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# 数学代写|表示论代写Representation Theory代考|MAT4270 Down-Up Representations

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## 数学代写|表示论代写Representation Theory代考|Down-Up Representations

Definition 1.5.1 Let $G$ be a Lie group, $H$ a closed subgroup of $G$ and $\pi$ a unitary representation of $G$. The unitary representation $\rho(G, H, \pi)=\operatorname{ind}H^G\left(\pi{\mid H}\right)$ of $G$ is called a down-up representation.
If $G$ is an exponential solvable Lie group, then the following holds [96].
Theorem 1.5.2 Let $G$ be an exponential solvable Lie group and let $H$ be a proper, connected, closed subgroup of $G$. If $\pi$ is a unitary representation of $G$, then the down-up representation $\rho(G, H, \pi)$ is reducible.

Remark 1.5.3 Let $G=\exp \mathfrak{g}$ be an exponential solvable Lie group, let $H=\exp \mathfrak{h}$ be an analytic subgroup, and let $\pi$ be an irreducible unitary representation of $G$.
(1) It follows from Theorem $1.5 .2$ that a down-up representation $\rho(G, H, \pi)$ is irreducible if and only if $\rho(G, H, \pi) \simeq \pi$, and this can occur only if $G=H$.
(2) By Kirillov’s theory there exist a linear form $f$ in $\mathfrak{g}^{\star}$ and a Pukanszky polarization $B=\exp (\mathfrak{b})$ for $f$ such that $\pi=\operatorname{ind}B^G \chi_f$. Suppose that $K=B H$ is a closed subgroup of $G$ and that $\pi{\mid K}$ is irreducible. In this case $\pi_{\mid K}=\operatorname{ind}_B^K \chi_f$.

In fact, since $\pi=\operatorname{ind}B^G \chi_f=\operatorname{ind}_K^G\left(\operatorname{ind}_B^K \chi_f\right)$, it follows that ind ${ }_B^K \chi_f$ is irreducible, and hence equivalent to $\pi{\mid K}$. Therefore $K=G$. Moreover,
$$\rho(G, H, \pi)=\operatorname{ind}H^G(\pi \mid H)=\operatorname{ind}_H^G\left(\operatorname{ind}_B^G \chi_f \mid H\right)=\operatorname{ind}_H^G\left(\operatorname{ind}{H \cap B}^H \chi_f\right)=\operatorname{ind}{H \cap B}^G \chi_f,$$ which follows from the Mackey subgroup theorem. In this situation we must only apply the results concerning $\operatorname{ind}{H \cap B}^G \chi_f$.
(3) Suppose that $\mathfrak{h}$ is a Pukanszky polarization common to elements of the orbit $G \cdot f=\Omega(\pi)$. Then one can readily see that
$$\pi_{\mid H} \simeq \int_{\Omega(\pi)}^{\oplus} \chi_l d \Omega(l),$$
where $d \Omega$ is the canonical measure on the orbit $\Omega(\pi)$. It follows that
$$\rho(G, H, \pi) \simeq \int_{\Omega(\pi)}^{\oplus} \operatorname{ind}_H^G \chi_l d \Omega(l) \simeq \infty \cdot \pi$$

## 数学代写|表示论代写Representation Theory代考|The Down-Up Formula

In this section we prove the orbital spectrum formula for down-up representations of exponential solvable Lie groups.

Theorem 1.5.4 Let $G$ be an exponential solvable Lie group, $H$ be a closed connected subgroup of $G$ with Lie algebra $\mathfrak{h}$, and $\pi$ be an irreducible unitary representation of $G$. Let $d \Omega$ be a canonical measure on the orbit $\Omega(\pi)$, and let $\lambda$ be the Lebesgue measure on $\mathfrak{h}^{\perp}$. Let $\mu_\pi^{G, H}$ be the push-forward of the measure $(d \Omega \times \lambda)$ on $\Omega(\pi) \times \mathfrak{h}^{\perp}$ under the map
$$\Omega(\pi) \times \mathfrak{h}^{\perp} \rightarrow p(\Omega(\pi)) \times \mathfrak{h}^{\perp} \rightarrow p(\Omega(\pi))+\mathfrak{h}^{\perp} \rightarrow G \cdot\left(p(\Omega(\pi))+\mathfrak{h}^{\perp}\right) / G=G \cdot\left(\Omega(\pi)+\mathfrak{h}^{\perp}\right) / G$$
The down-up orbital spectrum formula reads
$$\rho(G, H, \pi) \simeq \int_{G \cdot\left(\Omega(\pi)+\mathfrak{h}^{\perp}\right) / G}^{\oplus} m_\pi(\phi) \pi_\phi d \mu_\pi^{G, H}(\phi)$$
where
$$m_\pi(\phi)=\sum_{\Omega^H \in\left[p(\Omega(\pi)) \cap p\left(\Omega\left(\pi_\phi\right)\right)\right] / H} n_\pi^{\Omega^H} n_{\pi_\phi}^{\Omega^H}$$
with $n_\pi^{\Omega^H}=#\left[\Omega(\pi) \cap p^{-1}\left(\Omega^H\right)\right] / H$ and $n_{\pi_\phi}^{\Omega^H}=#\left[\Omega\left(\pi_\phi\right) \cap p^{-1}\left(\Omega^H\right)\right] / H$.

## 数学代写|表示论代写Representation Theory代考|Down-Up Representations

(1) 由定理1.5.2$可知，当且仅当$rho(G, H, \pi)\simeq \pi$时，一个下行上行表示$rho(G, H, \pi)$是不可还原的，而这只能在$G=H$时发生。 (2) 根据基里洛夫的理论，在$mathfrak{g}^{\star}$中存在一个线性形式$f$和对$f$的普坎斯基极化$B=\exp (b)$，使得$pi=$ind$B^G \chi_f$。假设$K=B H$是$G$的一个封闭子群，并且$pi\mid K$是不可还原的。在这种情况下，$pi_{mid K}=\operatorname{ind}B^K \chi_f$事实上，由于$pi=operatorname{ind}_B^G chi_f$B^G\chi_f=\operatorname{ind}_K^G\left(\operatorname{ind}_B^K \chi_f\right)$，因此Ind ${ }_B^K \chi_f$是不可还原的，因此等同于$pi \mid K$。因此$K=G$。此外。 $$\rho(G, H, \pi）=\operatorname{ind} H^G（\pi mid H）=\operatorname{ind}_H^G\left（\operatorname{ind}_B^G \chi_f \mid H\right）=\operatorname{ind}_H^G\left（\operatorname{ind} H\cap B^H \chi_fright）=\operatorname{ind} H \cap B^G \chi_f。$$ 这是从Mackey子群定理中得出的。在这种情况下，我们必须只应用关于ind $H\cap B^G \chi_f$的结果 (3) 假设$mathrm{h}$是$G\cdot f=\Omega(\pi)$轨道上的元素所共有的Pukanszky极化。那么我们可以很容易地看到 $$\pi{\mid H} \simeq \int_{\Omega(\pi)}^{\oplus} \chi_l d \Omega(l)。$$

$$\rho(G, H, \pi) \simeq\int_{\Omega(\pi)}^{oplus } \operatorname{ind}_H^G\chi_l d \Omega(l) \simeq \infty \cdot \pi$$

## 数学代写|表示论代写Representation Theory代考|The Down-Up Formula

$$\Omega(\pi) \times mathfrak{h}^{perp}下的映射Omega(\pi)mathfrak{h}^{perp}。\Rightarrow p(\Omega(\pi)) \times \mathfrak{h}^{perp} \rightarrow p(operp) \rightarrow p(\Omega(\pi))+\mathfrak{h}^{perp}。\G cdot\left(p(\Omega(\pi))+mathfrak{h}^{perp}\right) / G=G cdot\left(\Omega(\pi)+mathfrak{h}^{perp}\right) / G$$

$$\rho(G, H, \pi) \simeq\int_{G \cdot\left(\Omega(\pi)+\mathfrak{h}^{perp}\right) / G}^{\oplus} m_\pi(\phi) \pi_\phi d \mu_pi^{G, H}(\phi)$$

$$m_\pi(\phi)=\sum_{\Omega^H \in\left[p(\Omega(\pi)) \cap p\left(\Omega\left(\pi_phi\right)\right)] / H} n_{pi^{\Omega^H} n_{pi_phi}^{\Omega^H}。$$

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