Langlands duality and Poisson–Lie duality via cluster theory and tropicalization

Abstract

Let G be a connected semisimple Lie group. There are two natural duality constructions that assign to G: its Langlands dual group $G^\vee$, and its Poisson–Lie dual group $G^*$, respectively. The main result of this paper is the following relation between these two objects: the integral cone defined by the cluster structure and the Berenstein–Kazhdan potential on the double Bruhat cell $G^{\vee ; w_0, e} \subset G^\vee$ is isomorphic to the integral Bohr–Sommerfeld cone defined by the Poisson structure on the partial tropicalization of $K^* \subset G^*$ (the Poisson–Lie dual of the compact form $K \subset G$). By Berenstein and Kazhdan (in: Contemporary mathematics, vol. 433. American Mathematical Society, Providence, pp 13–88, 2007), the first cone parametrizes the canonical bases of irreducible G-modules. The corresponding points in the second cone belong to integral symplectic leaves of the partial tropicalization labeled by the highest weight of the representation. As a by-product of our construction, we show that symplectic volumes of generic symplectic leaves in the partial tropicalization of $K^*$ are equal to symplectic volumes of the corresponding coadjoint orbits in ${{\,\mathrm{Lie}\,}}(K)^*$. To achieve these goals, we make use of (Langlands dual) double cluster varieties defined by Fock and Goncharov (Ann Sci Ec Norm Supér (4) 42(6):865–930, 2009). These are pairs of cluster varieties whose seed matrices are transpose to each other. There is a naturally defined isomorphism between their tropicalizations. The isomorphism between the cones described above is a particular instance of such an isomorphism associated to the double Bruhat cells $G^{w_0, e} \subset G$ and $G^{\vee ; w_0, e} \subset G^\vee$.

Appendices

Appendices

Example: duality between $B_2$ and $C_2$

Note $\mathrm{SO}_{2n+1}^\vee =\mathrm{Sp}_{2n}$. Let us focus on the case $n=2$. Here we use an alternative description of $\mathrm{SO}_5$. Denote

The group $\mathrm{SO}_5$ is isomorphic to

$$\begin{aligned} G=\{X\in \mathrm{GL}_5\mid XJ_5X^T=J_5\}, \end{aligned}$$

with Lie algebra:

$$\begin{aligned} {\mathfrak {g}}=\{x\in \mathfrak {gl}(5)\mid x+J_5x^TJ_5=0\}. \end{aligned}$$

Cartan subalgebra:

$$\begin{aligned} {\mathfrak {h}}=\{{{\,\mathrm{diag}\,}}(x_1,x_2,0,-x_2,-x_1)\}. \end{aligned}$$

Borel subalgebra:

$$\begin{aligned} {\mathfrak {b}}={\mathfrak {g}}\cap \{\text {upper-triangular matrices}\}. \end{aligned}$$

Cartan matrix and a symmetrizer:

$$\begin{aligned} A= \begin{bmatrix} 2 &{}\quad -1 \\ -2 &{}\quad 2 \end{bmatrix}= \begin{bmatrix} 1 &{} \\ &{}\quad 2 \end{bmatrix} \begin{bmatrix} 2 &{}\quad -1 \\ -1 &{}\quad 1 \end{bmatrix}, \quad D= \begin{bmatrix} 1 &{} \\ &{}\quad 2 \end{bmatrix}, \end{aligned}$$

Orthonormal basis in ${\mathfrak {h}}^*$:

$$\begin{aligned} \zeta _i: {{\,\mathrm{diag}\,}}(x_1,x_2,0,-x_2,-x_1)\mapsto x_i. \end{aligned}$$

Simple roots:

$$\begin{aligned} \alpha _1=\zeta _1-\zeta _2,\quad \alpha _2=\zeta _2. \end{aligned}$$

Positive roots:

$$\begin{aligned} \alpha _1,\quad \alpha _2,\quad \alpha _3:=\alpha _1+\alpha _2,\quad \alpha _4:=\alpha _1+2\alpha _2. \end{aligned}$$

Simple coroots:

$$\begin{aligned} \alpha _1^\vee ={{\,\mathrm{diag}\,}}(1,-1,0,1,-1); \quad \alpha _2^\vee ={{\,\mathrm{diag}\,}}(0,2,0,-2,0). \end{aligned}$$

Simple root vectors:

$$\begin{aligned} F_1=E_{21}-E_{54}; \quad F_2=E_{32}-E_{43}. \end{aligned}$$

Fundamental weights:

$$\begin{aligned} \omega _1=\alpha _3, \quad \omega _2=\frac{1}{2}\alpha _4. \end{aligned}$$

Fundamental coweights:

$$\begin{aligned} \omega _1^\vee =\alpha _1^\vee +\frac{1}{2}\alpha _2^\vee , \quad \omega _2^\vee =\alpha _1^\vee +\alpha _2^\vee . \end{aligned}$$

Character lattice of the maximal torus:

$$\begin{aligned} X^*(H)={\mathbb {Z}}\{\alpha _1,\alpha _2\}. \end{aligned}$$

Cocharacter lattice of the maximal torus:

$$\begin{aligned} X_*(H)={\mathbb {Z}}\{\omega _1^\vee ,\omega _2^\vee \}. \end{aligned}$$

Weyl group:

$$\begin{aligned} W=S_2 < imes {\mathbb {Z}}_2, \, \text { with generator }\, s_1, s_2 \, \text {satisfying}\, (s_1s_2)^4=1. \end{aligned}$$

The longest element:

$$\begin{aligned} w_0=(s_1s_2)^2=(s_2s_1)^2. \end{aligned}$$

Now let us compute the BK potential and the BK cone. Note that the lift of $s_i$ to G is given by:

$$\begin{aligned} \overline{s_1}=P_1P_4;\quad \overline{s_2}=P_2P_3P_2; \quad \text {where}\, P_i = E_{i,i+1}-E_{i+1,i}. \end{aligned}$$

And note $(\overline{s_1}\overline{s_2})^2=P_1P_2P_3P_4P_1P_2P_3P_1P_2P_1$. Let

$$\begin{aligned} \varvec{x}=\exp \left( \ln (x_1)\omega _1^\vee +\ln (x_2)\omega _2^\vee \right) \end{aligned}$$

and

$$\begin{aligned} x_{-1}(t)= \begin{bmatrix} t^{-1} &{} &{} &{} &{} \\ 1 &{}\quad t &{} &{} &{}\\ &{} &{}\quad 1 &{}&{} \\ &{}&{}&{}\quad t^{-1} &{}\\ &{}&{}&{}\quad -1&{}\quad t \end{bmatrix};\quad x_{-2}(t)= \begin{bmatrix} 1 &{} &{} &{} &{} \\ &{}\quad t^{-2} &{} &{} &{}\\ &{}\quad t^{-1} &{}\quad 1 &{} &{} \\ &{}\quad -\frac{1}{2}&{}\quad -t &{}\quad t^2 &{}\\ &{}&{}&{}&{} 1 \end{bmatrix}. \end{aligned}$$

Then for the longest word $(s_1s_2)^2$, generic elements of the double Bruhat cell $G^{w_0,e}$ can be written:

$$\begin{aligned} \varvec{x}x_{-1}(t_1)x_{-2}(t_2)x_{-1}(t_3)x_{-2}(t_4)\in G^{w_0,e}. \end{aligned}$$

This element is equal to

$$\begin{aligned} \varvec{x} \begin{bmatrix} \dfrac{1}{t_1 t_3} &{} &{} &{} &{} \\ \dfrac{t_1}{t_2^2}+\dfrac{1}{t_3} &{}\quad \dfrac{t_1 t_3}{t_2^2 t_4^2} &{} &{} &{} \\ \dfrac{1}{t_2} &{}\quad \dfrac{t_3}{t_2t_4^2}+\dfrac{1}{t_4} &{}\quad 1 &{} &{} \\ -\dfrac{1}{2 t_1} &{}\quad -\dfrac{\left( t_3+t_2 t_4\right) {}^2}{2 t_1 t_3 t_4^2} &{}\quad -\dfrac{t_2 \left( t_3+t_2 t_4\right) }{t_1 t_3} &{}\quad \dfrac{t_2^2 t_4^2}{t_1 t_3} &{} \\ \dfrac{1}{2} &{}\quad \dfrac{1}{2} \left( \dfrac{\left( t_3+t_2 t_4\right) {}^2}{t_3 t_4^2}+t_1\right) &{}\quad \dfrac{t_4 t_2^2}{t_3}+t_2+t_1 t_4 &{}\quad -\dfrac{\left( t_2^2+t_1 t_3\right) t_4^2}{t_3} &{}\quad t_1 t_3 \\ \end{bmatrix}. \end{aligned}$$

Thus the potential is

$$\begin{aligned} \left( t_1+\frac{\left( t_3+t_2 t_4\right) ^2}{t_3t_4^2}\right) +t_4+ x_1\cdot \frac{1}{t_1}+x_2\left( \frac{1}{t_4}+\frac{t_2^2+t_1t_3}{t_2t_3}\right) , \end{aligned}$$

which gives us the cone cut out by the inequalities:

$$\begin{aligned} \begin{aligned} x_1\geqslant t_1&\geqslant 0;\\ x_2 \geqslant t_4&\geqslant 0;\\ 2t_2\geqslant t_3 \geqslant 2t_4&\geqslant 0;\\ x_2&\geqslant t_2-t_1;\\ x_2&\geqslant t_3-t_2. \end{aligned} \end{aligned}$$

(40)

Now let us describe $\mathrm{Sp}_4$ as the dual of $\mathrm{SO}_5$. Denote

$$\begin{aligned} J'_{2n}= \begin{bmatrix} &{} J_n \\ -J_n &{} \end{bmatrix}. \end{aligned}$$

The group $\mathrm{Sp}_4$ is isomorphic to

$$\begin{aligned} G^\vee =\{X\in \mathrm{GL}_4\mid XJ'_4X^T=J'_4\} \end{aligned}$$

with Lie algebra:

$$\begin{aligned} {\mathfrak {g}}^\vee =\{x\in \mathfrak {gl}(4)\mid x-J'x^TJ'=0\}. \end{aligned}$$

Cartan subalgebra:

$$\begin{aligned} {\mathfrak {h}}=\{{{\,\mathrm{diag}\,}}(x_1,x_2,-x_2,-x_1)\}. \end{aligned}$$

The Borel subalgebra:

$$\begin{aligned} {\mathfrak {b}}^\vee ={\mathfrak {g}}^\vee \cap \{\text {upper-triangular matrices}\}. \end{aligned}$$

Orthonormal basis in $({\mathfrak {h}}^\vee )^*$:

$$\begin{aligned} \zeta _i^\vee : {{\,\mathrm{diag}\,}}(x_1,x_2,-x_2,-x_1)\mapsto x_i. \end{aligned}$$

Simple roots:

$$\begin{aligned} \beta _1=\zeta _1^\vee -\zeta _2^\vee ,\quad \beta _2=2\zeta _2^\vee . \end{aligned}$$

Positive roots:

$$\begin{aligned} \beta _1,\quad \beta _2,\quad \beta _3:=2\beta _1+\beta _2,\quad \beta _4:=\beta _1+\beta _2. \end{aligned}$$

Simple coroots of ${\mathfrak {g}}^\vee$ are given by:

$$\begin{aligned} \beta _1^\vee ={{\,\mathrm{diag}\,}}(1,-1,1,-1); \quad \beta _2^\vee ={{\,\mathrm{diag}\,}}(0,1,-1,0); \end{aligned}$$

Simple root vectors:

$$\begin{aligned} F_1=E_{21}-E_{43}; \quad F_2=E_{32}. \end{aligned}$$

Fundamental weights:

$$\begin{aligned} \kappa _1=\frac{1}{2}\beta _3, \quad \kappa _2=\beta _4. \end{aligned}$$

Fundamental coweights:

$$\begin{aligned} \kappa _1^\vee =\beta _1^\vee +\beta _2^\vee , \quad \kappa _2^\vee =\frac{1}{2}\beta _1^\vee +\beta _2^\vee . \end{aligned}$$

Character lattice of the maximal torus:

$$\begin{aligned} X^*(H^\vee )={\mathbb {Z}}\{\kappa _1,\kappa _2\}. \end{aligned}$$

Cocharacter lattice of the maximal torus:

$$\begin{aligned} X_*(H)={\mathbb {Z}}\{\beta _1^\vee ,\beta _2^\vee \}. \end{aligned}$$

To calculate the potential for $G^\vee$, we need the lift of $s_i$ to $G^\vee$:

$$\begin{aligned} \overline{s_1}=P_1P_3;\quad \overline{s_2}=P_2; \quad \text { where }\, P_i = E_{i,i+1}-E_{i+1,i}. \end{aligned}$$

Note $(\overline{s_1}\overline{s_2})^2=P_1P_2P_3P_1P_2P_1$. Let

$$\begin{aligned} \varvec{y}^\vee =\exp \left( \ln (y_1)\beta _1^\vee +\ln (y_2)\beta _2^\vee \right) \end{aligned}$$

and

$$\begin{aligned} x_{-1}^\vee (t)= \begin{bmatrix} t^{-1} &{} &{} &{} \\ 1 &{}\quad t &{} &{} \\ &{}&{}\quad t^{-1} &{}\\ &{}&{}\quad -1&{}\quad t \end{bmatrix};\quad x_{-2}^\vee (t)= \begin{bmatrix} 1 &{} &{} &{} \\ &{}\quad t^{-1} &{} &{} \\ &{}\quad 1 &{}\quad t &{}\\ &{}&{}&{}\quad 1 \end{bmatrix}. \end{aligned}$$

Then for the longest word $(s_1s_2)^2$, generic elements of the double Bruhat cell $G^{\vee ;e,w_0}$ can be written:

$$\begin{aligned} \varvec{y}^\vee x_{-1}^\vee (t_1)x_{-2}^\vee (t_2)x_{-1}^\vee (t_3)x_{-2}^\vee (t_4)\in G^{\vee ;e,w_0}. \end{aligned}$$

This element is equal to

$$\begin{aligned} \varvec{y}^\vee \begin{bmatrix} \dfrac{1}{t_1 t_3} &{} &{} &{} \\ \dfrac{t_1}{t_2}+\dfrac{1}{t_3} &{}\quad \dfrac{t_1 t_3}{t_2 t_4} &{} &{} \\ \dfrac{1}{t_1} &{}\quad \dfrac{t_2}{t_1 t_3}+\dfrac{t_3}{t_1 t_4} &{}\quad \dfrac{t_2 t_4}{t_1 t_3} &{} \\ -1 &{} -t_1-\dfrac{t_2}{t_3}-\dfrac{t_3}{t_4} &{}\quad \left( -t_1-\dfrac{t_2}{t_3}\right) t_4 &{}\quad t_1 t_3 \\ \end{bmatrix}. \end{aligned}$$

Thus the potential is

$$\begin{aligned} \left( t_1+\frac{t_2}{t_3}+\frac{t_3}{t_4}\right) +t_4+ \frac{y_1^2}{y_2}\cdot \frac{1}{t_1}+ \frac{y_2^2}{y_1^2}\left( \frac{(t_1t_3+t_2)^2}{t_2t_3^2}+\frac{1}{t_4}\right) , \end{aligned}$$

which gives us the cone cut out by the following inequalities:

$$\begin{aligned} \begin{aligned} 2y_1-y_2\geqslant t_1&\geqslant 0;\\ 2y_2-2y_1 \geqslant t_4&\geqslant 0;\\ t_2\geqslant t_3 \geqslant t_4&\geqslant 0;\\ 2y_2-2y_1&\geqslant t_2-2t_1;\\ 2y_2-2y_1&\geqslant 2t_3-t_2. \end{aligned} \end{aligned}$$

(41)

Recall that $\psi : X_*(H)\rightarrow X^*(H)$ is given by:

$$\begin{aligned} x_1\omega _1^\vee +x_2\omega _2^\vee \mapsto (x_1+x_2)\alpha _1+(x_1+2x_2)\alpha _2. \end{aligned}$$

Then the map $\psi _{{\mathbf {i}}}: {\mathcal {L}} \rightarrow {\mathcal {L}}^\vee$ is given by:

$$\begin{aligned} (x_1,x_2;t_1,t_2,t_3,t_4)\mapsto (x_1+x_2,x_1+2x_2;t_1,2t_2,t_3,2t_4). \end{aligned}$$

Thus it easy to see, after replacing $(y_1,y_2;t_1,t_2,t_3,t_4)$ by $(x_1+x_2,x_1+2x_2;t_1,2t_2,t_3,2t_4)$, that the real cone defined by (41) is the real cone defined by (40).

Bohr–Sommerfeld lattices and tropical Poisson varieties

In this section we extend the notion of a Bohr–Sommerfeld lattice to the category of tropical Poisson varieties, which we called $\mathbf {PTrop}$ in [1]. We will actually consider the category $\mathbf {DecPTrop}$ of decorated tropical Poisson varieties. An object of $\mathbf {DecPTrop}$ is a tuple $({\mathcal {C}}\times T, X^t, \pi , {{\,\mathrm{hw}\,}}, P)$, where

• T is a connected subgroup of $(S^1)^n$;

• $X^t = {{\,\mathrm{Hom}\,}}(S^1,(S^1)^n)\cong {{\,\mathrm{Hom}\,}}({\mathbb {C}}^\times ,({\mathbb {C}}^\times )^n)=(({\mathbb {C}}^\times )^n)^t$;

• ${\mathcal {C}}\subset X^t\otimes {\mathbb {R}}$ is an open rational polyhedral cone in $X^t\otimes {\mathbb {R}}$;

• $\pi$ is a constant Poisson bivector on $(X^t\otimes {\mathbb {R}})\times T$, and the projections to $X^t \otimes {\mathbb {R}}$ and T (both equipped with the zero Poisson structure) are Poisson maps;

• $P\cong {\mathbb {Z}}^{n-\dim (T)}$ is a free abelian group;

• ${{\,\mathrm{hw}\,}}:X^t\rightarrow P$ is a ${\mathbb {Z}}$-linear map, so that fibers of the induced map ${{\,\mathrm{hw}\,}}_{{\mathbb {R}}}\circ {{\,\mathrm{pr}\,}}_1:{\mathcal {C}}\times T\rightarrow P\otimes {\mathbb {R}}$ are the symplectic leaves of $({\mathcal {C}}\times T,\pi )$.

An example of a decorated tropical Poisson variety is

$$\begin{aligned} \left( PT(K^*),(G^{w_0,e})^t, \pi _{PT}^{\varsigma ({\mathbf {i}})}, {{\,\mathrm{hw}\,}}^t, X_*(H)\right) . \end{aligned}$$

where tropicalization is taken with respect to the chart $\varsigma ({\mathbf {i}})$ given by (30). Here the formula for $\pi _{PT}^{\varsigma ({\mathbf {i}})}$ is given by Theorem 6.1, and we extend $\pi _{PT}^{\varsigma ({\mathbf {i}})}$ to be a constant bracket on all of $((G^{w_0,e}, \varsigma ({\mathbf {i}}))^t \otimes {\mathbb {R}}) \times (S^1)^m \cong {\mathbb {R}}^{m+r} \times (S^1)^m$.

An arrow in $\mathbf {DecPTrop}$ is defined to be a pair

$$\begin{aligned} (f,g):({\mathcal {C}}\times T,X^t, \pi , {{\,\mathrm{hw}\,}}, P) \rightarrow ({\mathcal {C}}'\times T',{X'}^t, \pi ', {{\,\mathrm{hw}\,}}', P') \end{aligned}$$

where $f:X^t\rightarrow {X'}^t$ is a piecewise ${\mathbb {Z}}$-linear map which is homogeneous in the sense that $f(nx)=nf(x)$ for $n\in {\mathbb {Z}}_{\geqslant 0}$, and $g:P\rightarrow P'$ is a ${\mathbb {Z}}$-linear map so that ${{\,\mathrm{hw}\,}}'\circ f=g\circ {{\,\mathrm{hw}\,}}$. We require that f induces a map of cones $f:{\mathcal {C}}\rightarrow {\mathcal {C}}'$, and, on each open linearity chamber $C\subset {\mathcal {C}}$ of f, the naturally induced map $C\times (S^1)^n\rightarrow f(C)\times (S^1)^{n'}$ restricts to a Poisson map $f:C\times T\rightarrow f(C)\times T'$.

There is an obvious forgetful functor from $\mathbf {DecPTrop}$ to $\mathbf {PTrop}$.

Definition B.1

For a point $\lambda \in P$, consider the fiber ${{\,\mathrm{hw}\,}}_{{\mathbb {R}}}^{-1}(\lambda ) \subset X^t\otimes {\mathbb {R}}$. For each $x\in {{\,\mathrm{hw}\,}}_{{\mathbb {R}}}^{-1}(\lambda )\cap X^t$, there is a lattice coming from $\pi$ in the tangent space $T_x ({{\,\mathrm{hw}\,}}_{{\mathbb {R}}}^{-1}(\lambda ))$, as in (34). Using the canonical identification $T_x ({{\,\mathrm{hw}\,}}_{{\mathbb {R}}}^{-1}(\lambda )) \cong {{\,\mathrm{hw}\,}}_{{\mathbb {R}}}^{-1}(\lambda )$, we can realize this lattice as a subset $\Lambda _x$ of ${{\,\mathrm{hw}\,}}_{{\mathbb {R}}}^{-1}(\lambda )$. A decorated tropical Poisson variety is quantizable if, for any $\lambda \in P$ and any $x,y\in {{\,\mathrm{hw}\,}}_{{\mathbb {R}}}^{-1}(\lambda )\cap X^t$, one has $\Lambda _x=\Lambda _y$. If ${\mathcal {C}}\times T$ is quantizable, define the Bohr–Sommerfeld lattice as

$$\begin{aligned} \Lambda := \bigcup _{x\in {{\,\mathrm{hw}\,}}^{-1}(P)} \Lambda _x. \end{aligned}$$

Our example $PT(K^*)$ is quantizable, and this definition of $\Lambda$ agrees with the one in (36), after forming the intersection with the cone ${\mathcal {C}}^G_{\varsigma ({\mathbf {i}})}({\mathbb {R}})$.

Because $\pi$ is assumed to be constant, to check that ${\mathcal {C}}\times T$ is quantizable it is enough to check that

$$\begin{aligned} X^t \cap {{\,\mathrm{hw}\,}}^{-1}_{\mathbb {R}}({{\,\mathrm{hw}\,}}(z))\subset \Lambda _z, \end{aligned}$$

(42)

for any one $z \in X^t$. Indeed, let $\lambda \in P$ and let $x,y\in {{\,\mathrm{hw}\,}}_{{\mathbb {R}}}^{-1}(\lambda )\cap X^t$. Then, because $\pi$ is constant, $\Lambda _x = \Lambda _z + (x-z)$ and so $X^t \cap {{\,\mathrm{hw}\,}}^{-1}_{\mathbb {R}}(\lambda )\subset \Lambda _x$. Again because $\pi$ is constant, $\Lambda _y = \Lambda _x + (x-y)$. So since $x-y\in X^t$, and since we have shown $X^t \cap {{\,\mathrm{hw}\,}}^{-1}_{\mathbb {R}}(\lambda )\subset \Lambda _x$, it follows that $\Lambda _y= \Lambda _x$.

Lemma B.2

Let (f, g) be an isomorphism of decorated tropical Poisson varieties:

$$\begin{aligned} (f,g):({\mathcal {C}}\times T,X^t, \pi , {{\,\mathrm{hw}\,}}, P) \rightarrow ({\mathcal {C}}'\times T',{X'}^t, \pi ', {{\,\mathrm{hw}\,}}', P'). \end{aligned}$$

Assume ${\mathcal {C}}\times T$ is quantizable. Then ${\mathcal {C}}'\times T'$ is quantizable. If $\Lambda '$ denotes the Bohr–Sommerfeld lattice of ${\mathcal {C}}'\times T'$, then $f(\Lambda \cap \overline{{\mathcal {C}}})=\Lambda ' \cap \overline{{\mathcal {C}}'}$.

Proof

Without loss of generality assume that $P=P'$ and $g={{\,\mathrm{Id}\,}}$. First, we show that ${\mathcal {C}}'\times T'$ is quantizable. Let $\lambda \in {{\,\mathrm{hw}\,}}({\mathcal {C}}\cap X^t)$ and let $x\in {{\,\mathrm{hw}\,}}_{{\mathbb {R}}}^{-1}(\lambda ) \cap X^t$ be a point which is inside an open linearity chamber C of f. Note that C is a cone because f is assumed to be homogeneous. We may assume we have chosen $\lambda$ and x such that the lattice generators of $\Lambda _x$ are contained in C.

Let $C_\lambda = C\cap {{\,\mathrm{hw}\,}}_{{\mathbb {R}}}^{-1}(\lambda )$, and let $L=C_\lambda \times T$. Then f induces a symplectomorphism of L onto its image, and we have $f(\Lambda _x\cap C_\lambda )=\Lambda _{f(x)}'\cap f(C_\lambda )$. Since f is an isomorphism, one has $f(X^t\cap C_\lambda )={X'}^t\cap f(C_\lambda )$. And since ${\mathcal {C}}\times T$ is quantizable, one has $X^t \cap C_\lambda \subset \Lambda _x \cap C_\lambda$. Therefore ${X'}^t \cap f(C_\lambda )\subset \Lambda _{f(x)}'\cap f(C_\lambda )$. Since we chose x so that the lattice generators of $\Lambda _x$ are contained in C, we can extend to all of ${{{\,\mathrm{hw}\,}}_{\mathbb {R}}'}^{-1}(\lambda )$. We then have

$$\begin{aligned} {X'}^t\cap {{{\,\mathrm{hw}\,}}'_{{\mathbb {R}}}}^{-1}(\lambda ) \subset \Lambda _{f(x)}'. \end{aligned}$$

By the criterion (42), since $\pi '$ is constant, this tells us that ${\mathcal {C}}'\times T$ is quantizable.

Now, let $\Lambda '$ be the Bohr–Sommerfeld lattice of ${\mathcal {C}}'\times T'$. We will show that $f(\Lambda )=\Lambda '$. It is enough to check, for each open linearity cone C of f, that $f(\Lambda \cap C)=\Lambda '\cap f(C)$; one can then extend to the boundary of C and f(C) by linearity. For this it is enough to check that $f(\Lambda _x\cap C)=\Lambda '_{f(x)}\cap f(C)$ for all $x\in C\cap X^t$. But this follows because f induces a Poisson isomorphism from $C\times T$ to its image. $\square$