### 数学代写|表示论代写Representation theory代考|MTH 4107

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• 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|The Presentation by Generators and Relations

We still assume that the interaction is weak. We start by some general Remarks that are proved exactly as in [AAH1, $\S 4.3]$.
Remark 4.6 Let
$$y_{2 k}=x_{21}^{k}, \quad y_{2 k+1}=x_{1} x_{21}^{k}, \quad k \in \mathbb{N}{0} .$$ By Lemma $4.2$ $$\partial{3}\left(z_{l}\right)=\mu_{t} y_{t}, \quad z_{t} y_{n}=\epsilon^{m t} q_{21}^{n} y_{n} z_{t}, \quad t, n \in \mathbb{N}{0}$$ Lemma 4.7 Assume that $\epsilon^{2}=q{22}^{2}=1$. In $\mathscr{B}\left(\mathfrak{L}\left(q_{22}, \mathscr{G}\right)\right)$, or correspondingly $B_{-}\left(\mathfrak{L}\left(q_{22}, \mathscr{G}\right)\right)$.
$$\begin{array}{rlrl} z_{|x|+1} & =0 \ z_{t} z_{t+1} & =q_{21} q_{22} z_{t+1} z_{t} & t \in \mathbb{N}{0}, t<|x| & \ z{t}^{2} & =0 & t \in \mathbb{N}{0}, \epsilon^{t} q{22}=-1 \ \partial_{3}\left(z_{t}^{n+1}\right) & =\mu_{t} q_{21}^{n t} q_{22}^{n} n y_{t} z_{t}^{n}, & n, t \in \mathbb{N}{0}, \epsilon^{t} q{22}=1 \end{array}$$

Lemma 4.8 Let $\mathscr{B}$ be a quotient algebra of $T(V)$. Assume that $x_{1} x_{3}=q_{12} x_{3} x_{1}$, and either
(a) (3.1), or else
(b) $(3.7), x_{21} x_{3}=q_{12}^{2} x_{3} x_{21}$
hold in $\mathscr{B}$. Then for all $n \in \mathbb{N}{0}$, \begin{aligned} x{1} z_{n} &=\epsilon^{n} q_{12} z_{n} x_{1} \ x_{21} z_{n} &=q_{12}^{2} z_{n} x_{21} \end{aligned}
Lemma 4.9 Let $\mathscr{B}$ be a quotient algebra of $T(V), \epsilon^{2}=q_{22}^{2}=1$.
(i) Assume that (4.20) and (4.21) hold in $\mathscr{B}$. Then for $0 \leq t<k \leq|r|$,
$$z_{t} z_{k}=\epsilon^{t k} q_{21}^{k-t} q_{22} z_{k} z_{t}$$
(ii) Assume that $z_{t}^{2}=0$ in $\mathscr{B}$ for $t \in \mathbb{N}{0}$ such that $\epsilon^{t} q{22}=-1$. Then $z_{t} z_{t+1}=$ $q_{21} q_{22} z_{I+1} z_{t}$ in $\mathscr{B}$.
In other words, (ii) says that (4.21) for a specific $t$ implies (4.20) for $t$.

## 数学代写|表示论代写Representation theory代考|The Setting and the Main Result

Let $\theta \in \mathbb{N}{\geq 3}, \mathbb{I}{2, \theta}=\mathbb{I}{\theta}-{1}, \mathbb{I}{\theta}^{\dagger}=\mathbb{I}{\theta} \cup\left{\frac{3}{2}\right}$. Let $\lfloor i\rfloor$ be the largest integer $\leq i$. We start from the data $$\left(q{i j}\right){i, j \in \mathbb{I}{\theta}} \in\left(\mathbb{k}^{\times}\right)^{\theta \times \theta}, \quad q_{11}^{2}=1 ; \quad\left(a_{2}, \ldots, a_{\theta}\right) \in \mathbb{k}^{\mathbb{I}_{2, \theta}}$$

We assume that $q_{11}=1=: a_{1}$. Let $(V, c)$ be the braided vector space of dimension $\theta+1$, with a basis $\left(x_{i}\right){i \in \mathbb{I}{\theta}^{\dagger}}$ and braiding given by
$$c\left(x_{i} \otimes x_{j}\right)= \begin{cases}q_{\lfloor i\rfloor j} x_{j} \otimes x_{i}, & i \in \mathbb{I}{\theta}^{\dagger}, j \in \mathbb{I}{\theta} \ q_{\lfloor i\rfloor 1}\left(x_{\frac{3}{2}}+a_{\lfloor i\rfloor} x_{1}\right) \otimes x_{i}, & i \in \mathbb{I}{\theta}^{\dagger}, j=\frac{3}{2}\end{cases}$$ We say that the block and the points have discrete ghost if $a{j} \in \mathbb{F}{p}^{\mathrm{I}{2, \theta}},\left(a_{j}\right) \neq 0$. When this is the case, we pick the representative $r_{j} \in \mathbb{Z}$ of $2 a_{j}$ by imposing $r_{j} \in$ ${1-p, \ldots,-1,0}$, and set $\mathscr{G}{j}=-r{j}$. The ghost between the block and the points is the vector $\mathscr{G}=\left(\mathscr{G}{j}\right){j \in \mathrm{I}{2, \theta}}$ given by $$\mathscr{G}=-\left(r{j}\right){j \in \mathbb{I}{2, \theta}} \in \mathbb{N}{0}^{\mathbb{I}{2, \theta}}$$
The braided subspace $V_{1}$ spanned by $x_{1}, x_{\frac{3}{2}}$ is $\simeq \mathcal{V}(1,2)$, while $V_{\text {diag }}$ spanned by $\left(x_{i}\right){i \in \mathbb{I}{2, \theta}}$ is of diagonal type. Obviously,
$$V=V_{1} \oplus V_{\text {diag }}$$
Let $\mathcal{X}$ be the set of connected components of the Dynkin diagram of the matrix $\mathbf{q}=\left(q_{i j}\right){i, j \in \mathbb{I}{2, \theta}}$. If $J \in \mathcal{X}$, then we set $J^{\prime}=\mathbb{I}{2, \theta}-J$, $$V{J}=\sum_{j \in J} \mathbb{k}{g j}^{\chi{j},} \quad \mathscr{G}{J}=\left(\mathscr{G}{j}\right)_{j \in J} .$$

## 数学代写|表示论代写Representation theory代考|The Presentation of the Nichols Algebras

We give defining relations and an explicit PBW-basis of $\mathscr{B}(V)$, for all $V$ as in Theorem $5.3$, assuming that the Dynkin diagram of $V_{\text {diag }}$ is connected, i.e., $V_{\text {diag }}=V_{J}$, where $J=\mathrm{I}{2, \theta}$. Essentially the relations are the same as in [AAH1] up to adding the suitable $p$-powers; we omit the proofs as they are minor variations of those in loc. cit. The passage from connected $V{\text {diag }}$ to the general case is standard, just add the quantum commutators between points in different components. Since the case $|J|=1$ was treated in Sect. 4 , we also suppose that $|J|>1$. These braided vector spaces have names given in [AAH1], see Table 2. The braided vector subspace $V_{1} \oplus \mathrm{k} x_{2}$ of such $V$ is of type

• $\mathfrak{L}(-1,2)$ when $V$ is of type $\mathfrak{L}\left(A_{2}, 2\right)$,
• $\mathfrak{L}(\omega, 1)$ when $V$ is of type $\mathfrak{L}\left(A(1 \mid 0)_{3} ; \omega\right)$, or
• $\mathfrak{L}(-1,1)$ for all the other cases.
Thus the subalgebra generated by $V_{1} \oplus \mathrm{kx} x_{2}$ is a Nichols algebra. We recall its relations up to the change of index with respect to Sect. 4 ; the 2 and 3 there are now $\frac{3}{2}$ and 2. As in (2.2), we set $x_{i_{1} i_{2} \ldots i_{M}}=\operatorname{ad}{c} x{i_{1}} x_{i_{2} \ldots i_{M}}$. Also, we have now $z_{n}=\left(\operatorname{ad}{c} x{\frac{3}{2}}\right)^{n} x_{2}, n \in \mathbb{N}{0} .$ First, the defining relations of $\mathscr{B}(\mathcal{L}(-1,1))$ are \begin{aligned} &x{\frac{2}{2}} x_{1}-x_{1} x_{\frac{1}{2}}+\frac{1}{2} x_{1}^{2}, \ &x_{1}^{p} \ &x_{\frac{3}{2}}^{p}, \ &x_{1} x_{2}-q_{12} x_{2} x_{1}, \ &\left(\operatorname{ad}{c} x{\frac{3}{2}}^{2}\right)^{2} x_{2}, \ &x_{2}^{2}, x_{\frac{3}{2}}^{2} 2^{\circ} \end{aligned}

## 数学代写|表示论代写Representation theory代考|The Presentation by Generators and Relations

∂3(和l)=μ吨是吨,和吨是n=ε米吨q21n是n和吨,吨,n∈ñ0引理 4.7 假设ε2=q222=1. 在乙(大号(q22,G))，或相应地乙−(大号(q22,G)).

(a) (3.1), 或者
(b)(3.7),X21X3=q122X3X21

X1和n=εnq12和nX1 X21和n=q122和nX21

(i) 假设 (4.20) 和 (4.21) 成立乙. 那么对于0≤吨<ķ≤|r|,

(ii) 假设和吨2=0在乙对于 $t \in \mathbb{N} {0}s在CH吨H一个吨\epsilon^{t}q {22}=-1.吨H和nz_{t} z_{t+1}=q_{21} q_{22} z_{I+1} z_{t}一世n\mathscr{B}.我n○吨H和r在○rds,(一世一世)s一个是s吨H一个吨(4.21)F○r一个sp和C一世F一世C吨一世米pl一世和s(4.20)F○rt$。

## 数学代写|表示论代写Representation theory代考|The Setting and the Main Result

(q一世j)一世,j∈我θ∈(ķ×)θ×θ,q112=1;(一个2,…,一个θ)∈ķ我2,θ

C(X一世⊗Xj)={q⌊一世⌋jXj⊗X一世,一世∈我θ†,j∈我θ q⌊一世⌋1(X32+一个⌊一世⌋X1)⊗X一世,一世∈我θ†,j=32我们说块和点有离散重影如果一个j∈Fp我2,θ,(一个j)≠0. 在这种情况下，我们选择代表rj∈从的2一个j通过强加rj∈ 1−p,…,−1,0, 并设置Gj=−rj. 块和点之间的幽灵是向量G=(Gj)j∈我2,θ由

G=−(rj)j∈我2,θ∈ñ0我2,θ

## 数学代写|表示论代写Representation theory代考|The Presentation of the Nichols Algebras

• 大号(−1,2)什么时候在是类型大号(一个2,2),
• 大号(ω,1)什么时候在是类型大号(一个(1∣0)3;ω)， 或者
• 大号(−1,1)对于所有其他情况。
因此生成的子代数在1⊕ķXX2是 Nichols 代数。我们回想起它与 Sect 指数变化的关系。4；现在有 2 和 332和 2. 如 (2.2) 中，我们设置X一世1一世2…一世米=广告⁡CX一世1X一世2…一世米. 另外，我们现在有和n=(广告⁡CX32)nX2,n∈ñ0.一、定义关系乙(大号(−1,1))是X22X1−X1X12+12X12, X1p X32p, X1X2−q12X2X1, (广告⁡CX322)2X2, X22,X3222∘

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