### 数学代写|表示论代写Representation theory代考|MATH4314

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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 General Case

Take $\mathfrak{h}{j} \in I(f, \mathfrak{g})$ and $H{j}=\exp \left(h_{j}\right)(j=1,2)$. We shall construct an intertwining operator $T_{\mathfrak{b}{2} \mathfrak{h}{1}}$ so that, given $\mathfrak{h}_{j} \in I(f, \mathfrak{g})(1 \leq j \leq 3)$, the composition formula
holds. In order to define this operator we use a third polarization, a Vergne polarization to be precise.

Theorem 2.5.1 Let $\mathfrak{h}{j} \in I(f, \mathfrak{g})(j=1,2)$ and take a Vergne polarization ho at $f \in \mathrm{g}^{*}$. Then the intertwining isometry $$T{\mathfrak{h}{2} \mathfrak{h}{1}}=a\left(\mathfrak{h}{1}, \mathfrak{h}{0}, \mathfrak{h}{2}\right) T{\mathfrak{h}{2} \mathfrak{h}{0}} \circ T_{\mathfrak{h}{0} \mathfrak{h}{1}},$$
where $T_{\mathfrak{h}{2} \mathfrak{h}{0}}, T_{\mathfrak{h}{0} \mathfrak{h}{1}}$ are the intertwining isometries defined in Theorem 2.4.2, does not depend on the choice of the Vergne polarization $\mathrm{h}{0}$. Moreover, $T{\mathrm{h}{2} \mathfrak{h}{1}}$ coincides with $I_{\mathfrak{h}{2} \mathfrak{h}{1}}$ if at least one of $\mathfrak{h}{1}, \mathfrak{h}{2}$ is a Vergne polarization.

Proof We proceed again by induction on $\operatorname{dim} G$. As in the previous proofs, we can at once suppose that there is no minimal ideal which is not central and $f$ does not vanish on any ideal of $\mathrm{g}$.
Let now $\mathbf{h}{3}, \mathbf{h}{4}$ be two Vergne polarizations at $f$. We claim
$$a\left(b_{1}, h_{3}, h_{2}\right) T_{h_{2} b_{3}} \circ T_{h_{3} h_{1}}=a\left(b_{1}, h_{4}, b_{2}\right) T_{b_{2} b_{4}} \circ T_{\mathfrak{h}{4} b{1}},$$
where $a\left(\mathfrak{h}{i}, \mathbf{h}{j}, \mathfrak{h}{k}\right)=e^{\frac{\mathbf{I}^{2}}{} \tau\left(\mathfrak{h}, \mathfrak{h}{j}, \mathbf{h}{k}\right)}$. If there is a minimal non-central ideal a contained in $\mathfrak{h}{3} \cap \mathbf{h}{4}$, then $\mathbf{h}{3} \cup \mathbf{h}{4} \subset a^{f}$ and we can apply the induction hypothesis to $\mathfrak{h}{i}^{\prime}=\mathfrak{h}{i} \cap \mathfrak{a}^{f}+\mathfrak{a}, i=1,2$. Thus $$a\left(\mathfrak{h}{1}^{\prime}, \mathfrak{h}{3}, \mathfrak{h}{2}^{\prime}\right) T_{\mathfrak{h}{2}^{\prime} \mathfrak{h}{3}} \circ T_{\mathfrak{h}{3} \mathfrak{h}{1}^{\prime}}=a\left(\mathfrak{h}{1}^{\prime}, \mathfrak{h}{4}, \mathfrak{h}{2}^{\prime}\right) T{\mathfrak{h}{2}^{\prime} \mathfrak{h}{4}} \circ T_{\mathfrak{h}{4} \mathfrak{h}{1}^{\prime}}$$
We deduce from this
$$a\left(\mathfrak{h}{1}^{\prime}, \mathfrak{h}{3}, \mathfrak{h}{2}^{\prime}\right) T{\mathfrak{h}{2} \mathfrak{h}{3}} \circ T_{\mathfrak{h}{3} \mathfrak{h}{1}}=a\left(\mathfrak{h}{1}^{\prime}, \mathfrak{h}{4}, \mathfrak{h}{2}^{\prime}\right) T{\mathfrak{h}{2} \mathfrak{h}{4}} \circ T_{\mathfrak{h}{4} \mathfrak{h}{1}}$$
Hence it suffices to show
$$a\left(h_{1}, h_{3}, h_{2}\right) a\left(h_{1}^{\prime}, h_{4}, h_{2}^{\prime}\right)=a\left(h_{1}, h_{4}, h_{2}\right) a\left(h_{1}^{\prime}, h_{3}, h_{2}^{\prime}\right)$$
But from Lemma 2.3.11 we have
$a\left(h_{1}, h_{3}, h_{2}\right)=a\left(h_{4}, h_{3}, h_{2}\right) a\left(h_{1}, h_{4}, h_{2}\right) a\left(h_{1}, h_{3}, b_{4}\right)$
$=a\left(h_{4}, \mathbf{h}{3}, \mathbf{h}{2}^{\prime}\right) a\left(h_{1}, \mathbf{h}{4}, \mathbf{h}{2}\right) a\left(\mathbf{h}{1}^{\prime}, \mathbf{h}{3}, \mathbf{h}_{4}\right)$

## 数学代写|表示论代写Representation theory代考|A Local Result

We denote by $\mathscr{H}{\rho}$ the Hilbert space of a unitary representation $\rho$ of $G$, by $\mathscr{H}{\rho}^{\infty}$ the space of $C^{\infty}$-vectors of $\rho$ equipped with the usual topology and by $\mathscr{H}{\rho}^{-\infty}$ the anti-dual space of $\mathscr{H}{\rho}^{\infty}$. Given a closed subgroup $K$ of $G$ and a character $\lambda$ of $K$, we set
$$\left(\mathscr{H}{\rho}^{-\infty}\right)^{K, \lambda}=\left{a \in \mathscr{H}{\rho}^{-\infty} ; \rho(k) a=\lambda(k) a, \forall k \in K\right}$$
Let $T_{\mathfrak{h}{2} \mathfrak{h}{1}}$ be the isometry defined above which intertwines $\pi_{1}$ and $\pi_{2}$. This section is devoted to the proof of the next theorem.
Theorem 2.6.1 There exists a positive form $v=v_{H_{2}, H_{1} \cap H_{2}}$ such that
$$T_{b_{2} h_{1}} \varphi(e)=\oint_{H_{2} /\left(H_{1} \cap H_{2}\right)} \varphi(h) \chi_{f}(h) \Delta_{H_{2}, G}^{-1 / 2}(h) d v(h)$$
for any function $\varphi \in \mathscr{H}{\pi{1}}^{\infty}$ whose support is sufficiently small modulo $H_{1}$.

## 数学代写|表示论代写Representation theory代考|The Case Where h1 + h2 Is a Subalgebra

Lemma 2.7.1 Let $\mathfrak{h}{j} \in I(f, \mathfrak{g})(j=1,2)$ be such that $\mathfrak{t}=\mathfrak{h}{1}+\mathfrak{h}{2}$ is a subalgebra of $\mathrm{g}$. Then $K=\exp \mathrm{E}=\mathrm{H}{2} \mathrm{H}{1}$ and there exists a coexponential basis of $\mathrm{h} 1 \mathrm{in} \mathrm{g}, a$ part of which is coexponential to $\mathfrak{h}{1} \cap \mathfrak{h}{2}$ in $\mathrm{h}{2}$. In particular, $H_{1} H_{2}$ is closed in $G$.
Proof Supposing $\mathfrak{g}=\mathfrak{h}{1}+\mathbf{h}{2}$, we shall verify $G=H_{1} H_{2}$. We already know that $H_{1} H_{2}$ is open in $G$. Consider a sequence of subalgebras
$$h_{1}=m_{0} \subset m_{1} \subset \cdots \subset m_{k}=\mathfrak{h}{1}+[\mathfrak{g}, \mathfrak{g}]$$ such that the adjoint action of $\mathrm{m}{j-1}$ on $\mathrm{m}{j} / \mathrm{m}{j-1}$ is irreducible for all $1 \leq j \leq k$. We put $M_{j}=\exp \left(m_{j}\right)$. Supposing $M_{j-1} \subset H_{1} H_{2}$, we shall prove the same inclusion for $M_{j}$. Suppose first $\operatorname{dim}\left(m_{j} / m_{j-1}\right)=1$. We take a coexponential basis ${X}$ of $\mathrm{m}{j-1}$ in $\mathrm{m}{j}$. By writing $g \in M_{j}$ in the form $g=\exp (x X) \cdot g_{0}$, where $x \in \mathbb{R}$ and $g_{0} \in M_{j-1}$, we immediately see that $g \in H_{2} H_{1}$ if and only if $\exp (x X) \in H_{2} H_{1}$.

On the other hand, the condition
$$\exp (x X) \in H_{2} H_{1}=\left{a \in G ; a \cdot\left(f+\left(\mathfrak{h}{1}\right)^{\perp}\right) \cap\left(f+\left(\mathbf{h}{2}\right)^{\perp}\right) \neq \emptyset\right}$$
is expressed by the vanishing of a real analytic function of $x$. Since $H_{2} H_{1}$ is open in $G$, we have rank $B(g) \leq \operatorname{rank} A(e)$ for all $g \in G$. Here, $B(g)$ and $A(e)$ denote the matrices introduced just after Lemma 2.3.11. On the other hand we have $\left(H_{2} H_{1}\right) \cap$ $M_{j}=\exp \left(\mathrm{h}{2} \cap \mathrm{m}{j}\right) H_{1}$, which is connected. These observations give us the desired inclusion $\mathrm{M}{j} \subset \mathrm{H}{2} \mathrm{H}_{1}$.

Suppose now $\operatorname{dim}\left(\mathrm{m}{j} / \mathrm{m}{j-1}\right)=2$. We take $\left{X, X^{\prime}\right}$ in $[\mathfrak{g}, \mathfrak{g}]$ so that $\left{X, X^{\prime}\right}$ is a coexponential basis of $\mathrm{m}{j-1}$ in $\mathrm{m}{j}$. When we write $X=Y_{1}+Y_{2}, X^{\prime}=Y_{1}^{\prime}+Y_{2}^{\prime}$ with $Y_{i}, Y_{i}^{\prime} \in \mathfrak{h}{i}(i=1,2)$, we may for instance assume $\left[Y{i}^{\prime}, \mathrm{m}{j}\right] \subset \mathrm{m}{j-1}(i=1,2)$. If, further, $\left[Y_{i}, \mathrm{~m}{j}\right] \subset \mathrm{m}{j-1}(i=1,2)$, then $\left{Y_{2}, Y_{2}^{\prime}\right}$ becomes a coexponential basis of $\mathrm{m}{j-1}$ in $\mathrm{m}{j}$. For otherwise $\left[Y_{i}, \mathrm{~m}{j}\right] \not \subset \mathrm{m}{j-1}$ and it suffices to replace $X$ by $\left[Y_{1}, X^{\prime}\right]-\left[X, Y_{2}^{\prime}\right]=\left[Y_{1}, Y_{1}^{\prime}\right]-\left[Y_{2}, Y_{2}^{\prime}\right]$ to go fall back into the previous case.

Hence $H_{1} \subset M_{k} \subset H_{2} H_{1}$. Since $\mathrm{m}{k} \supset[\mathrm{g}, \mathrm{g}]$, we can finally choose elements of $\mathfrak{h}{2}$ which constitute a coexponential basis of $m_{k}$ in $\mathfrak{g}$.

## 数学代写|表示论代写Representation theory代考|The General Case

=一个(H4,H3,H2′)一个(H1,H4,H2)一个(H1′,H3,H4)

## 数学代写|表示论代写Representation theory代考|A Local Result

\left(\mathscr{H}{\rho}^{-\infty}\right)^{K, \lambda}=\left{a \in \mathscr{H}{\rho}^{-\infty} ; \rho(k) a=\lambda(k) a, \forall k \in K\right}\left(\mathscr{H}{\rho}^{-\infty}\right)^{K, \lambda}=\left{a \in \mathscr{H}{\rho}^{-\infty} ; \rho(k) a=\lambda(k) a, \forall k \in K\right}

## 数学代写|表示论代写Representation theory代考|The Case Where h1 + h2 Is a Subalgebra

H1=米0⊂米1⊂⋯⊂米ķ=H1+[G,G]这样的伴随动作米j−1上米j/米j−1对所有人都是不可约的1≤j≤ķ. 我们把米j=经验⁡(米j). 假如米j−1⊂H1H2，我们将证明相同的包含米j. 假设首先暗淡⁡(米j/米j−1)=1. 我们以共指数为基础X的米j−1在米j. 通过写作G∈米j在表格中G=经验⁡(XX)⋅G0， 在哪里X∈R和G0∈米j−1，我们立即看到G∈H2H1当且仅当经验⁡(XX)∈H2H1.

\exp (x X) \in H_{2} H_{1}=\left{a \in G ; 一个 \cdot\left(f+\left(\mathfrak{h}{1}\right)^{\perp}\right) \cap\left(f+\left(\mathbf{h}{2}\right)^ {\perp}\right) \neq \emptyset\right}\exp (x X) \in H_{2} H_{1}=\left{a \in G ; 一个 \cdot\left(f+\left(\mathfrak{h}{1}\right)^{\perp}\right) \cap\left(f+\left(\mathbf{h}{2}\right)^ {\perp}\right) \neq \emptyset\right}

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