### 数学代写|表示论代写Representation theory代考|Intertwining Operators for Irreducible Representations

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## 数学代写|表示论代写Representation theory代考|A Trace Relation

Let $G=\exp \mathfrak{g}$ be an exponential solvable Lie group with Lie algebra $\mathfrak{g}$. If h,s and $\mathfrak{h}{2}$ are two polarizations of $\mathfrak{g}$ at $f \in \mathfrak{g}^{*}$ which satisfy the Pukanszky condition, the orbit method asserts that the monomial representations $\pi{i}=$ ind $_{H_{i}}^{G} \chi_{f}\left(H_{i}=\right.$ $\exp \left(h_{i}\right), 1 \leq i \leq 2$ ) of $G$ are irreducible and mutually equivalent. We are interested in constructing an explicit intertwining operator between these representations. Let us sketch the idea, which goes back to Vergne [151]. If we have
$$\text { Tr } \mathrm{ad}{\mathfrak{h}{1} /\left(\mathfrak{h}{1} \cap \mathfrak{h}{2}\right)} X+\operatorname{Tr} \mathrm{ad}{\mathbf{h}{2} /\left(\mathfrak{h}{1} \cap \mathfrak{h}{2}\right)} X=0$$ for any $X \in \mathfrak{h}{1} \cap \mathfrak{h}{2}$, then
$$\Delta_{H_{1}, G}(h)=\Delta_{H_{2}, G}(h) \Delta_{H_{1} \cap H_{2}, H_{2}}(h)^{2}$$
for all $h \in H_{1} \cap H_{2}$, and, for $\phi$ in the Hilbert space $\mathscr{H}{\pi{1}}$ of $\pi_{1}$ and $g \in G$, the function $\Phi_{g}$ on $H_{2}$ given by
$$\Phi_{g}(h)=\phi(g h) \chi_{f}(h) \Delta_{H_{2}, G}^{-1 / 2}(h)$$
verifies the relation
$$\Phi_{g}(h x)=\Delta_{H_{1} \cap H_{2}, H_{2}}(x) \Phi_{g}(h)\left(h \in H_{2}, x \in H_{1} \cap H_{2}\right)$$
Thus we can, at least formally, consider the integral
$$\left(I_{b_{2} \mathfrak{b}} \phi\right)(g)=\oint_{H_{2} /\left(H_{1} \cap H_{2}\right)} \phi(g h) \chi_{f}(h) \Delta_{H_{2}, G}^{-1 / 2}(h) d v_{H_{2}, H_{1} \cap H_{2}}(h)$$
for $\phi \in \mathscr{H}{\Pi}$ and $g \in G$. If this integral converges for any $g \in G$, it is clear that $I{\mathfrak{h}{2} \mathfrak{h}{1}} \phi$ verifies the covariance relation required on the elements of the space $\mathscr{H}{\pi{2}}$, and $I_{\left.\mathfrak{h}_{2} \mathfrak{h}\right) 1}$ commutes with the action of $G$ by left translations. In fact, Vergne proved the following proposition. An ideal of $g$ is said to be minimal non-central if it is minimal among all non-central ideals of $\mathfrak{g}$. For an ideal $a$ of $\mathfrak{g}$, we put
$$\mathfrak{a}^{f}={X \in \mathfrak{g} ; f([X, \mathfrak{a}])={0}} .$$

## 数学代写|表示论代写Representation theory代考|Relations Between Two Polarizations

We keep the notations.
Remark 2.3.1 Without assuming the Pukanzsy condition Theorem $2.2 .2$ might fail, and we cannot write the integral of $(2.2 .1)$ even if $\pi_{1} \simeq \pi_{2}$. Indeed, let $\mathfrak{g}$ be the completely solvable Lie algebra of dimension 4 with basis $(T, P, Q, Z)$ satisfying
$$[T, P]=\frac{1}{2} P,[T, Q]=\frac{1}{2} Q,[T, Z]=[P, Q]=Z .$$

Let $f=Z^{} \in \mathfrak{g}^{}$. Then $\mathfrak{g}(f)={0}$ and
$$\mathbf{h}{1}=\mathbb{R} T+\mathbb{R} P+\mathbb{R} Z, \mathbf{h}{2}=\mathbb{R} T+\mathbb{R} Q+\mathbb{R} Z$$
belong to $M(f, \mathfrak{g})$, but neither $\mathfrak{h}{1}$ nor $\mathfrak{h}{2}$ verifies the Pukanszky condition.
There exist two open coadjoint orbits $O_{\pm}$of $G=\exp g$ :
$$O_{+}=\left{l \in \mathfrak{g}^{} ; l(Z)>0\right}, O_{-}=\left{l \in \mathfrak{g}^{} ; l(Z)<0\right}$$
Denoting by $\rho(\Omega)$ the irreducible unitary representation of $G$ corresponding to the orbit $\Omega \in \mathfrak{g}^{*} / G$, we know that $\pi_{1} \simeq \pi_{2} \simeq \rho\left(O_{+}\right) \oplus \rho\left(O_{-}\right)$(cf. [151]). However, it is evident that $\operatorname{Tr} \mathrm{ad}{\mathfrak{h}{1} /\left(\mathfrak{h}{1} \cap \mathfrak{h}{2}\right)} T=\operatorname{Tr} \mathrm{ad}{\mathfrak{h}{2} /\left(\mathfrak{h}{1} \cap \mathfrak{h}{2}\right)} T=1 / 2$.

When $G=\exp g$ is nilpotent and $K_{1}, K_{2}$ are analytic subgroups, the product $K_{1} K_{2}=\left{k_{1} k_{2} ; k_{1} \in K_{1}, k_{2} \in K_{2}\right}$ is always a closed subset of $G$ (cf. [97]). We denote by $\mathscr{H}\left(f, \mathfrak{h}{1}, G\right)$ the dense subspace of $\mathscr{H}{\pi_{1}}$ consisting of continuous functions with compact support modulo $H_{1}$. It follows that the integral (2.2.1) converges for $\phi \in \mathscr{H}\left(f, \mathbf{h}{1}, G\right)$. In fact, Lion proved that (2.2.1) gives an intertwining operator between $\pi{1}$ and $\pi_{2}$. On the other hand, when we pass to the exponential case $K_{1} K_{2}$ might not be closed, thus making the convergence of integral (2.2.1) a serious issue.
Example 2.3.2 Take
$$G_{2}=\exp g_{2}=\left{\left(\begin{array}{ll} a & b \ 0 & 1 \end{array}\right) \in M_{2}(\mathbb{R}) ; a>0\right}(a x+b \text { group })$$
The elements
$$e_{1}=\left(\begin{array}{ll} 1 & 0 \ 0 & 0 \end{array}\right), e_{2}=\left(\begin{array}{ll} 0 & 1 \ 0 & 0 \end{array}\right)$$
form a basis of $\mathfrak{g}{2}$ such that $\left[e{1}, e_{2}\right]=e_{2}$. Let $f=e_{2}^{} \in \mathfrak{g}{2}^{}, \mathfrak{t}{1}=\mathbb{R} e_{1}, \mathfrak{e}{2}=\mathbb{R}\left(e{1}+\right.$ $\left.e_{2}\right)$ and $K_{i}=\exp \mathfrak{t}{i}(i=1,2)$. Then $\mathfrak{t}{i} \in M\left(f, \mathfrak{g}{2}\right)$ but $\mathfrak{t}{i} \notin I\left(f, \mathfrak{g}{2}\right)(i=1,2)$. We immediately see $$K{2} K_{1}=\left{g=\left(\begin{array}{ll} a & b \ 0 & 1 \end{array}\right) \in G_{2} ; b>-1\right}$$
and $\mathfrak{t}{2}=g{0} \cdot \mathfrak{t}{1}$ with $g{0}=\left(\begin{array}{cc}1 & -1 \ 0 & 1\end{array}\right) \in G_{2}$.

## 数学代写|表示论代写Representation theory代考|Vergne Polarizations

We retain the notations $G=\exp g, f \in \mathfrak{g}^{*}, \mathfrak{h}{j} \in I(f, \mathfrak{g})$ and $H{j}=\exp \left(\mathfrak{h}_{j}\right)$ for $j=1,2$.

Lemma 2.4.1 If $h_{1}$ is a Vergne polarization, then there exists a coexponential basis of $\mathrm{h}{1} \cap \mathrm{h}{2}$ in $\mathrm{h}{2}$ which is a part of a coexponential basis of $\mathrm{h}{1}$ in $\mathrm{g}$. Likewise, there exists a coexponential basis of $\mathrm{h}{1} \cap \mathrm{h}{2}$ in $\mathrm{h}{1}$ which is a part of a coexponential basis of $h{2}$ in $\mathrm{g}$. In particular, $\mathrm{H}{2} \mathrm{H}{1}$ (hence $\mathrm{H}{1} \mathrm{H}{2}$, too) is closed in $G$.

Proof We proceed by induction on $\operatorname{dim} G$. If there exists a non-trivial ideal a of $\mathfrak{g}$ on which $f$ vanishes, everything passes to the quotient $G / A$ with $A=\exp a$. This case will be excluded in what follows. If there exists a minimal ideal $a$ which is not central, we know (cf. [24, Chap. VI]) that a is contained in any element of $I(f, \mathfrak{g})$. Hence for $j=1,2, \mathfrak{h}{j} \subset \mathfrak{a}^{f} \neq \mathfrak{g}$. Suppose that the Vergne polarization $\mathfrak{h}{1}$ is constructed starting from a good sequence $\mathfrak{s}=\left(a_{j}\right){0 \leq j \leq n}$ of subalgebras, namely: $$a{j-1} \subset a_{j}, \operatorname{dim}\left(a_{j}\right)=j, \mathbf{h}{1}=\sum{j=1}^{n} a_{j}\left(f_{j}\right), f_{j}=\left.f\right|{a{j}}(1 \leq j \leq n) .$$
Put $\mathfrak{b}{j}=\mathfrak{a}{j} \cap \mathfrak{a}^{f}$ and $f_{j}^{\prime}=\left.f\right|{\mathfrak{b}{j}}(0 \leq j \leq n)$. There exists an index $j_{0}\left(1 \leq j_{0} \leq n\right)$ such that $\mathbf{b}{j 0}=\mathfrak{b}{j_{0}-1}$ and it is clear that $\mathfrak{s}^{\prime}=\left(\mathfrak{b}{j}\right){0 \leq j \leq n, j \neq j_{0}}$ is a good sequence of subalgebras of $\mathfrak{a}^{f}$. Since $\mathbf{h}{1}=\sum{j=1}^{n} a_{j}\left(f_{j}\right) \subset \mathbf{a}^{f}$, we see $a_{j}\left(f_{j}\right) \subset b_{j}\left(f_{j}^{\prime}\right)$ for every $j$ and hence
$$\mathfrak{h}{1} \subset \sum{j=1}^{n} \mathfrak{b}{j}\left(f{j}^{\prime}\right)$$
Well, $\mathfrak{h}_{1}$ being a Lagrangian subspace, we necessarily have equality in (2.4.1). Now it suffices to apply the induction hypothesis to $A^{f}=\exp \left(a^{f}\right)$.

Suppose now that there exists no minimal ideal of $\mathfrak{g}$ which is not central. Take the good sequence $s$ which defines $\mathfrak{h}{1}$. With our hypothesis, we necessarily have $\mathfrak{a}{1}=\mathfrak{z}$ and $f_{1} \neq 0$. If $\mathfrak{a}{2}$ is an ideal of $\mathfrak{g}$, the above reasoning shows that $\mathfrak{h}{1}$ is a Vergne polarization of the subalgebra $\left(a_{2}\right)^{f}$. If $h_{2} \subset\left(a_{2}\right)^{f}$, it suffices to apply the induction hypothesis. If not, we modify $\mathfrak{h}{2}$ to $\mathfrak{h}{2}^{\prime}=\left(\mathfrak{h}{2} \cap\left(\mathfrak{a}{2}\right)^{f}\right)+\mathfrak{a}{2}$ and $H{2}$ to $H_{2}^{\prime}=\exp \left(\mathfrak{h}{2}^{\prime}\right)$. From the Pukanszky condition we can take (cf. [24, Chap. VI]) a coexponential basis ${X}$ of $\left(a{2}\right)^{f}$ in $g$ in such a way that $X$ belongs to $h_{2}$. Take also $Y$ in $a_{2} \backslash a_{1}$

On the other hand, the induction hypothesis says that there exists a coexponential basis $\left{X_{1}^{\prime}, \ldots, X_{k}^{\prime}\right}$ to $\mathfrak{h}{1}$ in $\left(a{2}\right)^{f}$ which contains a coexponential basis $\left{X_{i_{1}}^{\prime}, \ldots, X_{i_{m}}^{\prime}\right}$ of $\mathfrak{h}{1} \cap \mathfrak{h}{2}^{\prime}$ in $\mathfrak{h}{2}^{\prime}$ and even in $\mathfrak{h}{2} \cap\left(a_{2}\right)^{f}$. Then $\left{X, X_{1}^{\prime}, \ldots, X_{k}^{\prime}\right}$ is a coexponential basis of $h_{1}$ in $\mathfrak{g}$, whose part $\left{X_{i_{1}}^{\prime}, \ldots, X_{i_{m}}^{\prime}\right}$ is coexponential for $\mathfrak{h}{1} \cap \mathfrak{h}{2}$ in $\boldsymbol{h}_{2}$.

## 数学代写|表示论代写Representation theory代考|A Trace Relation

Tr 一个dH1/(H1∩H2)X+Tr⁡一个dH2/(H1∩H2)X=0对于任何X∈H1∩H2， 然后

ΔH1,G(H)=ΔH2,G(H)ΔH1∩H2,H2(H)2

(我b2bφ)(G)=∮H2/(H1∩H2)φ(GH)χF(H)ΔH2,G−1/2(H)d在H2,H1∩H2(H)

## 数学代写|表示论代写Representation theory代考|Relations Between Two Polarizations

[吨,磷]=12磷,[吨,问]=12问,[吨,从]=[磷,问]=从.

H1=R吨+R磷+R从,H2=R吨+R问+R从

O_{+}=\left{l \in \mathfrak{g}^{} ; l(Z)>0\right}, O_{-}=\left{l \in \mathfrak{g}^{} ; l(Z)<0\右}O_{+}=\left{l \in \mathfrak{g}^{} ; l(Z)>0\right}, O_{-}=\left{l \in \mathfrak{g}^{} ; l(Z)<0\右}

G_{2}=\exp g_{2}=\left{\left(\begin{array}{ll} a & b \ 0 & 1 \end{array}\right) \in M_{2}(\mathbb {R}); a>0\right}(a x+b \text { group })G_{2}=\exp g_{2}=\left{\left(\begin{array}{ll} a & b \ 0 & 1 \end{array}\right) \in M_{2}(\mathbb {R}); a>0\right}(a x+b \text { group })

K{2} K_{1}=\left{g=\left(\begin{array}{ll} a & b \ 0 & 1 \end{array}\right) \in G_{2} ; b>-1\右}K{2} K_{1}=\left{g=\left(\begin{array}{ll} a & b \ 0 & 1 \end{array}\right) \in G_{2} ; b>-1\右}

H1⊂∑j=1nbj(Fj′)

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