'_1,\cdots,X'_m\}$ is given by
\begin{align}
X_k'&=X^{\sum_{1\leq i\leq m}[b_{ik}]_{+} {\bf e}_i -{\bf e}_k}+X^{\sum_{1\leq i\leq m}[-b_{ik}]_{+} {\bf e}_i -{\bf e}_k},\nonumber\\
X_i'&=X_i ,\quad 1\leq i\leq m,\quad i\neq k,\nonumber
\end{align}
where for each integer $a$ we set $[a]_{+}:=max\{0,a\}$.

Two quantum seeds  $(\Lambda, \tilde{B}, X)$ and  $(\Lambda', \tilde{B}', X')$ are
called {\em mutation-equivalent}, denoted by $(\Lambda, \tilde{B}, X)\sim(\Lambda', \tilde{B}', X')$, if they can be obtained from each other
by a sequence of mutations. Let $\mathcal{C}=\{X'_i~|~ (\Lambda', \tilde{B}', X')\sim(\Lambda, \tilde{B}, X),1\leq i\leq n\}$, and
the elements in $\mathcal{C}$ are called the {\em quantum cluster
variables}. Let $\mathbb{P}=\{X_i~|~n+1\leq i\leq m\}$, and the
elements in $\mathbb{P}$ are called the {\em coefficients}. Denote by
$\ZZ\mathbb{P}$ the ring of  Laurent polynomials in the elements of $\mathbb{P}$ with coefficients in $\ZZ[\mathfrak{q}^{\pm\frac{1}{2}}]$. Then the
{\em quantum cluster algebra}
$\mathcal{A}_{\mathfrak{q}}(\Lambda,\tilde{B})$ is defined to be the
$\ZZ\mathbb{P}$-subalgebra of $\mathcal{F}_{\mathfrak{q}}$ generated by all
quantum cluster variables.

\subsection{Quantum Caldero--Chapoton map and cluster multiplication formulas }\label{quantum_cluster_char}
Fix a positive integer $n$ and let $Q=\{Q_0, Q_1, s, t\}$ be an acyclic quiver with the vertex set $Q_0=
\{1, 2,\cdots, n\}$ and arrow set $Q_1$. We denote by $s(\rho)$ and $t(\rho)$ the source and
target of an arrow $\rho\in Q_1$, respectively. Take $\A=\A_{Q}$ to be the category of finite dimensional $kQ$-modules. Then the Grothendieck group $K(\A)$ is isomorphic to $\mathbb{Z}Q_0\cong\mathbb{Z}^n$. There is a bilinear form $\lr{-,-}:\mathbb{Z}^n\times\mathbb{Z}^n\to\mathbb{Z}$ defined by
$$\lr{{\bf x},{\bf y}}=\sum\limits_{i\in Q_0}x_iy_i-\sum\limits_{\rho\in Q_1}x_{s(\rho)}y_{t(\rho)}$$
for ${\bf x}=(x_i),{\bf y}=(y_i)\in\mathbb{Z}^n$,
which is called the {\em Euler form} of $Q$. It is well known that this form coincides with the (homological) Euler form defined in (\ref{Euler form}).

Let $B(Q)$ and $R(Q)$ be the $n\times n$ matrixes with the $i$-th row and $j$-th column element given respectively by
$$b_{ij}=|\{\mathrm{arrows}\, i\longrightarrow
j\}|-|\{\mathrm{arrows}\, j\longrightarrow i\}|$$
and
$$r_{ij}=|\{\mathrm{arrows}\, j\longrightarrow i\}|.$$

By definition, it is easy to see that for any ${\bf x}, {\bf y}\in\mathbb{Z}^n$
$$\lr{{\bf x},{\bf y}}={\bf x}^{tr}(I_n-R(Q)^{tr}){\bf y},$$
where $I_n$ is the $n\times n$ identity matrix. Moreover, for each projective $kQ$-module $P$,
$$(I_n-R(Q))\Dim P=\Dim (P/\rad P).$$

For each fixed integer $m\geq n$, we choose a quiver $\widetilde{Q}$ with the vertex set $\{1,2,\cdots,m\}$ such that $Q$ is a full subquiver of $\widetilde{Q}$.
Let $\widetilde{I}$, $\widetilde{B}$, $\widetilde{R}$ and $\widetilde{R}'$ be the
left $m\times n$ submatrixes of the matrixes $I_{m}$, $B(\widetilde{Q})$, $R(\widetilde{Q})$ and $R(\widetilde{Q})^{tr}$, respectively. For concision, we write $B$ and $R$ for $B(Q)$ and $R(Q)$, respectively.
Note that $\widetilde{B}=\widetilde{R}'-\widetilde{R}$ and $B=R^{tr}-R$.
We always assume that there exists a skew-symmetric $m\times m$ integral
matrix $\Lambda$ such that
\begin{align}\label{eq:simply_laced_compatible}
\Lambda(-\widetilde{B})={I_n\choose0}.\end{align}We remark that such $\widetilde{Q}$ and $\Lambda$ exist for a given quiver $Q$ (cf. \cite{rupel}).

Let $\mathcal C_{\widetilde{Q}}$ be the cluster category of $k
\widetilde{Q}$, i.e., the orbit category of the bounded derived category
$\mathcal{D}^b(k\widetilde{Q})$ under the action of  the functor
$F=\tau^{-1}\circ[1]$ (cf. \cite{BMRRT}). For each $1\leq i \leq
m$, let $S_i$ be the simple $k \widetilde{Q}$-module corresponding to the vertex $i$, and let $P_i$ be the projective cover of $S_i$.  Then the indecomposable $k \widetilde{Q}$-modules and all $P_i[1]$
exhaust all indecomposable objects of $\mathcal C_{\widetilde{Q}}$. Each object $X$ in
$\mathcal C_{\widetilde{Q}}$ can be uniquely decomposed as
$$X=M\oplus P[1]$$
where $M$ is a $k \widetilde{Q}$-module and $P$ is a projective $k \widetilde{Q}$-module.

In what follows, we adopt the convention that for each given module we will use the corresponding lowercase boldface letter to denote its dimension vector. For example, given a $kQ$-module $X$, its dimension vector is denoted by ${\bf x}$, and the dimension vector of $X$ viewed as a $k\widetilde{Q}$-module is also denoted by ${\bf x}$ if there is no confusion. Thus, for each $kQ$-module $X$ we have that $(\widetilde{I}-\widetilde{R}){\bf x}=(I_m-R(\widetilde{Q})){\bf x}$.

The quantum Caldero--Chapoton map associated to an acyclic quiver
$Q$ has been defined in \cite{rupel} and \cite{fanqin}.
In \cite{rupel}, the author defined the quantum Caldero--Chapoton map
for $kQ$-modules while in \cite{fanqin} for
coefficient-free rigid objects in $\mathcal C_{\widetilde{Q}}$.
For our purpose, we need to modify  the definition as follows:

Let $M$ be a $kQ$-module and $P$ be a projective $k\widetilde{Q}$-module. We define the \emph{quantum cluster character}
\begin{equation}X_{M\oplus P[1]}=\sum_{\bf{e}} |\mathrm{Gr}_{\bf{e}} M|q^{\frac{1}{2}
\langle
\bf{p}-\bf{e},\bf{m}-\bf{e}\rangle}X^{-\widetilde{B}{\bf e}-(\widetilde{I}-\widetilde{R}'){\bf m}+{\bf t}_P},\end{equation}
where ${\bf t}_P=\Dim (P/\rad P)$ and
$\mathrm{Gr}_{\bf{e}}M$ denotes the set of all submodules $V$
of $M$ with $\Dim V= \bf{e}$. In particular, $X_{P[1]}=X^{{\bf t}_P}$.

Given a finite acyclic quiver $Q$, denote by $\mathcal{AH}_{q}(Q)$ the
$\mathbb{ZP}$-subalgebra of $\mathcal{F}_q$ generated by
all the quantum cluster characters $X_{M\oplus P[1]}$
with $M$ being a $k Q$-module and $P$ being a projective $k
\widetilde{Q}$-module, where we appoint that $P$ can be taken to be zero.

The following lemma is needed for the proof of the subsequent theorem.
\begin{lemma}\label{AE}
For any $kQ$-module $M$ and projective $k\widetilde{Q}$-module $P$, we have that
\begin{equation}
\Lambda(\widetilde{B}{\bf m},{\bf t}_P)=\lr{{\bf p},\bf{m}}.
\end{equation}
\begin{proof}
Noting that ${\bf t}_P=(I_{m}-R(\widetilde{Q})){\bf p}$, by definition, we have that
\begin{flalign*}\Lambda(\widetilde{B}{\bf m},{\bf t}_P)&=\Lambda(\widetilde{B}{\bf m},(I_{m}-R(\widetilde{Q})){\bf p})\\
&={\bf m}^{tr}{\widetilde{B}}^{tr}\Lambda(I_{m}-R(\widetilde{Q})){\bf p}\\
&={\bf m}^{tr}(I_n,0)(I_{m}-R(\widetilde{Q})){\bf p}\\
&={\bf p}^{tr}(I_{m}-R(\widetilde{Q})^{tr}){I_n\choose0
}{\bf m}\\
&={\bf p}^{tr}(I_{m}-R(\widetilde{Q})^{tr}){\bf m}\\
&=\lr{{\bf p},{\bf m}},\end{flalign*}
where we should be reminded that ${I_n\choose0}
{\bf m}=\widetilde{I}{\bf m}$ is the dimension vector of $M$ viewed as a $k\widetilde{Q}$-module, and it is still denoted by ${\bf m}$.
\end{proof}
\end{lemma}
\begin{theorem}\label{exchange2}
Let $M$ be any $kQ$-module  and $P$ any projective
$k\widetilde{Q}$-module. Then
\vspace{0.2cm}
\begin{itemize}
  \item [(1)] $\ X_{M}X_{P[1]}=q^{-\frac{1}{2}\Lambda((\widetilde{I}-\widetilde{R}){\bf m},{\bf t}_P)}X_{M\oplus
P[1]};$
\vspace{0.4cm}
\item [(2)] $\ X_{P[1]}X_{M}=q^{\frac{1}{2}\Lambda((\widetilde{I}-\widetilde{R}){\bf m},{\bf t}_P)-\lr{{\bf p},{\bf m}}}
\sum\limits_{[B],[Q]}|{}_Q\Hom_{k\widetilde{Q}}(P,M)_B| X_{B\oplus Q[1]}.$
\end{itemize}
\end{theorem}

\begin{proof}
$(1)$~~By definition,
\begin{flalign*}
   X_{M}X_{P[1]}\nonumber
   &=\sum_{\bf{e}} |\mathrm{Gr}_{\bf{e}} M|q^{-\frac{1}{2}\langle
\bf{e},\bf{m}-\bf{e}\rangle}X^{-\widetilde{B}{\bf e}-(\widetilde{I}-\widetilde{R}')\bf{m}} X^{{\bf t}_P}\\
  &=\sum_{\bf{e}} |\mathrm{Gr}_{\bf{e}} M|q^{-\frac{1}{2}
\langle{\bf e},\bf{m}-\bf{e}\rangle}q^{\frac{1}{2}\Lambda(-\widetilde{B}{\bf e}
-(\widetilde{I}-\widetilde{R}'){\bf m},{\bf t}_P)}X^{-\widetilde{B}{\bf e}-(\widetilde{I}-\widetilde{R}'){\bf m}+{\bf t}_P}.
\end{flalign*}
Noting that
\begin{flalign*}
\Lambda(-\widetilde{B}{\bf e}
-(\widetilde{I}-\widetilde{R}'){\bf m},{\bf t}_P)
   &=\Lambda(-\widetilde{B}{\bf e}+\widetilde{B}{\bf m}
-(\widetilde{I}-\widetilde{R}){\bf m},{\bf t}_P)\nonumber\\
&=\Lambda(-(\widetilde{I}-\widetilde{R}){\bf m},{\bf t}_P)+\Lambda(\widetilde{B}({\bf m}-{\bf e}),{\bf t}_P)\\
 &=-\Lambda((\widetilde{I}-\widetilde{R}){\bf m},{\bf t}_P)+\langle \bf{p},\bf{m}-\bf{e}\rangle,
\end{flalign*}
we obtain that
\begin{flalign*}
X_{M}X_{P[1]}
&=q^{-\frac{1}{2}\Lambda((\widetilde{I}-\widetilde{R}){\bf m},{\bf t}_P)}\sum_{\bf{e}} |\mathrm{Gr}_{\bf{e}} M|q^{-\frac{1}{2}\langle
\bf{e},\bf{m}-\bf{e}\rangle}q^{\frac{1}{2}\langle
\bf{p},\bf{m}-\bf{e}\rangle}X^{-\widetilde{B}{\bf e}-(\widetilde{I}-\widetilde{R}'){\bf m}+{\bf t}_P}\\
&=q^{-\frac{1}{2}\Lambda((\widetilde{I}-\widetilde{R}){\bf m},{\bf t}_P)}X_{M\oplus
P[1]}.
\end{flalign*}

$(2)$~~On the one hand, by definition,
\begin{flalign*}
X_{P[1]}X_{M}&=\sum_{[H],[G]}q^{-\frac{1}{2}\langle {\bf h},{\bf g}\rangle}F^{M}_{GH}X^{{\bf t}_P}X^{-\widetilde{B}{\bf h}-(\widetilde{I}-\widetilde{R}')\bf{m}}\\
&=\sum_{[H],[G]}q^{-\frac{1}{2}\langle {\bf h},{\bf g}\rangle}F^{M}_{GH}q^{\frac{1}{2}\Lambda({\bf t}_P,-\widetilde{B}{\bf h}-(\widetilde{I}-\widetilde{R}')\bf{m})}X^{-\widetilde{B}{\bf h}-(\widetilde{I}-\widetilde{R}'){\bf m}
  +{\bf t}_P}\\
&=q^{\frac{1}{2}\Lambda((\widetilde{I}-\widetilde{R}){\bf m},{\bf t}_P)}\sum_{[H],[G]}q^{-\frac{1}{2}\langle
{\bf p}+{\bf h},{\bf g}\rangle}F^{M}_{GH}X^{-\widetilde{B}{\bf h}-(\widetilde{I}-\widetilde{R}'){\bf m}+{\bf t}_P}.
\end{flalign*}

On the other hand, by definition,
\begin{flalign*}&\sum_{[B],[Q]}|{}_Q\Hom_{k\widetilde{Q}}(P,M)_B|X_{B\oplus
Q[1]}\\&=\sum_{[B],[Q],[L],[G],[Y]}a_LF^{P}_{LQ}F^{M}_{BL}q^{\frac{1}{2}\langle
{\bf q}-{\bf y},{\bf g}\rangle}F^{B}_{GY}X^{-\widetilde{B}{\bf y}-(\widetilde{I}-\widetilde{R}'){\bf b}+{\bf t}_Q}\\
&=\sum_{[H],[Q],[L],[G],[Y]}q^{\frac{1}{2}\langle
{\bf q}-{\bf y},{\bf g}\rangle}a_LF^{P}_{LQ}F^{H}_{YL}F^{M}_{GH}X^{-\widetilde{B}{\bf y}-(\widetilde{I}-\widetilde{R}')\bf{b}+{\bf t}_Q}\\
&=\sum_{[H],[Q],[G],[Y]}q^{\frac{1}{2}\langle
{\bf q}-{\bf y},{\bf g}\rangle}|{}_Q\Hom_{k\widetilde{Q}}(P,H)_Y|F^{M}_{GH}X^{-\widetilde{B}{\bf y}-(\widetilde{I}-\widetilde{R}')\bf{b}+{\bf t}_Q}
\end{flalign*}where we have used the associativity formula (\ref{jiehe}) for the second equation.

Noting that
$$\bf{b}+\bf{l}=\bf{m},~~\bf{q}+\bf{l}=\bf{p},~~\bf{y}+\bf{l}=\bf{h},~~\text{and~~thus}~~\bf{q}-\bf{y}=\bf{p}-\bf{h},$$
we have that
\begin{flalign*}
&-\widetilde{B}{\bf y}-(\widetilde{I}-\widetilde{R}'){\bf b}+{\bf t}_Q\\
&=-\widetilde{B}{\bf h}+\widetilde{B}{\bf l}-(\widetilde{I}-\widetilde{R}'){\bf b}+(I_{m}-R(\widetilde{Q})){\bf q}\\
&=-\widetilde{B}{\bf h}+\widetilde{R}'{\bf l}-\widetilde{R}{\bf l}
-\widetilde{I}{\bf b}+\widetilde{R}'{\bf b}+(I_{m}-R(\widetilde{Q})){\bf q}\\
&=-\widetilde{B}{\bf h}+\widetilde{R}'{\bf m}-\widetilde{R}{\bf l}
-\widetilde{I}({\bf m}-{\bf l})+(I_{m}-R(\widetilde{Q})){\bf q}\\
&=-\widetilde{B}{\bf h}-(\widetilde{I}-\widetilde{R}'){\bf m}
  +(\widetilde{I}-\widetilde{R}){\bf l}+(I_{m}-R(\widetilde{Q})){\bf q}\\
&=-\widetilde{B}{\bf h}-(\widetilde{I}-\widetilde{R}'){\bf m}
  +(I_{m}-R(\widetilde{Q})){\bf l}+(I_{m}-R(\widetilde{Q})){\bf q}\\
&=-\widetilde{B}{\bf h}-(\widetilde{I}-\widetilde{R}'){\bf m}
  +(I_{m}-R(\widetilde{Q})){\bf p}\\
  &=-\widetilde{B}{\bf h}-(\widetilde{I}-\widetilde{R}'){\bf m}
  +{\bf t}_P.
\end{flalign*}
Hence,
\begin{flalign*}
&\sum_{[B],[Q]}|{}_Q\Hom_{k\widetilde{Q}}(P,M)_B|X_{B\oplus
Q[1]}\\
&=\sum_{[H],[Q],[G],[Y]}q^{\frac{1}{2}\langle
{\bf p}-{\bf h},{\bf g}\rangle}|{}_Q\Hom_{k\widetilde{Q}}(P,H)_Y|F^{M}_{GH}X^{-\widetilde{B}{\bf h}-(\widetilde{I}-\widetilde{R}'){\bf m}
  +{\bf t}_P}\\
  &=\sum_{[H],[G]}q^{\frac{1}{2}\langle
{\bf p}-{\bf h},{\bf g}\rangle+\lr{{\bf p},{\bf h}}}F^{M}_{GH}X^{-\widetilde{B}{\bf h}-(\widetilde{I}-\widetilde{R}'){\bf m}
  +{\bf t}_P}\\
&=q^{\lr{{\bf p},{\bf m}}}\sum_{[H],[G]}q^{-\frac{1}{2}\lr{{\bf p}+{\bf h},{\bf g}}}F^{M}_{GH}X^{-\widetilde{B}{\bf h}-(\widetilde{I}-\widetilde{R}'){\bf m}
  +{\bf t}_P}.
\end{flalign*}
\end{proof}

\section{Quantum cluster algebras via Hall algebras}
Let $Q$ be an acyclic quiver with the vertex set $\{1,2,\cdots,n\}$. As stated in \cite{Rup}, a result of Fomin and Zelevinsky in \cite{FZ4} asserts that the cluster variables
are completely determined by the cluster variables of the principal coefficient quantum cluster algebra.
So we consider the quiver $\widetilde{Q}$ associated to $Q$ as follows: add the vertices $\{n+1,\dots,2n\}$ to the quiver $Q$ with the additional arrows
$n+i\longrightarrow i$ for any $1\leq i\leq n$. By \cite{BZ05}, we can take a skew-symmetric $2n\times 2n$ integral
matrix $\Lambda$ such that
\begin{align}\label{eq:simply_laced_compatible2}
\Lambda(-\widetilde{B})={I_n\choose0}.
\end{align}

In order to relate the Hall algebra $\M\H(\A_Q)$ to the algebra $\mathcal{AH}_{q}(Q)$, we twist the multiplication on $\M\H(\A_Q)$, and define $\mathcal{MH}_{\Lambda}(\A_Q)$ to be the same module as $\mathcal{MH}(\A_Q)$ but with the twisted
multiplication defined on basis elements by
{\begin{equation}\begin{split}
   &(K_{\alpha}\ast\mathbb{X}_{M\oplus P[1]})\star (K_{\beta}\ast\mathbb{X}_{N\oplus Q[1]})=\\&
   q^{\frac{1}{2}\Lambda((\widetilde{I}-\widetilde{R})({\bf m}-{\bf p})-\widetilde{\alpha},(\widetilde{I}-\widetilde{R})(\bf{n}
   -\bf{q})-\widetilde{\beta})}
   (K_{\alpha}\ast\mathbb{X}_{M\oplus P[1]})\ast (K_{\beta}\ast\mathbb{X}_{N\oplus Q[1]}),\end{split}
\end{equation}}
where $M, N\in\A_Q$, $P, Q\in\P$, $\alpha,\beta\in \mathbb{Z}^{n}$, and for each $\alpha\in \mathbb{Z}^{n}$, $\widetilde{\alpha}:={0\choose\alpha}
\in \mathbb{Z}^{2n}$.

For any $\alpha\in \mathbb{Z}^{n}$, $M\in\A_Q$ and $P\in\P$, we define
$$\grad(K_{\alpha}\ast\mathbb{X}_{M\oplus P[1]}):=(\widetilde{I}-\widetilde{R})({\bf m}-{\bf p})-\widetilde{\alpha},$$
and for each $\varpi\in\mathbb{Z}^{2n}$ we define $\mathcal{MH}(\A_Q)_\varpi$ to be the submodule of $\mathcal{MH}(\A_Q)$ spanned by the elements $K_{\alpha}\ast\mathbb{X}_{M\oplus P[1]}$ with
$\grad(K_{\alpha}\ast\mathbb{X}_{M\oplus P[1]})=\varpi$.

The following lemma shows that the functions $\grad$ provide a $\mathbb{Z}^{2n}$-grading on the Hall algebra $\mathcal{MH}(\A_Q)$.
\begin{lemma}
For any $\varpi_1,\varpi_2\in\mathbb{Z}^{2n}$, we have that
\begin{equation*}\mathcal{MH}(\A_Q)_{\varpi_1}\ast\mathcal{MH}(\A_Q)_{\varpi_2}\subseteq\mathcal{MH}(\A_Q)_{\varpi_1+\varpi_2}.\end{equation*}
That is, $\mathcal{MH}(\A_Q)$ is a $\mathbb{Z}^{2n}$-graded algebra.
\end{lemma}
\begin{proof}
Let $M, N\in\A_Q$, $P, Q\in\P$, $\alpha,\beta\in \mathbb{Z}^{n}$ such that $\grad(K_{\alpha}\ast\mathbb{X}_{M\oplus P[1]})=\varpi_1$ and $\grad(K_{\beta}\ast\mathbb{X}_{N\oplus Q[1]})=\varpi_2$. Then
\begin{flalign*}
&(K_{\alpha}\ast\mathbb{X}_{M\oplus P[1]})\ast (K_{\beta}\ast\mathbb{X}_{N\oplus Q[1]})\\&=
K_{\alpha}\ast\mathbb{X}_{M}\ast\X_{P[1]}\ast K_{\beta}\ast\mathbb{X}_{N}\ast\X_{Q[1]}\\
&=K_{\alpha+\beta}\ast\mathbb{X}_{M}\ast\X_{P[1]} \ast\mathbb{X}_{N}\ast\X_{Q[1]}\\
&=\sum\limits_{[B],[Q']}q^{-\lr{P,N}}|{}_{Q'}\Hom_{\A}(P,N)_B|K_{\alpha+\beta}\ast\X_M\ast\X_B\ast\X_{(Q'\oplus Q)[1]}\\
&=\sum\limits_{[B],[Q'],[L]}q^{-\lr{P,N}+\lr{M,B}}|{}_{Q'}\Hom_{\A}(P,N)_B|\cdot\frac{|\mathrm{Ext}_{\A}^{1}(M,B)_{L}|}{|\mathrm{Hom}_{\A}(M,B)|}K_{\alpha+\beta}\ast\X_{L\oplus(Q'\oplus Q)[1]}.
\end{flalign*}
For each $K_{\alpha+\beta}\ast\X_{L\oplus(Q'\oplus Q)[1]}$ in the summation above, since ${\bf l}={\bf m}+{\bf b}={\bf m}+{\bf q'}-{\bf p}+{\bf n}$, we obtain that \begin{flalign*}\grad(K_{\alpha+\beta}\ast\X_{L\oplus(Q'\oplus Q)[1]})&=(\widetilde{I}-\widetilde{R})({\bf l}-({\bf q'}+{\bf q}))-(\widetilde{\alpha}+\widetilde{\beta})\\&=(\widetilde{I}-\widetilde{R})({\bf m}+{\bf n}-{\bf p}-{\bf q})-(\widetilde{\alpha}+\widetilde{\beta})\\
&=\grad(K_{\alpha}\ast\mathbb{X}_{M\oplus P[1]})+\grad (K_{\beta}\ast\mathbb{X}_{N\oplus Q[1]}).
\end{flalign*}
\end{proof}

\begin{lemma}
The algebra $\mathcal{MH}_{\Lambda}(\A_Q)$ is still associative.
\end{lemma}
\begin{proof}
For any $L,M,N\in\A$, $P,Q,R\in\P$ and $\alpha,\beta,\gamma\in\mathbb{Z}^{n}$, it is easy to see that
\begin{flalign*}&((K_{\alpha}\ast\mathbb{X}_{M\oplus P[1]})\star (K_{\beta}\ast\mathbb{X}_{N\oplus Q[1]}))\star (K_{\gamma}\ast\mathbb{X}_{L\oplus R[1]})\\&=
q^{\frac{1}{2}e}(K_{\alpha}\ast\mathbb{X}_{M\oplus P[1]})\ast (K_{\beta}\ast\mathbb{X}_{N\oplus Q[1]})\ast (K_{\gamma}\ast\mathbb{X}_{L\oplus R[1]})\\&
=(K_{\alpha}\ast\mathbb{X}_{M\oplus P[1]})\star ((K_{\beta}\ast\mathbb{X}_{N\oplus Q[1]})\star (K_{\gamma}\ast\mathbb{X}_{L\oplus R[1]}))\end{flalign*}
where $e=\Lambda((\widetilde{I}-\widetilde{R})({\bf m}-{\bf p})-\widetilde{\alpha},(\widetilde{I}-\widetilde{R})({\bf n}
-{\bf q})-\widetilde{\beta})+\Lambda((\widetilde{I}-\widetilde{R})({\bf m}-{\bf p})-\widetilde{\alpha},(\widetilde{I}-\widetilde{R})
({\bf l}-{\bf r})-\widetilde{\gamma})+\Lambda((\widetilde{I}-\widetilde{R})({\bf n}-{\bf q})-\widetilde{\beta},
(\widetilde{I}-\widetilde{R})({\bf l}
-{\bf r})-\widetilde{\gamma})$. That is, the algebra $\mathcal{MH}_{\Lambda}(\A_Q)$ is associative.\end{proof}

Using Theorem \ref{ydygx}, we write down the defining relations of $\mathcal{MH}_{\Lambda}(\A_Q)$ in the following
\begin{proposition}
The algebra $\mathcal{MH}_{\Lambda}(\A_Q)$ is generated by all $K_{\alpha}$ and $\mathbb{X}_{M\oplus P[1]}$ $($with $\alpha\in\mathbb{Z}^{n}$, $M\in\A_Q$ and $P\in\P$$)$,
which are subject to the following relations
\begin{equation}\label{x1}
\begin{split}K_{\alpha}\star K_{\beta}=q^{\frac{1}{2}\Lambda(\widetilde{\alpha},\widetilde{\beta})}K_{\alpha+\beta}=q^{\Lambda(\widetilde{\alpha},\widetilde{\beta})}K_{\beta}\star K_{\alpha};\end{split}\end{equation}
\begin{equation}\label{x2}
\begin{split}K_{\alpha}\star\mathbb{X}_{M\oplus P[1]}
    =q^{-\Lambda(\widetilde{\alpha},(\widetilde{I}-\widetilde{R})({\bf m}-{\bf p}))}\mathbb{X}_{M\oplus P[1]}\star K_{\alpha};\end{split}\end{equation}
\begin{equation}\label{x3}
\begin{split}
\mathbb{X}_{P[1]}\star \mathbb{X}_{Q[1]}&=
q^{\frac{1}{2}\Lambda((\widetilde{I}-\widetilde{R}){\bf p},(\widetilde{I}-\widetilde{R})\bf{q}
)}\mathbb{X}_{(P\oplus Q)[1]}\\
&=q^{\Lambda((\widetilde{I}-\widetilde{R}){\bf p},(\widetilde{I}-\widetilde{R})\bf{q})}\mathbb{X}_{Q[1]}\star\mathbb{X}_{P[1]};
\end{split}\end{equation}
\begin{equation}\label{x4}\mathbb{X}_{M}\star\mathbb{X}_{N}=q^{\frac{1}{2}\Lambda((\widetilde{I}-\widetilde{R}){\bf m},
(\widetilde{I}-\widetilde{R})\bf{n})+\langle\bf{m},\bf{n}\rangle}\sum_{[L]}\frac{|\mathrm{Ext}_{\A}^{1}(M,N)_{L}|}{|\mathrm{Hom}_{\A}(M,N)|}\mathbb{X}_L;\end{equation}
\begin{equation}\label{x5}\mathbb{X}_{M}\star\mathbb{X}_{P[1]}=q^{-\frac{1}{2}\Lambda((\widetilde{I}-\widetilde{R}){\bf m},(\widetilde{I}-\widetilde{R}){\bf p})}\mathbb{X}_{M\oplus
P[1]};\end{equation}
\begin{equation}\label{x6}\mathbb{X}_{P[1]}\star\mathbb{X}_{M}=q^{-\frac{1}{2}\Lambda((\widetilde{I}-\widetilde{R}){\bf p},(\widetilde{I}-\widetilde{R}){\bf m})-\lr{{\bf p},{\bf m}}}
\sum\limits_{[B],[Q]}|{}_Q\Hom_{\A}(P,M)_B|\mathbb{X}_{B\oplus Q[1]};\end{equation}
for any $\alpha,\beta\in\mathbb{Z}^n$, $M,N\in\A_Q$ and $P,Q\in\P$.
\end{proposition}
\begin{theorem}\label{alg-homo}
There exists a surjective algebra homomorphism $$\xymatrix{\Psi: \mathcal{MH}_{\Lambda}(\A_Q)\ar@{->>}[r]& \mathcal{AH}_{q}(Q)}$$ defined on generators by
$$K_{\alpha}\mapsto X^{\widetilde{\alpha}}\quad\text{and}\quad\mathbb{X}_{M\oplus P[1]}\mapsto X_{M\oplus P[1]}$$
for any $\alpha\in\mathbb{Z}^n$, $M\in\A_Q$ and $P\in\P$.
\end{theorem}
\begin{proof}
In order to prove $\Psi$ is a homomorphism of algebras, it suffices to prove that $\Psi$ preserves the relations $(\ref{x1}$-$\ref{x6})$. Clearly, the relation $(\ref{x1})$ is preserved.

By definition,
\begin{flalign*}
X^{\widetilde{\alpha}}X_{M\oplus P[1]}&=\sum\limits_{[H],[G]}q^{\frac{1}{2}\lr{{\bf p}-{\bf h},{\bf g}}}F_{GH}^MX^{\widetilde{\alpha}}X^{-\widetilde{B}{\bf h}-(\widetilde{I}-\widetilde{R}'){\bf m}+{\bf t}_P}\\
&=\sum\limits_{[H],[G]}q^{\frac{1}{2}\lr{{\bf p}-{\bf h},{\bf g}}}F_{GH}^Mq^{\Lambda(\widetilde{\alpha},-\widetilde{B}{\bf h}-(\widetilde{I}-\widetilde{R}'){\bf m}+{\bf t}_P)}X^{-\widetilde{B}{\bf h}-(\widetilde{I}-\widetilde{R}'){\bf m}+{\bf t}_P}X^{\widetilde{\alpha}}
\end{flalign*}
and
\begin{flalign*}
\Lambda(\widetilde{\alpha},-\widetilde{B}{\bf h}-(\widetilde{I}-\widetilde{R}'){\bf m}+{\bf t}_P)&=\Lambda(\widetilde{\alpha},\widetilde{B}({{\bf m}-{\bf h}})-(\widetilde{I}-\widetilde{R}){\bf m}+{\bf t}_P)\\
&=-\Lambda(\widetilde{B}{\bf g},\widetilde{\alpha})-\Lambda(\widetilde{\alpha},(\widetilde{I}-\widetilde{R})({\bf m}-{\bf p})).
\end{flalign*}
We claim that $\Lambda(\widetilde{B}{\bf g},\widetilde{\alpha})=0$. In fact, for any $\alpha=(\alpha_1,\cdots,\alpha_n)\in\mathbb{Z}^n$, $\widetilde{\alpha}=\sum\limits_{i=1}^n\alpha_i{\bf t}_{P_{n+i}}$.
Then by Lemma \ref{AE},
\begin{flalign*}
\Lambda(\widetilde{B}{\bf g},\widetilde{\alpha})=\sum\limits_{i=1}^n\alpha_i\Lambda(\widetilde{B}{\bf g},{\bf t}_{P_{n+i}})
=\sum\limits_{i=1}^n\alpha_i\lr{{\bf p}_{n+i},{\bf g}}=0.
\end{flalign*}
Hence, \begin{equation*}X^{\widetilde{\alpha}}X_{M\oplus P[1]}=q^{-\Lambda(\widetilde{\alpha},(\widetilde{I}-\widetilde{R})({\bf m}-{\bf p}))}X_{M\oplus P[1]}X^{\widetilde{\alpha}}.\end{equation*} That is, the relation $(\ref{x2})$ is preserved.

By definition,
\begin{flalign*}
{X}_{P[1]}{X}_{Q[1]}=X^{{\bf t}_P}X^{{\bf t}_Q}&=q^{\frac{1}{2}\Lambda((\widetilde{I}-\widetilde{R}){\bf p},(\widetilde{I}-\widetilde{R})\bf{q}
)}{X}^{{\bf t}_P+{\bf t}_Q}\\&=
q^{\frac{1}{2}\Lambda((\widetilde{I}-\widetilde{R}){\bf p},(\widetilde{I}-\widetilde{R})\bf{q}
)}{X}_{(P\oplus Q)[1]}
\end{flalign*}
and
\begin{flalign*}
{X}_{P[1]}{X}_{Q[1]}=X^{{\bf t}_P}X^{{\bf t}_Q}&=q^{\Lambda((\widetilde{I}-\widetilde{R}){\bf p},(\widetilde{I}-\widetilde{R})\bf{q}
)}X^{{\bf t}_Q}X^{{\bf t}_P}\\
&=q^{\Lambda((\widetilde{I}-\widetilde{R}){\bf p},(\widetilde{I}-\widetilde{R})\bf{q}
)}{X}_{Q[1]}{X}_{P[1]}.\end{flalign*}Thus, the relation $(\ref{x3})$ is preserved.

On the one hand, the following equation has been proved in \cite{DSC,DX}
\begin{equation}{X}_{M}{X}_{N}=q^{\frac{1}{2}\Lambda((\widetilde{I}-\widetilde{R}'){\bf m},
(\widetilde{I}-\widetilde{R}')\bf{n})+\langle\bf{m},\bf{n}\rangle}\sum_{[L]}\frac{|\mathrm{Ext}_{\A}^{1}(M,N)_{L}|}{|\mathrm{Hom}_{\A}(M,N)|}X_L.\end{equation}
On the other hand, by \cite[Lemma 3.1]{DSC},
\begin{flalign*}
&\Lambda((\widetilde{I}-\widetilde{R}'){\bf m},
(\widetilde{I}-\widetilde{R}')\bf{n})=\\&\Lambda((\widetilde{I}-\widetilde{R}){\bf m},
(\widetilde{I}-\widetilde{R}){\bf n})-\Lambda((\widetilde{I}-\widetilde{R}'){\bf m},\widetilde{B}{\bf n})-\Lambda(\widetilde{B}{\bf m},(\widetilde{I}-\widetilde{R}'){\bf n})-\Lambda(\widetilde{B}{\bf m},\widetilde{B}{\bf n})\\
&=\Lambda((\widetilde{I}-\widetilde{R}){\bf m},
(\widetilde{I}-\widetilde{R}){\bf n})+\lr{{\bf n},{\bf m}}-\lr{{\bf m},{\bf n}}+\lr{{\bf m},{\bf n}}-\lr{{\bf n},{\bf m}}\\
&=\Lambda((\widetilde{I}-\widetilde{R}){\bf m},
(\widetilde{I}-\widetilde{R}){\bf n}).
\end{flalign*}
Thus, the relation $(\ref{x4})$ is preserved.

By Theorem \ref{exchange2}, we know that the relations $(\ref{x5}$-$\ref{x6})$ are preserved. Hence, $\Psi$ is a homomorphism of algebras.

For each $1\leq i\leq n$, $X_{P_{n+i}[1]}=X^{{\bf t}_{P_{n+i}}}=X^{e_{n+i}}$, and thus $\Psi(K_{\hat{S}_i})=X_{P_{n+i}[1]}$. Then it is clear that $\Psi$ is surjective. Therefore, we complete the proof.
\end{proof}

Let $\A_q(Q)$ be the quantum cluster algebra with principal coefficients corresponding to the quiver $Q$, which is the subalgebra of $\mathcal{AH}_{q}(Q)$ generated by
\begin{equation*}
\{X^{\widetilde{\alpha}}, X_{M\oplus P[1]}~|~\alpha\in\mathbb{Z}^n, M\in\Ind(\A_Q)~~\text{is rigid and}~~P\in\Ind(\P)\}.
\end{equation*}
Parallelly, we define $\mathcal{MC}_{\Lambda}(\A_Q)$ to be the subalgebra of $\mathcal{MH}_{\Lambda}(\A_Q)$ generated by $$\{K_{\alpha},\X_{M\oplus P[1]}~|~\alpha\in\mathbb{Z}^n, M\in\Ind(\A_Q)~~\text{is rigid and}~~P\in\Ind(\P)\}.$$
Then we have the following

\begin{corollary}
There exists a surjective algebra homomorphism $$\xymatrix{\psi: \mathcal{MC}_{\Lambda}(\A_Q)\ar@{->>}[r]& \mathcal{A}_{q}(Q).}$$\end{corollary}
\begin{proof}
Taking $\psi$ to be the restriction of $\Psi$ in Theorem \ref{alg-homo} to $\mathcal{MC}_{\Lambda}(\A_Q)$ gives the proof.
\end{proof}
\begin{remark}
By Proposition \ref{generators},
we know that the set $$S:=\{K_{\alpha},\mathbb{X}_M,\mathbb{X}_{P[1]}~|~\alpha\in\mathbb{Z}^{n},~M\in\Ind(\A_Q)~~\text{and}~~P\in\Ind(\P)\}$$ is a generating set of the algebra $\mathcal{MH}_{\Lambda}(\A_Q)$. For a Dynkin quiver $Q$, each indecomposable $kQ$-module is rigid. Thus, in this case, $\mathcal{MC}_{\Lambda}(\A_Q)=\mathcal{MH}_{\Lambda}(\A_Q)$. By Theorem \ref{alg-homo}, the images of all the elements in $S$ under the map $\Psi$, which constitute the set $\{X^{\widetilde{\alpha}},{X}_M,{X}_{P[1]}~|~\alpha\in\mathbb{Z}^{n},~M\in\Ind(\A_Q)~~\text{and}~~P\in\Ind(\P)\}$, generate $\mathcal{AH}_{q}(Q)$.
Also, for a Dynkin quiver $Q$, since each indecomposable $kQ$-module is rigid, we have that $\mathcal{AH}_{q}(Q)=\mathcal{A}_{q}(Q)$.
\end{remark}

\begin{corollary}
Let $Q$ be a Dynkin quiver.
There exists a surjective algebra homomorphism $$\xymatrix{\psi: \mathcal{MH}_{\Lambda}(\A_Q)\ar@{->>}[r]& \mathcal{A}_{q}(Q).}$$\end{corollary}

\section*{Acknowledgments}
The authors are grateful to the anonymous referees for their valuable suggestions and comments,
and supported partially by the National Natural Science Foundation of China (No.s 11771217, 11471177, 11801273), Natural Science Foundation of Jiangsu Province of China (No.BK20180722)
and Natural Science Foundation of Jiangsu Higher Education Institutions of China  (No.18KJB110017).

redpajama_set_name: RedPajamaArXiv


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Earthen concrete walls, planted patios and pools feature in this house in Mexican state Hidalgo, which was designed by Mexico City firm Centro de Colaboración Arquitectónica.
"We worked hand-in-hand with the suppliers to obtain a unique colour that corresponds to the warm tones of the site at sunset," CCA architect Bernardo Quinzaños told Dezeen.
"Right now, the architecture in Mexico is about using raw materials," the studio said. "In a way that creates comfortable spaces, as with these earthy colours and getting back to the roots."
"The foundation of the house stands out from the ground level, generating a skirting board that continues throughout the length and breadth of the house," said the studio.
Described by the firm as "large horizontal monoliths", the two wings of the house host bedrooms. The volumes are arranged around walled-in gardens with arid plants, and a central patio with swimming pools and lounge areas.
"The project is based on integrating the construction into the natural landscape," the studio said.
For Casa Moulat, CCA has accompanied each of the home's bedrooms with patios, while the open-plan living and dining room is flanked by sliding glass doors. One side accesses a small courtyard, and the other side joins the pool.
"The house explores the duality of open and closed spaces, inside and outside," the studio said.
A lofted atrium is nestled within the roof's gable with views to the lounge. Two black armchairs furnish the elevated nook.


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OK ! I hope be hre for this great show !
Hi everybody, hi Cloud 88 !
Last edited by Dan Reitar; Aug 4th, 2008 at 00:17.
I hope you will enjoy this second edition of Uplifting Moment tonight !
Hi Picolli, Squal and Although !


Over the last 24 hours more than 390,000 transactions have been made in the BitShares network, which exceeds the number of Bitcoin transactions.
Only in the last 24 hours BTS has grown by 37.5%, and over the last week - by 246.63%. Now BitShares occupies the 15th place in terms of market capitalization.


text:
New Fanfare! Co-Chai...
New Fanfare! Co-Chairs Lift Arts Affinity Group to New Heights
Supporting the Arts
New Co-Chairs Aim to Double Membership of Group for Emerging Leaders for the Arts
This fall kicks off the second year of Fanfare! at The Smith Center, and this affinity group for emerging leaders for the arts is being taken to a whole new level, thanks to four new co-chairs with big plans to give members a deep connection with the arts in Southern Nevada.
All accomplished community leaders, Fanfare! Co-Chairs Kady Casullo, Travis Turner, Alan Walker and Sean Walker have announced goals to double the group's membership this year, as well as organize more events exploring Southern Nevada's arts and culture scene.
They already kicked off the latter plan in October, when they arranged a special behind-the-scenes tour of Cirque du Soleil show "KA" for Fanfare! members.
"We want to give Fanfare members a fun experience, on top of the opportunity to build a relationship with The Smith Center," Turner says. "We hope to celebrate and advocate for the arts in Las Vegas as a whole."
Bringing Performing Arts Lovers Together
Open to Southern Nevada professionals ages 21 to 40, Fanfare! serves as a pipeline to new generations of arts lovers.
Members have the opportunity to network with like-minded community leaders, experience the arts in Southern Nevada by participating in exclusive events, and serve as community ambassadors for The Smith Center.
The new co-chairs see great potential for Fanfare! to grow.
They are already planning upcoming events involving Nevada Ballet Theatre, Las Vegas-based companies such as Zappos and a variety of performing arts-related activities.
"We want to do things that incorporate the arts across the valley that tie back into The Smith Center," Casullo says.
Community Leaders Take the Helm
The new Fanfare! co-chairs bring wide-ranging experience and a passion for community service to their new roles.
A Las Vegas native, Casullo serves as director of firm relations for The Powell Law Firm. She remains closely involved with the Make-A-Wish Foundation, where she serves as a wish granter, ambassador and is involved in resource development. She was named the organization's Volunteer of the Year in 2016.
Turner works as director of administration and oversees community and philanthropic initiatives for a Las Vegas FinTech organization. Committed to supporting the community, he serves on the boards of several nonprofits, including Noah's Animal House, Executive Pride, the Mark E. Curry Family Foundation, Club Curry, and, until his term recently finished, The Shade Tree.
Alan has built a career in marketing at top hospitality companies in Las Vegas, including Caesars Entertainment Corporation and Hakkasan Group, where he is currently corporate senior marketing manager. He has served on boards for several nonprofits across the valley, including Relay for Life, the Human Rights Campaign and other LGBT and homeless youth organizations.
Sean serves as co-manager of R&R Partners Foundation, as well as manager of culture and communications with the organization. He has also worked at leading Las Vegas corporations, such as Caesars Entertainment Corporation and The Cosmopolitan of Las Vegas. He has invested more than a decade serving with many nonprofits, including the Human Rights Campaign and Relay for Life.
All look forward to building Fanfare! membership together.
"I think we can blow Fanfare! events out of the park and make people want to get involved," Turner says.
Community members can join Fanfare! with a tax-deductible, minimum contribution of $250. For more information, visit TheSmithCenter.com/Support-Us/Fanfare.

redpajama_set_name: RedPajamaCommonCrawl


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GLuint
glShader::getProgramAtIndex(int32 progIndex) const {
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} // namespace _priv
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redpajama_set_name: RedPajamaGithub


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How to Hustle Like a Mom from a Mompreneur
Pamela Pekerman, on-air expert and creator of Hustle Like a Mom, shares the best parenting advice she's ever received.
By Pamela Pekerman March 12, 2020
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Meet Pamela Pekerman, the Westchester mom of 2 behind Hustle Like a Mom. She is a mompreneur, on-air lifestyle expert, and an Influencer We Love.
You might recognize Pekerman from morning shows, where she appeared for a decade talking about style trends. And then she had kids and decided she didn't want to travel as much. "So, I began to write about the real-life moments I was experiencing as a mom of two under two!" she says. Soon Pekerman was blogging about time-saving tips for busy moms, creating videos for Parents.com, and hosting events for moms. Her focus evolved into content and networking events that appeal to hustling mompreneurs. (Find out when the next Hustle Like a Mom mompreneur mingle and panel is going to be.) While she still does school pickup and drop-off every day, she says, "I've never felt more fulfilled by my work than I am now, as I grow a community of women that are redefining what it means to be a working mom."
Why follow Pamela Pekerman? This fierce and friendly mama tells it like it is—and is on a mission to get moms to feel good about themselves and the work that they do.
Instagram | @pamelapekerman
Website | pamelapekerman.com
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What is typical day for you?
No two days are alike, and that's common for most mompreneurs. But, typically, I aim to be up around 6 or 6:30am so I can have some tea in silence—power of the pause is key, and I believe my best ideas for mom life and entrepreneur life come when I'm in stillness. Around 7am, I'll do my makeup and get dressed. Sometimes my daughter wakes up and I'll get in bed for a bit of cuddles. By 7:20am it's time to get her and her brother up and dressed. We eat breakfast, read, and then it's time for drop-off. From 8:45am-3pm, I switch into mompreneur mode, and it is a mix of content creating; meeting with sponsors for upcoming Hustle Like a Mom mompreneur mingles; Instagram engagement, which is my main social platform; virtual meetings with my all-female team; and pitching TV producers and/or prepping for upcoming segments. By 3pm it's time for pickup. Some days I pick the kids up, some days I ask a sitter to do it, so I have an extra 30 minutes to work. Then it's eating, activities with kids, and once a week I give them date night with mom. By 8pm it's bedtime and books. I rarely have energy to work at 8:30pm, plus I prefer to spend time with my hubs. But, sometimes, I am back at the computer until 9:30pm to wrap up loose ends, make sure I engage with my hustling mom tribe, and prep for the next day. This year, I'm really making a point to be in bed by 10pm, sleeping by 10:30pm.
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Listen to the crowd but go with your gut. Don't silence your inner mama bear, as she truly knows what is best for her cubs. It's great to breast-feed, but that's not happy making for every mama. Sure, it's nice to be a PTA mom, but that might not be on your priority list. The only advice I would ever offer is to dig into your gut, listen to your deepest desires, and do whatever you can to align your personal and professional aspirations. Parenting is hard, and you need to ensure you are coming at it from a place of inner contentment and confidence in your choices—because so many moments will challenge your sanity, and your sense of self.
What famous parent would you like to have dinner with and why?
Busy Philipps is my celeb mom crush. I've interviewed her a few times, and she's such a fun gal. I'm sure we would laugh our butts off about the highs, lows, and sanity-sucking, moments-of-madness that is mom life. She also gets entrepreneur life and hustle as a mom life, so I'd imagine we would dish on our professional dreams, as well, all while enjoying cotton candy and champagne.
What do you think is a parent's hardest job?
Being present. It's a gift to the person you are speaking with, in this case your children, to be present, to not have your mind wander into what you need to do once they go to bed, or what you could be doing if you didn't have to practice piano with them. Being present is a constant practice of focusing on the person you are with, and on your breath (if you need to ground yourself back in the moment). In 2019, I found that replacing the phrase "I have to" with "I get to" gives you a boost into the present, as well as a dose of gratitude for that which is before you. Wow, I get to practice piano with my kids. What a privilege. Wow, I get to pick up my kids from school, my healthy kids. When kids know you are there, really there, they will return the love with their focus and heart. And, if we as parents are present and in the moment, we will teach our kids to be calmer, better listeners, and more mindful human beings.
RELATED: Meet Mike Julianelle, aka Dad and Buried
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text:
Helena Alexandrovna Timofeeff-Ressovsky (June 21, 1898 – April 29, 1973) was a Russian biologist known for her work in developmental and population genetics and radiation ecology.

Early and personal life

Helena's father, Aleksandr Fidler, ran a private girls' school. She and her siblings (two brothers and six sisters) were well-educated; some of her sisters studied chemistry and musicology. Helena met her future husband, Nikolai Vladimirovich, while studying biology and zoology in Moscow.

The couple had two sons, Dmitry (born 1923) and Andrei (born 1927). Accused of antifascist activities in Germany, Dmitry was arrested in 1943 and died in the concentration camp in Mauthausen. Andrei became a physicist.

Scientific career

Helena and Nikolai co-authored a dozen papers that shaped developmental genetics. Their studies of Drosophila funebris in the 1920s explored the concepts of penetrance and expressivity. In response to Muller's X-ray experiments in the late 1920s, they turned their attention to mutation and population genetics.<ref>Jonathan Harwood, Styles of Scientific Thought: The German Genetics Community, 1900-1933" (Chicago: University of Chicago Press, 1993), https://books.google.com/books?id=dygUxfsyaTwC (accessed 10 June 2015), 55-56.</ref> A 1927 paper provided the first experimental proof of concealed genetic variability.

On the recommendation of N. K. Koltsov, director of Moscow's Institute of Experimental Biology, the Timofeef-Ressovskys were recruited by Oskar Vogt to work at Berlin's Kaiser Wilhelm Institute for Brain Research in 1925. Official reluctance to let the couple leave Russia and enter Germany were overcome in part due to Helena's German ancestry. A number of other Russians, including Sergei R. Zarapkin and his family, technical assistant Natalie Kromm, and the Timofeeff-Ressovskys' nanny, were in residence at the Institute. Berlin was a comfortable place for women in the sciences: Cécile and Marthe Vogt, Estera Tenenbaum, Stella Rose, Rosa Schragenheim, Irmgard Leux, and Gertrud Soeken also worked at the Institute.

Following the Nazi rise to power in 1933, the political climate became increasingly uncomfortable in Germany. The Institute was pressured to dismiss foreigners, women, and Jews. Helena officially retired, though she continued to collaborate with her husband. Despite the efforts of Nikolai and Oskar Vogt, Tenenbaum was forced out of the Institute (and country) in 1934; facilitated by a research grant, Marthe Vogt left Germany in 1935. The situation in the Soviet Union was perilous: members of Nikolai and Helena's families were arrested and some were killed. Colleagues warned against returning home; when ordered to do so in 1937, Nikolai refused. He used competing offers from the United States to negotiate improved working conditions in his Berlin laboratory.

Helena's eldest son was arrested by the Germans in 1943 and her husband by the Russians in 1945. Their fate remained uncertain, but Helena was able to continue working in Berlin while her younger son resumed his studies. From 1946-1947, she worked under Hans Nachtsheim at Berlin University's Zoological Institute. In 1947, she received word that Nikolai was alive, and in August Helena and Andrei joined him in the Urals.

Helena once more worked as a researcher in a department headed by her husband, first in the Urals and then, for the last decade of her life, the Institute of Medical Radiology near Moscow. She continued her scientific work, though she received no salary and opportunities for publication were often limited by the secret nature of her work. During this period, she worked on problems of radiation ecology and population genetics, though the study of genetics was generally disallowed in the Soviet Union until the mid-60s. She was part of the first group in the USSR to study Arabidopsis'' genetics.

Helena died in 1973; the loss was a serious blow for the husband whose scientific work and political travails had always overshadowed hers. The Max Delbrück Center for Molecular Medicine hosts a lecture series named in Helena's honor.

References

Women geneticists
Russian geneticists
1898 births
1973 deaths
Scientists from Moscow
Russian women scientists
20th-century women scientists

redpajama_set_name: RedPajamaWikipedia


text:
Vinyl records make a comeback, are your old records worth thousands?
By Sara Otto
Today is Vinyl Record Day! Whether caused by nostalgia or a newfound appreciation for the distinctly unique sound of a vinyl record, vinyl sales are rapidly increasing, despite the availability of nearly free music online.
Over 13 million vinyl records were sold in the U.S. in 2016, up from 11,900,000 in 2015 and 9,200,000 in 2014, according to Nielsen reports.
Although the classics are still popular, modern hits make up a large portion of vinyl record sales today. With 49,004 albums sold, Twenty One Pilots' Blurryface was the highest selling vinyl record in 2016 according to BuzzAngle's 2016 music report. The top five rank as follows:
1. Blurryface - Twenty One Pilots: (49,004)
2. Back to Black - Amy Winehouse (41,087)
3. A Moon Shaped Pool - Radiohead (39,861)
4. Abbey Road - The Beatles (39, 615)
5. 25 - Adele (39,512)
Record sales are up 260% from five years ago, but the numbers selling today still pale in comparison to the number of vinyl records sold in their heyday.
In 1973, vinyl records made up 73% of total music sales. Today, vinyl record sales only make up .2% of all music sold.
Any of the top selling records are probably not worth much, but rarer tracks go for a pretty penny. If you have boxes of old records sitting around in your basement, they could be worth a fortune.
Ebay released a list of vinyl albums worth upwards of $25,000 sold on their site.
Elvis Presley's Good Luck Charm 33-rpm single released in 1962 is valued at $24,000. When it initially came out in 1962, it stayed at the top of the Billboard singles chart for two weeks.
A copy of the Beatles' Yesterday and Today album sold for nearly $40,000 in 1996. The copy was unique because the band was posing in butcher smocks on the cover, with pieces of meat and parts of plastic dolls hanging from them.
At John Lennon's request, this photo was used for the U.S. album cover. It was only on sale for one day.
Everyone hated the cover then, but it sure is valuable now.
Coming in on top, the vinyl copy of John Lennon and Yoko Ono's "Double Fantasy" album Lennon signed just five hours before he was fatally shot in 1980 sold for over $400,000 in 1999.
If you happen to find that one in your basement, don't let it out of your sight!
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At Audi Flatirons, we are committed to providing a world-class customer service experience to each and every guest. We believe this commitment to service results in happier customers and employees! If you have had an experience with Audi Flatirons that's worth sharing, we encourage you to do just that by posting a review of Audi Flatirons on the site of your choice. Better yet, post it on them all!
To assist you in posting your review, we have outlined the process for the most popular review sites below.
Google is a leading review site for local business reviews. Google uses the ranking of a business to determine how that business show up in search results. Posting a review on Google isn't difficult. It simply requires you be signed in to an existing Google or Gmail account, and to locate Audi Flatirons on Google.
Once there, simply click the button labeled "Write a review." A small box will open and allow you to select your rating and post your review.
Yelp! is another popular local review site. Originally focused on restaurants and bars, Yelp's influence has extended to almost every industry, including Automotive. Yelp! is a site based solely on reviews, so posting a review is easy. If you do not already have an active Yelp! account, you will need to create one.
DealerRater was started with the simple idea of providing a platform for reviews of automotive dealers. The site allows consumers to share their experience with almost any dealership. The process is designed to be very easy. DealerRater also offers a certification program. A certified dealership must meet certain criteria regarding the number and quality of reviews. You will not need an account to post your review on DealerRater.