### 数学代写|表示论代写Representation theory代考|MATH7333

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• Statistical Inference 统计推断
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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|表示论代写Representation theory代考|A Pale Block and a Point

Let $V$ be a braided vector space of dimension 3 with braiding given in the basis $\left(x_{i}\right){i \in I{3}}$ by
$$\left(c\left(x_{i} \otimes x_{j}\right)\right){i, j \in I{3}}=\left(\begin{array}{ccc} \epsilon x_{1} \otimes x_{1} & \epsilon x_{2} \otimes x_{1} & q_{12} x_{3} \otimes x_{1} \ \epsilon x_{1} \otimes x_{2} & \epsilon x_{2} \otimes x_{2} & q_{12} x_{3} \otimes x_{2} \ q_{21} x_{1} \otimes x_{3} & q_{21}\left(x_{2}+x_{1}\right) \otimes x_{3} & q_{22} x_{3} \otimes x_{3} \end{array}\right)$$
Let $V_{1}=\left\langle x_{1}, x_{2}\right\rangle, V_{2}=\left\langle x_{3}\right\rangle$. Let $\Gamma=\mathbb{Z}^{2}$ with a basis $g_{1}, g_{2}$. We realize $V$ in $\mathrm{k} \Gamma \Gamma \Gamma D$ $g_{2} \cdot x_{2}=q_{21}\left(x_{2}+x_{1}\right), g_{i} \cdot x_{3}=q_{i 2} x_{3}$.

As usual, let $\tilde{q}{12}=q{12} q_{21}$; in particular the Dynkin diagram of the braided subspace $\left\langle x_{1}, x_{3}\right\rangle$ is $\overbrace{0}^{\epsilon} \widetilde{q}_{12} \quad q 22 .$

As for other cases, we consider $K=\mathscr{B}(V)^{\operatorname{co}} \mathscr{B}\left(V_{1}\right)$; then $K=\oplus_{n \geq 0} K^{n}$ inherits the grading of $\mathscr{B}(V) ; \mathscr{B}(V) \simeq K # \mathscr{B}\left(V_{1}\right)$ and $K$ is the Nichols algebra of $K^{1}=$ ad $_{c} \cdot \mathscr{B}\left(V_{1}\right)\left(V_{2}\right)$. Now $K^{1} \in \mathscr{A ( V _ { 1 } ) \pm \mathbb { k } \Gamma} \mathcal{D}\left(V_{1}\right) \pm \mathbb{D} \Gamma$ with the adjoint action and the coaction given by $(4.4)$, i.e., $\delta=\left(\pi_{\mathscr{B}\left(V_{1}\right) # f \in \Gamma} \otimes\right.$ id) $\Delta_{\mathscr{B}(V) # \mathrm{lk} \Gamma}$. Next we introduce $\Pi_{m, n}=$

$\left(\mathrm{ad}{c} x{1}\right)^{m}\left(\operatorname{ad}{c} x{2}\right)^{n} x_{3}$; we distinguish two cases:
By direct computation,
\begin{aligned} g_{1} \cdot \mathrm{II}{m, n}=q{12} \epsilon^{m+n} \mathrm{III}{m, n}, & g{2} \cdot w_{m} &=q_{21}^{m} q_{22} w_{m} \ z_{n+1}=x_{2} z_{n}-q_{12} \epsilon^{n} z_{n} x_{2}, & \mathrm{mI}{m+1, n} &=x{1} \mathrm{II}{m, n}-q{12} \epsilon^{m+n} \mathrm{II}{m, n} x{1} \ \partial_{1}\left(\mathrm{mI}{m, n}\right)=0, & \partial{2}\left(\mathrm{mI}{m, n}\right) &=0 \ \partial{3}\left(w_{m}\right) &=\prod_{0 \leq j \leq m-1}\left(1-\epsilon^{j} \tilde{q}{12}\right) x{1}^{m} \end{aligned}

## 数学代写|表示论代写Representation theory代考|The Block Has =1

Here $\mathscr{B}\left(V_{1}\right) \simeq S\left(V_{1}\right)$ is a polynomial algebra, so that $x_{1}$ and $x_{2}$ commute, and
$$\left(\operatorname{ad}{c} x{2}\right)^{s} \mathrm{II}{m, n}=\mathrm{III}{m, n+s} \quad \text { for all } m, n, s \in \mathbb{N}{0}$$ Thus $\mathrm{mI}{m, n}, m, n \in \mathbb{N}{0}$ generate $K^{1}$. As in [AAH1, $\S 8.1$ ], we have that $$g{2} \cdot \mathrm{II}{m, n}=q{21}^{m+n} q_{22} \sum_{0 \leq j \leq n}\left(\begin{array}{c} n \ j \end{array}\right) \mathrm{m}{m+j, n-j}$$ For $q \in \mathbb{k}^{\times}$, let $\in{p}(q)=V$ be the braided vector space as in (7.1) under the assumptions that $\epsilon=1, q_{12}=q=q_{21}^{-1}, q_{22}=-1$. We call $\mathscr{B}\left(\mathbb{E}{p}(q)\right)$ and the Nichols algebras $\mathscr{B B}\left(\mathfrak{E}{\pm}(q)\right), \mathscr{B}\left(\mathfrak{E}_{\star}(q)\right)$ studied in Propositions $7.2-7.4$ the Endymion algebras.

Proposition 7.1 The algebra $\mathcal{B}\left(\mathfrak{E}{p}(q)\right)$ is presented by generators $x{1}, x_{2}, x_{3}$ and relations
\begin{aligned} x_{1}^{p} &=0, \quad x_{2}^{p}=0, \quad x_{1} x_{2}=x_{2} x_{1}, \ x_{1} x_{3} &=q x_{3} x_{1}, \ z_{t}^{2} &=0, \quad t \in \mathbb{I}{0, p-1} \end{aligned} The dimension of $\mathscr{B}\left(\mathfrak{E}{p}(q)\right)$ is $2^{p} p^{2}$, since it has a $P B W$-basis
$$B=\left{x_{1}^{m_{1}} x_{2}^{m_{2}} z_{p-1}^{n_{p-1}} \ldots z_{0}^{n_{0}}: n_{i} \in{0,1}, m_{j} \in \mathbb{I}_{0, p-1}\right}$$

## 数学代写|表示论代写Representation theory代考|The Block Has=-1

Here $\mathscr{B}\left(V_{1}\right) \simeq \Lambda\left(V_{1}\right)$ is an exterior algebra and consequently $m_{m, n}, m, n \in{0,1}$ generates $K^{1}$. By direct computation,
\begin{aligned} &g_{2} \cdot z_{1}=q_{21} q_{22}\left(z_{1}+w_{1}\right), \quad \partial_{3}\left(z_{1}\right)=\left(1-\tilde{q}{12}\right) x{2}-\tilde{q}{12} x{1}, \ &\delta\left(z_{1}\right)=g_{1} g_{2} \otimes z_{1}+\left(\left(1-\tilde{q}{12}\right) x{2}-\tilde{q}{12} x{1}\right) g_{2} \otimes x_{3} \end{aligned}

Proposition 7.2 The algebra $\mathcal{B}\left(\mathfrak{E}{+}(q)\right)$ is presented by generators $x{1}, x_{2}, x_{3}$ and relations
\begin{aligned} x_{1}^{2} &=0, \quad x_{2}^{2}=0, \quad x_{1} x_{2}=-x_{2} x_{1}, \ \left(x_{2} x_{3}-q x_{3} x_{2}\right)^{2} &=0, \quad x_{3}^{p}=0, \ x_{3}\left(x_{2} x_{3}-q x_{3} x_{2}\right) &=q{ }^{-1}\left(x_{2} x_{3}-q x_{3} x_{2}\right) x_{3}, \ x_{1} x_{3} &=q x_{3} x_{1} . \end{aligned}
Let $z_{1}=x_{2} x_{3}-q x_{3} x_{2}$. Then $\mathcal{B}\left(\mathfrak{E}{+}(q)\right)$ has a PBW-basis $$B=\left{x{1}^{m_{1}} x_{2}^{m_{1}} x_{3}^{m_{3}} z_{1}^{m_{4}}: m_{1}, m_{2}, m_{4} \in{0,1}, m_{3} \in \mathbb{I}{0, p-1}\right}$$ hence $\operatorname{dim} \mathcal{B}\left(\mathfrak{E}{+}(q)\right)=2^{3} p$.
Proof Notice that $x_{3}^{p}=0$ since $x_{3}$ is a point labeled with $q_{22}=1$ in $K^{1}$. Also, $B$ is a basis thanks to the isomorphism $\mathscr{B}\left(\mathfrak{E}{+}(q)\right) \simeq \mathscr{B}\left(K^{1}\right) # \mathscr{B}\left(V{1}\right)$. The rest of the proof follows as in [AAH1, Proposition 8.1.6].

Proposition $7.3$ The algebra $\mathcal{B}\left(\mathfrak{E}{-}(q)\right)$ is presented by generators $x{1}, x_{2}, x_{3}$ and relations $(7.17),(7.20)$,
\begin{aligned} x_{3}^{2} &=0, \quad\left(x_{2} x_{3}-q x_{3} x_{2}\right)^{p}=0, \ x_{3}\left(x_{2} x_{3}-q x_{3} x_{2}\right) &=-q^{-1}\left(x_{2} x_{3}-q x_{3} x_{2}\right) x_{3} . \end{aligned}
Let $z_{1}=x_{2} x_{3}-q x_{3} x_{2}$. Then $\mathcal{B}\left(\mathfrak{E}{-}(q)\right)$ has a $P B W$-basis $$\boldsymbol{B}=\left{x{1}^{m_{1}} x_{2}^{m_{2}} x_{3}^{m_{3}} z_{1}^{m_{4}}: m_{1}, m_{2}, m_{3} \in{0,1}, m_{4} \in \mathbb{I}{0, p-1}\right}$$ hence $\operatorname{dim} \mathcal{B}\left(E{-}(q)\right)=2^{3} p$.

## 数学代写|表示论代写Representation theory代考|A Pale Block and a Point

\left(c\left(x_{i} \otimes x_{j}\right)\right) {i, j \in I {3}}=\left(

εX1⊗X1εX2⊗X1q12X3⊗X1 εX1⊗X2εX2⊗X2q12X3⊗X2 q21X1⊗X3q21(X2+X1)⊗X3q22X3⊗X3\right)


(一个dCX1)米(广告⁡CX2)nX3; 我们区分两种情况：

G1⋅我我米,n=q12ε米+n我我我米,n,G2⋅在米=q21米q22在米 和n+1=X2和n−q12εn和nX2,米我米+1,n=X1我我米,n−q12ε米+n我我米,nX1 ∂1(米我米,n)=0,∂2(米我米,n)=0 ∂3(在米)=∏0≤j≤米−1(1−εjq~12)X1米

## 数学代写|表示论代写Representation theory代考|The Block Has =1

(广告⁡CX2)s我我米,n=我我我米,n+s 对所有人 米,n,s∈ñ0因此米我米,n,米,n∈ñ0产生ķ1. 如 [AAH1,§§8.1]，我们有

G2⋅我我米,n=q21米+nq22∑0≤j≤n(n j)米米+j,n−j为了q∈ķ×， 让∈p(q)=在是（7.1）中的编织向量空间，假设如下：ε=1,q12=q=q21−1,q22=−1. 我们称之为乙(和p(q))和 Nichols 代数乙乙(和±(q)),乙(和⋆(q))在命题中学习7.2−7.4Endymion 代数。

X1p=0,X2p=0,X1X2=X2X1, X1X3=qX3X1, 和吨2=0,吨∈我0,p−1的维度乙(和p(q))是2pp2，因为它有一个磷乙在-基础

B=\left{x_{1}^{m_{1}} x_{2}^{m_{2}} z_{p-1}^{n_{p-1}} \ldots z_{0}^{ n_{0}}: n_{i} \in{0,1}, m_{j} \in \mathbb{I}_{0, p-1}\right}B=\left{x_{1}^{m_{1}} x_{2}^{m_{2}} z_{p-1}^{n_{p-1}} \ldots z_{0}^{ n_{0}}: n_{i} \in{0,1}, m_{j} \in \mathbb{I}_{0, p-1}\right}

## 数学代写|表示论代写Representation theory代考|The Block Has=-1

G2⋅和1=q21q22(和1+在1),∂3(和1)=(1−q~12)X2−q~12X1, d(和1)=G1G2⊗和1+((1−q~12)X2−q~12X1)G2⊗X3

X32=0,(X2X3−qX3X2)p=0, X3(X2X3−qX3X2)=−q−1(X2X3−qX3X2)X3.

\boldsymbol{B}=\left{x{1}^{m_{1}} x_{2}^{m_{2}} x_{3}^{m_{3}} z_{1}^{m_{ 4}}: m_{1}, m_{2}, m_{3} \in{0,1}, m_{4} \in \mathbb{I}{0, p-1}\right}\boldsymbol{B}=\left{x{1}^{m_{1}} x_{2}^{m_{2}} x_{3}^{m_{3}} z_{1}^{m_{ 4}}: m_{1}, m_{2}, m_{3} \in{0,1}, m_{4} \in \mathbb{I}{0, p-1}\right}因此暗淡⁡乙(和−(q))=23p.

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