Symplectic Geometry and Spectral Properties of Classical and Quantum Coupled Angular Momenta

Abstract

We give a detailed study of the symplectic geometry of a family of integrable systems obtained by coupling two angular momenta in a non-trivial way. These systems depend on a parameter $t \in [0,1]$ and exhibit different behaviors according to its value. For a certain range of values, the system is semitoric, and we compute some of its symplectic invariants. Even though these invariants have been known for almost a decade, this is to our knowledge the first example of their computation in the case of a non-toric semitoric system on a compact manifold. (The only invariant of toric systems is the image of the momentum map.) In the second part of the paper, we quantize this system, compute its joint spectrum and describe how to use this joint spectrum to recover information about the symplectic invariants.

Appendices

Appendix A: Proofs of Technical Results

Proof of Lemma 3.3

A standard computation shows that the symplectic form becomes

$$\begin{aligned} (\Psi ^{-1})^* \omega = \frac{2 i R_1 \mathrm{d}z \wedge d{\bar{z}}}{(1+|z|^2)^2} + \frac{2 i R_2 \mathrm{d}w \wedge d{\bar{w}}}{(1+|w|^2)^2}. \end{aligned}$$

Moreover, we deduce from Eq. (7) that

$$\begin{aligned} \frac{\partial J}{\partial z} = \frac{-2R_1 {\bar{z}}}{(1+|z|^2)^2}, \qquad \frac{\partial J}{\partial w} = \frac{2R_2 {\bar{w}}}{(1+|w|^2)^2}. \end{aligned}$$

Since J is real-valued, $\frac{\partial J}{\partial {\bar{z}}} = \overline{\frac{\partial J}{\partial z}}$ and $\frac{\partial J}{\partial {\bar{w}}} = \overline{\frac{\partial J}{\partial w}}$, and we obtain the desired result. Furthermore,

$$\begin{aligned} \frac{\partial H}{\partial z}= & {} \frac{1}{2} \left( \frac{-2{\bar{z}}}{(1+|z|^2)^2} \right. \\&\left. + \frac{(2w-{\bar{z}}|w|^2+{\bar{z}})(1+|z|^2) - {\bar{z}}(2(zw + {\bar{z}}{\bar{w}}) + (1-|z|^2)(|w|^2-1))}{(1+|z|^2)^2(1+|w|^2)} \right) , \end{aligned}$$

which yields, after simplification

$$\begin{aligned} \frac{\partial H}{\partial z} = \frac{w-2{\bar{z}}|w|^2-{\bar{z}}^2{\bar{w}}}{(1+|z|^2)^2(1+|w|^2)}. \end{aligned}$$

A similar computation shows that

$$\begin{aligned} \frac{\partial H}{\partial w} = \frac{z + {\bar{w}}-{\bar{w}}|z|^2-{\bar{z}}{\bar{w}}^2}{(1+|z|^2)(1+|w|^2)^2}, \end{aligned}$$

and we conclude by using the same argument as above. $\square$

Proof of Lemma 3.4

Let $\lambda _1(z,w), \lambda _2(z,w)$ be as in the previous lemma, and let $(z,w) \in \Lambda _0 \setminus \{(0,0)\}$. Since $z \ne 0$, we get, by multiplying both the numerator and the denominator in $\lambda _1$ by ${\bar{z}}$, that

$$\begin{aligned} \lambda _1(z,w) = \frac{{\bar{z}} {\bar{w}} - 2 |z|^2 |w|^2 - zw |z|^2}{R_1 {\bar{z}}(1+|w|^2)}. \end{aligned}$$

But the second equation in (8) yields ${\bar{z}} {\bar{w}} = - zw - |w|^2 + |z|^2 |w|^2$, hence

$$\begin{aligned} \lambda _1(z,w) = \frac{-zw - |w|^2 - |z|^2 |w|^2 - zw |z|^2}{R_1 {\bar{z}}(1+|w|^2)} = -\frac{w(z+{\bar{w}})(1+|z|^2)}{R_1 {\bar{z}}(1+|w|^2)}. \end{aligned}$$

The first equation in (8) allows us to further simplify this expression and to obtain

$$\begin{aligned} \lambda _1(z,w) = -\frac{zw(z+{\bar{w}})(1+|z|^2)}{R_2 |w|^2 (1+|z|^2)} = -\frac{z(z+{\bar{w}})}{R_2 {\bar{w}}}. \end{aligned}$$

Now, since $w \ne 0$, we have that

$$\begin{aligned} \lambda _2(z,w)= & {} \frac{{\bar{z}} |w|^2 + w |w|^2 - w |z|^2 |w|^2- z w^2 |w|^2}{R_2 |w|^2 (1+|z|^2)}\\= & {} \frac{{\bar{z}} |w|^2 - z w^2 |w|^2 + w |w|^2 (1 -|z|^2)}{R_2 |w|^2 (1+|z|^2)}. \end{aligned}$$

The second equation in (8) gives

$$\begin{aligned}&{\bar{z}} |w|^2 - z w^2 |w|^2 + w |w|^2 (1 -|z|^2) = {\bar{z}} |w|^2\\&\quad - z w^2 |w|^2 - z w^2 - {\bar{z}} |w|^2 = -zw^2(1+|w|^2), \end{aligned}$$

and thus, using the first one again, we finally obtain that $\lambda _2(z,w) = -\frac{w^2}{R_1 {\bar{z}}}$. $\square$

Proof of Proposition 3.7

We start by diagonalizing A over ${\mathbb {C}}$. With our choice of parameters, it follows from Eq. (3) that

$$\begin{aligned} A = \begin{pmatrix} 0 &{} 0 &{} 0 &{} \frac{1}{2} \\ 0 &{} 0 &{} -\frac{1}{2} &{} 0 \\ 0 &{} -\frac{1}{5} &{} 0 &{} -\frac{1}{5} \\ \frac{1}{5} &{} 0 &{} \frac{1}{5} &{} 0 \end{pmatrix}. \end{aligned}$$

A has eigenvalues $\lambda _1 = \frac{3+i}{10}, {\bar{\lambda }}_1, \lambda _2 = - \lambda _1, {\bar{\lambda }}_2$ with respective eigenvectors $X_1, \overline{X_1}, X_2, \overline{X_2}$ where

$$\begin{aligned} X_1 = \frac{1}{2} \begin{pmatrix} 3 - i \\ -1 - 3i \\ 2i \\ 2 \end{pmatrix} , \qquad X_2 = \frac{1}{2} \begin{pmatrix} -3 + i \\ - 1 - 3i \\ -2i \\ 2 \end{pmatrix}. \end{aligned}$$

Let $E_{\lambda }$ be the eigenspace associated with $\lambda$, and let $F = E_{\lambda _1} \oplus E_{{\bar{\lambda }}_1}$ and $G = E_{\lambda _2} \oplus E_{{\bar{\lambda }}_2}$; then $T_{m_0}M = F \oplus G$ and a real basis of F is given by

$$\begin{aligned} Y_1 = X_1 + \overline{X_1} = \begin{pmatrix} 3 \\ -1 \\ 0 \\ 2 \end{pmatrix}, \qquad Y_2 = -i(X_1 - \overline{X_1}) = \begin{pmatrix} -1 \\ -3 \\ 2 \\ 0 \end{pmatrix}. \end{aligned}$$

There exists a unique basis $(Z_1,Z_2)$ of G such that $(Y_1,Y_2,Z_1,Z_2)$ is a symplectic basis of $T_{m_0}M$ [see for instance (Meyer et al. 2009, Lemma 3.2.3)]; in the latter, A will take the form displayed in Eq. (12). Since

$$\begin{aligned} G = \mathrm {Span}\left( \begin{pmatrix} -3 \\ -1 \\ 0 \\ 2 \end{pmatrix}, \begin{pmatrix} 1 \\ -3 \\ -2 \\ 0 \end{pmatrix} \right) , \end{aligned}$$

there exists $a,b,c,d \in {\mathbb {R}}$ such that

$$\begin{aligned} Z_1 = \begin{pmatrix} -3a + b \\ -a - 3b \\ -2b \\ 2a \end{pmatrix} , \qquad Z_2 = \begin{pmatrix} -3c + d \\ -c - 3d \\ -2d \\ 2c \end{pmatrix}. \end{aligned}$$

Recall that $\omega _{m_0} = - \mathrm{d}x_1 \wedge \mathrm{d}y_1 + \frac{5}{2} \mathrm{d}x_2 \wedge \mathrm{d}y_2$. Hence, we need to solve the system

$$\begin{aligned} {\left\{ \begin{array}{ll} 1 = \omega _{m_0}(Y_1,Z_1) = 6a + 18b \\ 0 = \omega _{m_0}(Y_1,Z_2) = 6c + 18d \\ 0 = \omega _{m_0}(Y_2,Z_1) = 18a - 6b \\ 1 = \omega _{m_0}(Y_2,Z_2) = 18c - 6d \end{array}\right. }. \end{aligned}$$

We find $a = \frac{1}{60}, b = \frac{1}{20}, c = \frac{1}{20}$ and $d = -\frac{1}{60}$, hence

$$\begin{aligned} Z_1 = \frac{1}{30} \begin{pmatrix} 0 \\ -5 \\ -3 \\ 1 \end{pmatrix} , \qquad Z_2 = \frac{1}{30} \begin{pmatrix} -5 \\ 0 \\ 1 \\ 3 \end{pmatrix}. \end{aligned}$$

Consequently, the matrix P of change of basis from the basis ${\mathcal {B}}$ associated with $(x_1,y_1,x_2,y_2)$ to the basis $(Y_1,Y_2,Z_1,Z_2)$ satisfies

$$\begin{aligned} P = \begin{pmatrix} 3 &{} -1 &{} 0 &{} -\frac{1}{6} \\ -1 &{} -3 &{} - \frac{1}{6} &{} 0 \\ 0 &{} 2 &{} -\frac{1}{10} &{} \frac{1}{30} \\ 2 &{} 0 &{} \frac{1}{30} &{} \frac{1}{10} \end{pmatrix}, \quad P^{-1} = \begin{pmatrix} \frac{1}{6} &{} 0 &{} \frac{1}{12} &{} \frac{1}{4} \\ 0 &{} -\frac{1}{6} &{} \frac{1}{4} &{} -\frac{1}{12} \\ -1 &{} -3 &{} -5 &{} 0 \\ -3 &{} 1 &{} 0 &{} 5 \end{pmatrix}, \end{aligned}$$

which yields the desired expression for the coordinates $u_1,u_2,\xi _1,\xi _2$. Moreover, $P^{-1}AP$ is as in Eq. (12), with $\alpha = \frac{-3}{10}$ and $\beta =\frac{1}{10}$; this gives

$$\begin{aligned} B = \begin{pmatrix} 1 &{} 0 \\ \frac{1}{10} &{} -\frac{3}{10} \end{pmatrix}^{-1} = \begin{pmatrix} 1 &{} 0 \\ \frac{1}{3} &{} -\frac{10}{3} \end{pmatrix}. \end{aligned}$$

This is not satisfactory since we want the lower right coefficient in this matrix to be positive. In order to obtain a B satisfying this requirement, it suffices to perform the symplectic change of coordinates $(u_1,u_2,\xi _1,\xi _2) \mapsto (-\xi _1,-\xi _2, u_1, u_2)$. $\square$

Appendix B: Critical Points of Corank One

We explain how to prove the claim about the critical points of corank one of $F=(J,H_t)$ in the proof of Corollary 2.11, namely that they are non-degenerate for every $t \in ]0,1]$. (The case $t=0$ is clear since the system is toric up to vertical scaling.) It suffices to prove that for every E in the image of J, except the ones corresponding to critical points of corank two (which are the only ones for which $dJ = 0$ for this system), the critical points of the restriction of $H_t$ to the symplectic quotient $M^{\text {red}}_E = J^{-1}(E) / S^1$ with respect to the action generated by J are non-degenerate, see Toth and Zelditch (2003, Definition 3) for a general explanation or Hohloch and Palmer (2018, Corollary 2.5) for the particular case of dimension 4. Coming back to our particular case, let $E \in (-(R_1 + R_2), R_1 + R_2) \setminus \{ R_1 - R_2, R_2 - R_1 \}$; since the poles do not give rise to critical points on $J^{-1}(E)$, we may work with cylindrical coordinates as in Sect. 3.5 [or as in Hohloch and Palmer (2018, Section 3.3)] where a similar computation is performed)

$$\begin{aligned} (x_j, y_j, z_j) = \left( \sqrt{1-z_j^2} \cos \theta _j, \sqrt{1-z_j^2} \sin \theta _j, z_j \right) , \quad j = 1,2. \end{aligned}$$

In these coordinates, $H_t$ reads

$$\begin{aligned} H_t(\theta _1, z_1, \theta _2, z_2) = (1-t) z_1 + t \left( \sqrt{(1-z_1^2)(1-z_2^2)} \cos (\theta _1 - \theta _2) + z_1 z_2 \right) . \end{aligned}$$

Since $z_2$ can be deduced from $z_1$ on $J^{-1}(E)$, namely

$$\begin{aligned} z_2 = \frac{E-R_1 z_1}{R_2}, \qquad \max \left( -1, \frac{E - R_2}{R_1} \right)< z_1 < \min \left( 1, \frac{E + R_2}{R_1} \right) \end{aligned}$$

and since the action of J preserves the angle $\theta = \theta _1 - \theta _2$, we can use $(z_1, \theta )$ as coordinates on $M^{\text {red}}_E$:

$$\begin{aligned} H_t(z_1,\theta )= & {} (1-t) z_1 + \frac{t z_1(E-R_1 z_1)}{R_2} + \frac{t \cos \theta }{R_2} \sqrt{P_E(z_1)}, \\ P_E(z_1)= & {} (1-z_1^2)\left( R_2^2 - (E - R_1 z_1)^2 \right) . \end{aligned}$$

The first derivatives of $H_t$ read

$$\begin{aligned} \frac{\partial H_t}{\partial \theta }(z_1,\theta )= & {} \frac{-t \sqrt{P_E(z_1)}\sin \theta }{R_2}, \quad \frac{\partial H_t}{\partial z_1}(z_1,\theta )\\= & {} 1-t + \frac{tE}{R_2} - \frac{2 t R_1 z_1}{R_2} + \frac{t P_E'(z_1) \cos \theta }{2 R_2 \sqrt{P_E(z_1)}}. \end{aligned}$$

Hence if $(z_1^*,\theta ^*)$ is a critical point, then necessarily $\theta ^* \in \{0, \pi \}$, and $\frac{\partial ^2 H_t}{\partial \theta \partial z_1}(z_1^*,\theta ^*) = 0$. Let $\varepsilon = \cos \theta ^* \in \{-1,1\}$; then one readily checks that

$$\begin{aligned} \frac{\partial ^2 H_t}{\partial \theta ^2}(z_1^*,\theta ^*)= & {} \frac{-\varepsilon t \sqrt{P_E(z_1^*)}}{R_2}, \quad \frac{\partial ^2 H_t}{\partial z_1^2}(z_1^*,\theta ^*) \\= & {} \frac{t}{4 R_2} \left( - 8 R_1 + \varepsilon \left( \frac{2 P_E''(z_1^*) P_E(z_1^*) - P_E'(z_1^*)^2}{P_E(z_1^*)^{3/2}} \right) \right) \end{aligned}$$

We claim that this last quantity has the sign of $-\varepsilon$; this follows from the fact that

$$\begin{aligned} Q(E,z_1) = \frac{2 P_E''(z_1) P_E(z_1) - P_E'(z_1)^2}{P_E(z_1)^{3/2}} < - 8 R_1 \end{aligned}$$

for any $E, z_1$ satisfying the above bounds. In order to prove this, one may check that Q is minimal at $(E,z_1) = (0,0)$; since $Q(0,0) = -4 \left( R_2 + \frac{R_1^2}{R_2} \right)$ and since the function $x > 0 \mapsto x + \frac{R_1^2}{x}$ is minimal at $x = R_1$ with value $2 R_1$, we obtain the desired result because $R_2 > R_1$.

In fact, this analysis gives us the sign of the determinant of the Hessian of $H_t$ at a critical point, so we can deduce from it that the corank one critical points are of elliptic–transverse type. Hence, if one is only interested in proving this, and not in finding a parametrization of the boundary of the image of the momentum map, this appendix constitutes a faster way to obtain Corollary 2.11.