Properties of the primary spin variables
In this appendix, we first elaborate the properties of the primary spin variables that were summarized in Proposition 2.1. As was already mentioned, the pertinent Poisson structure on \({\mathbb {C}}^n \simeq {\mathbb {R}}^{2n}\) is a special case of the \({\mathrm{U}}(n)\) covariant Poisson structures due to Zakrzewski [56]. Nevertheless, to make our text self-contained, we shall also verify its Jacobi identity and covariance property. Then, we present the corresponding moment map and symplectic form, which have not been considered in previous work.
For any real function \(F\in C^\infty ({\mathbb {C}}^n)\), we define its \({\mathbb {C}}^n\)-valued ‘gradient’ \(\nabla F\) by the equalityFootnote 6
$$\begin{aligned} \mathfrak {I}\left( (\nabla F(w))^\dagger V\right) := \left. \frac{\hbox {d}}{\hbox {d}t}\right| _{t=0}F(w + t V), \qquad \forall w, V\in {\mathbb {C}}^n, \end{aligned}$$
(A.1)
where the elements of \({\mathbb {C}}^n\) are viewed as column vectors. We note that any real linear function on the real vector space \({\mathbb {C}}^n\) is of the form
$$\begin{aligned} F_\xi (w):= \mathfrak {I}(\xi ^\dagger w), \end{aligned}$$
(A.2)
for some \(\xi \in {\mathbb {C}}^n\), and for such function \(\nabla F_\xi = \xi \). Next we give a convenient presentation of Zakrzewski’s Poisson bracket.
Proposition A.1
For real functions \(F, H\in C^\infty ({\mathbb {C}}^n)\), let \(\xi (w):= \nabla F(w)\) and \(\eta (w):= \nabla H(w)\). Then, the following formula
$$\begin{aligned} \{F, H\}(w)= & {} \mathfrak {I}\left( \xi (w)^{\dagger }(w \eta (w)^{\dagger })_{{\mathfrak {u}}(n)} w - {\textstyle {\frac{1}{2}}}\xi (w)^{\dagger }w\eta (w)^{\dagger }w - {\textstyle {\frac{1}{2}}}\xi (w)^{\dagger }w w^{\dagger }\eta (w)\right. \nonumber \\&\left. - \xi (w)^{\dagger }\eta (w) \right) , \end{aligned}$$
(A.3)
where the notation (2.4) is used, defines a Poisson bracket on \(C^\infty ({\mathbb {C}}^n)\). Equivalently to the formula (A.3), the Hamiltonian vector field \(V_H\) associated with \(H\in C^\infty ({\mathbb {C}}^n)\) is given by
$$\begin{aligned} V_H(w) = (w \eta (w)^{\dagger })_{{\mathfrak {u}}(n)} w - \eta (w) -{\textstyle {\frac{1}{2}}}(\eta (w)^{\dagger }w + w^{\dagger }\eta (w))w, \qquad \eta (w)= \nabla H(w). \nonumber \\ \end{aligned}$$
(A.4)
Extending the real Poisson bracket to complex functions by complex bilinearity, the Poisson brackets of the component functions \(w\mapsto w_i\) satisfy the explicit formulae (2.28) and (2.29).
Proof
The anti-symmetry of the last two terms of (A.3) is obvious, while the anti-symmetry of the sum of the first and second terms is seen from the identity
$$\begin{aligned} \mathfrak {I}\left( \xi ^{\dagger }(w \eta ^{\dagger })_{{\mathfrak {u}}} w - {\textstyle {\frac{1}{2}}}\xi ^{\dagger }w\eta ^{\dagger }w\right) = {\textstyle {\frac{1}{2}}}\mathfrak {I}\,\text {tr}\left( (w \eta ^{\dagger })_{{\mathfrak {u}}} (w \xi ^{\dagger })_{\mathfrak {b}}-(w \xi ^{\dagger })_{{\mathfrak {u}}} (w \eta ^{\dagger })_{\mathfrak {b}}\right) ,\nonumber \\ \end{aligned}$$
(A.5)
where we used constant \(\xi \) and \(\eta \) for simplicity. Here and below, the subscripts \({\mathfrak {u}}\) and \({\mathfrak {b}}\) stand for \({\mathfrak {u}}(n)\) and \({\mathfrak {b}}(n)\).
Regarding the Jacobi identity, it is enough to verify it for linear functions \(F_\xi \), \(F_\eta \) and \(F_\zeta \) for arbitrary \(\xi ,\eta ,\zeta \in {\mathbb {C}}^n\). In this verification, we may use the formula (A.4), since this expresses the identity \(\{F,H\}(w) = \mathfrak {I}(\xi (w)^\dagger V_H(w))\) and does not rely on the Jacobi identity.
We start by calculating the gradient of \(\{ F_\xi , F_\eta \}\) from (A.3) and find
$$\begin{aligned} \left( \nabla \{ F_\xi , F_\eta \}(w)\right) ^\dagger = \xi ^{\dagger }(w\eta ^{\dagger })_{\mathfrak {u}}- \eta ^{\dagger }(w\xi ^{\dagger })_{\mathfrak {u}}+ {\textstyle {\frac{1}{2}}}\eta ^{\dagger }w\xi ^{\dagger }- {\textstyle {\frac{1}{2}}}\xi ^{\dagger }w \eta ^{\dagger }- {\textstyle {\frac{1}{2}}}(w^{\dagger }\eta )\xi ^{\dagger }+ {\textstyle {\frac{1}{2}}}(w^{\dagger }\xi )\eta ^{\dagger }.\nonumber \\ \end{aligned}$$
(A.6)
Combining this with \(V_{{F_\zeta }}(w)\) from (A.4), we have to inspect
$$\begin{aligned} \begin{aligned} {{\mathcal {J}}}(w):&= \{\{ F_\xi , F_\eta \}, F_\zeta \}(w) + \hbox {cycl. perm.} \\&=\mathfrak {I}\left[ \left( \nabla \{ F_\xi , F_\eta \}(w)\right) ^\dagger \left( (w\zeta ^{\dagger })_{\mathfrak {u}}w - \zeta - {\textstyle {\frac{1}{2}}}(\zeta ^{\dagger }w)w - {\textstyle {\frac{1}{2}}}(w^{\dagger }\zeta )w\right) \right] + \hbox {c.p.} \end{aligned}\nonumber \\ \end{aligned}$$
(A.7)
By spelling this out, we obtain
$$\begin{aligned} \begin{aligned}&{{\mathcal {J}}(w)}= \mathfrak {I}\Bigl [ \eta ^{\dagger }(w\xi ^{\dagger })_{\mathfrak {u}}\zeta - \xi ^{\dagger }(w\eta ^{\dagger })_{\mathfrak {u}}\zeta + {\textstyle {\frac{1}{2}}}(\xi ^{\dagger }w)\eta ^{\dagger }\zeta \\&\quad - {\textstyle {\frac{1}{2}}}(\eta ^{\dagger }w)\xi ^{\dagger }\zeta + {\textstyle {\frac{1}{2}}}(w^{\dagger }\eta )\xi ^{\dagger }\zeta - {\textstyle {\frac{1}{2}}}(w^{\dagger }\xi )\eta ^{\dagger }\zeta + \hbox {c.p.}\Bigr ] \\&\quad + \mathfrak {I}\Bigl [\xi ^{\dagger }(w\eta ^{\dagger })_{\mathfrak {u}}(w\zeta ^{\dagger })_{\mathfrak {u}}w - \eta ^{\dagger }(w\xi ^{\dagger })_{\mathfrak {u}}(w\zeta ^{\dagger })_{\mathfrak {u}}w + {\textstyle {\frac{1}{2}}}\eta ^{\dagger }w\xi ^{\dagger }(w\zeta ^{\dagger })_{\mathfrak {u}}w - {\textstyle {\frac{1}{2}}}\xi ^{\dagger }w \eta ^{\dagger }(w\zeta ^{\dagger })_{\mathfrak {u}}w \\&\quad + {\textstyle {\frac{1}{2}}}(w^{\dagger }\xi )\eta ^{\dagger }(w\zeta ^{\dagger })_{\mathfrak {u}}w - {\textstyle {\frac{1}{2}}}(w^{\dagger }\eta ) \xi ^{\dagger }(w\zeta ^{\dagger })_{\mathfrak {u}}w +\hbox { c.p.} \\&\quad - {\textstyle {\frac{1}{2}}}(\zeta ^{\dagger }w)\xi ^{\dagger }(w\eta ^{\dagger })_{\mathfrak {u}}w + {\textstyle {\frac{1}{2}}}(\zeta ^{\dagger }w)\eta ^{\dagger }(w\xi ^{\dagger })_{\mathfrak {u}}w - \textstyle {\frac{1}{4}}(\eta ^{\dagger }w)(\xi ^{\dagger }w)(\zeta ^{\dagger }w) + \textstyle {\frac{1}{4}}(\xi ^{\dagger }w) (\eta ^{\dagger }w)(\zeta ^{\dagger }w) \\&\quad + \textstyle {\frac{1}{4}}(w^{\dagger }\eta )(\xi ^{\dagger }w)(\zeta ^{\dagger }w) - \textstyle {\frac{1}{4}}(w^{\dagger }\xi )(\eta ^{\dagger }w)(\zeta ^{\dagger }w) + \hbox {c.p.} \\&\qquad - {\textstyle {\frac{1}{2}}}(w^{\dagger }\zeta )\xi ^{\dagger }(w\eta ^{\dagger })_{\mathfrak {u}}w + {\textstyle {\frac{1}{2}}}(w^{\dagger }\zeta )\eta ^{\dagger }(w\xi ^{\dagger })_{\mathfrak {u}}w - \textstyle {\frac{1}{4}}(w^{\dagger }\zeta )(\eta ^{\dagger }w)(\xi ^{\dagger }w) + \textstyle {\frac{1}{4}}(w^{\dagger }\zeta ) (\xi ^{\dagger }w)(\eta ^{\dagger }w) \\&\quad + \textstyle {\frac{1}{4}}(w^{\dagger }\eta )(w^{\dagger }\zeta )(\xi ^{\dagger }w) - \textstyle {\frac{1}{4}}(w^{\dagger }\xi )(w^{\dagger }\zeta )(\eta ^{\dagger }w) + \hbox {c.p.}\Bigr ] \end{aligned} \end{aligned}$$
After making several self-evident cancellations, and using cyclic permutations to reorganize terms in a convenient way, we get
$$\begin{aligned} \begin{aligned} {{\mathcal {J}}(w)}&=\mathfrak {I}\Bigl (\zeta ^{\dagger }(w\eta ^{\dagger })_{\mathfrak {u}}\xi - \xi ^{\dagger }(w\eta ^{\dagger })_{\mathfrak {u}}\zeta + {\textstyle {\frac{1}{2}}}(\xi ^{\dagger }w - w^{\dagger }\xi )\eta ^{\dagger }\zeta - {\textstyle {\frac{1}{2}}}(\eta ^{\dagger }w - w^{\dagger }\eta )\xi ^{\dagger }\zeta + \hbox {c.p.}\Bigr )\\&\quad +\,\mathfrak {I}\,\text {tr}\Bigl (w\xi ^{\dagger }\bigl [(w\eta ^{\dagger })_{\mathfrak {u}},(w\zeta ^{\dagger })_{\mathfrak {u}}\bigr ] + (\eta ^{\dagger }w) w\xi ^{\dagger }(w\zeta ^{\dagger })_{\mathfrak {u}}- (\xi ^{\dagger }w) w\eta ^{\dagger }(w\zeta ^{\dagger })_{\mathfrak {u}}+ \hbox {c.p.}\Bigr ). \end{aligned} \end{aligned}$$
It is not difficult to see that the first line gives zero. Rearranging the second line, we have
$$\begin{aligned} {{\mathcal {J}}(w)}= & {} \mathfrak {I}\,\text {tr}\Bigl (w\xi ^{\dagger }\bigl [(w\eta ^{\dagger })_{\mathfrak {u}},(w\zeta ^{\dagger })_{\mathfrak {u}}\bigr ] + w\eta ^{\dagger }w\xi ^{\dagger }(w\zeta ^{\dagger })_{\mathfrak {u}}- w\xi ^{\dagger }w\eta ^{\dagger }(w\zeta ^{\dagger })_{\mathfrak {u}}+ \hbox {c.p.}\Bigr ) \\= & {} \mathfrak {I}\,\text {tr}\Bigl (-w\xi ^{\dagger }\bigl [(w\eta ^{\dagger })_{\mathfrak {b}},(w\zeta ^{\dagger })_{\mathfrak {u}}\bigr ] + \hbox {c.p.}\Bigr )\\= & {} - \mathfrak {I}\,\text {tr}\Bigl ((w\xi ^{\dagger })_{\mathfrak {u}}\bigl ( [(w\eta ^{\dagger })_{\mathfrak {b}},(w\zeta ^{\dagger })_{\mathfrak {u}}] + [w\eta ^{\dagger },(w\zeta ^{\dagger })_{\mathfrak {b}}]\bigr ) \\&+ (w\xi ^{\dagger })_{\mathfrak {b}}\bigl ( [(w\eta ^{\dagger })_{\mathfrak {b}},(w\zeta ^{\dagger })_{\mathfrak {u}}] + [(w\eta ^{\dagger })_{\mathfrak {u}}, w\zeta ^{\dagger }]\bigr )\Bigr )\\= & {} -\mathfrak {I}\,\text {tr}\Bigl ( (w\xi ^{\dagger })_{\mathfrak {u}}\bigl ( [(w\eta ^{\dagger })_{\mathfrak {b}},(w\zeta ^{\dagger })_{\mathfrak {u}}] + [(w\eta ^{\dagger })_{\mathfrak {u}},(w\zeta ^{\dagger })_{\mathfrak {b}}] + [(w\eta ^{\dagger })_{\mathfrak {b}},(w\zeta ^{\dagger })_{\mathfrak {b}}]\bigr )\\&+ (w\xi ^{\dagger })_{\mathfrak {b}}\bigl ( [(w\eta ^{\dagger })_{\mathfrak {b}},(w\zeta ^{\dagger })_{\mathfrak {u}}] + [(w\eta ^{\dagger })_{\mathfrak {u}},(w\zeta ^{\dagger })_{\mathfrak {u}}] + [(w\eta ^{\dagger })_{\mathfrak {u}},(w\zeta ^{\dagger })_{\mathfrak {b}}]\bigr )\Bigr )\\= & {} \mathfrak {I}\,\text {tr}\bigl (w\xi ^{\dagger }[ w\eta ^{\dagger }, w\zeta ^{\dagger }]\bigr ) +\hbox {cycl. perm.} =0.\\ \end{aligned}$$
Having verified the Jacobi identity, it remains to calculate the Poisson brackets of the components of w and their complex conjugates. Let \(e_k\) \((k=1,\dots , n)\) denote the canonical basis of \({\mathbb {C}}^n\). One obtains by tedious calculation that the Hamiltonian vector fields of the linear functions given by the real and imaginary parts of the components \(w_k\) have the following form:
$$\begin{aligned} V_{\mathfrak {R}w_k}(w) = \left\{ \begin{aligned}&{\mathrm{i}}\mathfrak {R}(w_k) w_k e_k+ {\mathrm{i}}\sum _{r>k}(w_kw_re_r+ |w_r|^2e_k) + {\mathrm{i}}e_k- {\textstyle {\frac{1}{2}}}{\mathrm{i}}(w_k-{\overline{w}}_k)w\qquad k<n,\\&{\mathrm{i}}\mathfrak {R}(w_n)w_ne_n+ {\mathrm{i}}e_n- {\textstyle {\frac{1}{2}}}{\mathrm{i}}(w_n-{\overline{w}}_n)w \quad \qquad \qquad \qquad \qquad \qquad \qquad k=n. \end{aligned}\right. \end{aligned}$$
and
$$\begin{aligned} V_{\mathfrak {I}w_k}(w) = \left\{ \begin{aligned}&{\mathrm{i}}\mathfrak {I}(w_k) w_ke_k+ \sum _{r>k}(w_kw_re_r- |w_r|^2e_k) - e_k- {\textstyle {\frac{1}{2}}}(w_k+{\overline{w}}_k)w\qquad k<n,\\&{\mathrm{i}}\mathfrak {I}( w_n) w_ne_n- e_n- {\textstyle {\frac{1}{2}}}(w_n+{\overline{w}}_n)w \,\,\,\, \qquad \qquad \qquad \qquad \qquad \qquad k=n. \end{aligned}\right. \end{aligned}$$
By using these, one can check that the formulae (2.28) and (2.29) follow. If desired, the reader can supply the details. \(\square \)
The bracket (A.3) has the nice property that the natural action of \({\mathrm{U}}(n)\) on \({\mathbb {C}}^n\) is Poisson [56], and this can also be checked using linear functions \(F_\xi \). To this end, for any \(g\in {\mathrm{U}}(n)\) and \(w\in {\mathbb {C}}^n\) we define the functions \(F_\xi (g\,\cdot \,)\in C^\infty ({\mathbb {C}}^n)\) and \(F_\xi (\,\cdot \,w)\in C^\infty ({\mathrm{U}}(n))\) by
$$\begin{aligned} F_\xi (g\,\cdot \,)(w) = F_\xi (gw) = F_\xi (\,\cdot \,w)(g). \end{aligned}$$
(A.8)
Then, an easy calculation gives that
$$\begin{aligned} \{F_\xi ,F_\eta \}(gw) - \{F_\xi (g\,\cdot \, ),F_\eta (g\,\cdot \, )\}(w) \end{aligned}$$
(A.9)
is equal to
$$\begin{aligned} \mathfrak {I}\text {tr}\bigl (gw\xi ^{\dagger }(gw\eta ^{\dagger })_{{\mathfrak {u}}(n)} - w\xi ^{\dagger }g (w\eta ^{\dagger }g)_{{\mathfrak {u}}(n)}\bigr ), \end{aligned}$$
(A.10)
which in turn is equal to the value at g of the Poisson bracket (2.10) of the functions \(F_\xi (\,\cdot \, w)\) and \(F_\eta (\,\cdot \,w)\) on \({\mathrm{U}}(n)\). The last equality follows using \(D F_\xi (\,\cdot \,w)(g)= (g w \xi ^{\dagger })_{{\mathfrak {b}}(n)}\) and elementary manipulations. Thus, we have
$$\begin{aligned} \{F_\xi ,F_\eta \}(gw) = \{F_\xi (g\,\cdot \, ),F_\eta (g\,\cdot \, )\}(w) + \{F_\xi (\,\cdot \,w),F_\eta (\,\cdot \,w)\}_U(g), \end{aligned}$$
(A.11)
which means that the map \({\mathrm{U}}(n) \times {\mathbb {C}}^n \ni (g,w) \mapsto gw \in {\mathbb {C}}^n\) is indeed a Poisson map.
Let us recall the diffeomorphism
$$\begin{aligned} b \mapsto b b^\dagger \end{aligned}$$
(A.12)
from the group \(\text {B}(n)\) to the space \({{\mathfrak {P}}}(n)\) of positive definite Hermitian matrices. By this diffeomorphism, the Poisson structure (2.11) on \(\text {B}(n)\) is converted into a Poisson structure on \({{\mathfrak {P}}}(n)\), which is given by the first term of (2.23), i.e. ,
$$\begin{aligned} \{f, h\}_{{\mathfrak {P}}}(L)=4 \langle L d f(L), \left( L d h (L)\right) _{{\mathfrak {u}}(n)} \rangle \end{aligned}$$
(A.13)
for all \(f,h \in C^\infty ({{\mathfrak {P}}}(n))\). Here, the \({\mathfrak {u}}(n)\)-valued derivatives df and dh are defined by (2.22).
Proposition A.2
With respect to the brackets (A.3) and (A.13), the map
$$\begin{aligned} \Phi :w\mapsto {{\varvec{1}}}_n+ ww^{\dagger } \end{aligned}$$
(A.14)
from \({\mathbb {C}}^n\) to \({{\mathfrak {P}}}(n)\) is Poisson.
Proof
Let \(X,Y\in {\mathfrak {u}}(n)\) and consider the pull-backs \(\Phi ^*(f_X)\) and \(\Phi ^*(f_Y)\) of the functions \(f_X(L):=\langle X,L\rangle \) and \(f_Y(L) := \langle Y,L\rangle \). We have
$$\begin{aligned} \Phi ^*(f_X)(w) = \mathfrak {I}(w^{\dagger }Xw) + \mathfrak {I}\text {tr}(X)\quad \hbox {and}\quad \Phi ^*(f_Y)(w) = \mathfrak {I}(w^{\dagger }Yw) + \mathfrak {I}\text {tr}(Y).\nonumber \\ \end{aligned}$$
(A.15)
Using the formula (A.3) with \((\nabla \Phi ^* (f_X))(w) = - 2 Xw\) and similar for \(f_Y\), we can compute
$$\begin{aligned} \{\Phi ^*(f_X),\Phi ^*(f_Y)\}(w)= & {} 4 \mathfrak {I}\Bigl ( w^{\dagger }X(ww^{\dagger }Y)_{\mathfrak {u}}w + w^{\dagger }X Yw \nonumber \\&- {\textstyle {\frac{1}{2}}}w^{\dagger }Xww^{\dagger }Yw + {\textstyle {\frac{1}{2}}}w^{\dagger }X ww^{\dagger }Y w\Bigr ) \nonumber \\= & {} 4\mathfrak {I}\text {tr}\Bigl ( ({{\varvec{1}}}_n+ww^{\dagger })X\bigl (({{\varvec{1}}}_n +ww^{\dagger })Y\bigr )_{{\mathfrak {u}}(n)} \Bigr ) = \{ f_X, f_Y\}_{{\mathfrak {P}}}(\Phi (w)).\nonumber \\ \end{aligned}$$
(A.16)
Here, we have taken into account that, for example, \(\mathfrak {I}\text {tr}(XY) =0\) for \(X,Y\in {\mathfrak {u}}(n)\). The statement follows since the linear functions of the form \(f_X\) can serve as coordinates on \({{\mathfrak {P}}}(n)\). \(\square \)
Let \({{\mathbf {b}}}: {\mathbb {C}}^n \rightarrow \text {B}(n)\) be the map determined by the condition
$$\begin{aligned} \Phi = {{\mathbf {b}}}{{\mathbf {b}}}^\dagger . \end{aligned}$$
(A.17)
It follows from Proposition A.2 that this is a Poisson map with respect to the Poisson brackets (A.3) on \({\mathbb {C}}^n\) and (2.11) on \(\text {B}(n)\).
Proposition A.3
The map \({{\mathbf {b}}}\) defined by (A.17) with (A.14) is the moment map for the Poisson action (2.26) of \({\mathrm{U}}(n)\) on \({\mathbb {C}}^n\). According to (2.27), this means that we have
$$\begin{aligned} \mathfrak {I}\left( (\nabla F(w))^\dagger Xw\right) = \mathfrak {I}\mathrm{tr} \left( X \{F, {{\mathbf {b}}}\}(w) {{\mathbf {b}}}(w)^{-1}\right) , \quad \forall X\in {\mathfrak {u}}(n),\,w\in {\mathbb {C}}^n,\, F\in C^\infty ({\mathbb {C}}^n). \nonumber \\ \end{aligned}$$
(A.18)
Proof
For ease of notation, we verify the relation for linear functions \(F_\xi \) on \({\mathbb {C}}^n\), which is sufficient. For this, we have to calculate the \({\mathfrak {b}}(n)\)-valued function
$$\begin{aligned} \beta _F := \{ {{\mathbf {b}}}, F\} {{\mathbf {b}}}^{-1}\,, \quad F:=F_\xi . \end{aligned}$$
(A.19)
Since (A.12) is a diffeomorphism, \(\beta _F\) is uniquely determined by
$$\begin{aligned} \{ \Phi , F\} = \beta _F \Phi + \Phi \beta _F^\dagger , \end{aligned}$$
(A.20)
and this can be calculated as follows. First, we rearrange the expression (A.4) of the Hamiltonian vector field in the form
$$\begin{aligned} V_F(w)= {\textstyle {\frac{1}{2}}}(\xi ^{\dagger }w - w^{\dagger }\xi )w -\xi - (w\xi ^{\dagger })_{{\mathfrak {b}}(n)} w. \end{aligned}$$
(A.21)
Then, as \((\xi ^{\dagger }w - w^{\dagger }\xi )\in {\mathrm{i}}{\mathbb {R}}\), we obtain
$$\begin{aligned} \{ \Phi , F\}(w)= & {} V_F(w) w^{\dagger }+ w (V_F(w))^{\dagger }\\= & {} -(w\xi ^{\dagger })_{{\mathfrak {b}}(n)} ww^{\dagger }- ww^{\dagger }(w\xi ^{\dagger })_{{\mathfrak {b}}(n)}^{\dagger }- \xi w^{\dagger }- w\xi ^{\dagger }\nonumber \\= & {} -(w\xi ^{\dagger })_{{\mathfrak {b}}(n)}\Phi (w) - \Phi (w)(w\xi ^{\dagger })_{{\mathfrak {b}}(n)}^{\dagger }+ \bigl ((w\xi ^{\dagger })_{{\mathfrak {b}}(n)}-w\xi ^{\dagger }\bigr )\nonumber \\&+ \bigl ((w\xi ^{\dagger })_{{\mathfrak {b}}(n)}^{\dagger }-\xi w^{\dagger }\bigr ).\nonumber \end{aligned}$$
(A.22)
But the last two terms cancel, and hence we see that
$$\begin{aligned} \beta _F(w) = - \left( w \xi ^\dagger \right) _{{\mathfrak {b}}(n)}. \end{aligned}$$
(A.23)
By using this, the right-hand-side of (A.18) becomes
$$\begin{aligned} - \mathfrak {I}\text {tr}( X \beta _F(w)) = \mathfrak {I}\text {tr}(X w \xi ^\dagger ) = \mathfrak {I}(\xi ^\dagger X w), \end{aligned}$$
(A.24)
whereby the proof is complete. \(\square \)
Remark A.4
We had no need for the explicit formula of \({{\mathbf {b}}}(w)\) in the above, but in some other calculations it is needed. The reader can verify directly that it obeys equation (2.31).
Remark A.5
The maximal torus \({\mathbb {T}}^n < {\mathrm{U}}(n)\) is a Poisson subgroup with vanishing Poisson bracket, and therefore, the restriction of the \({\mathrm{U}}(n)\) action to \({\mathbb {T}}^n\) gives an ordinary Hamiltonian action. One can identify the dual Poisson–Lie group of \({\mathbb {T}}^n\) with \(\text {B}(n)_0\), the group of positive diagonal matrices, with zero Poisson bracket. Then, the corresponding group valued moment map is provided by \(w \mapsto {{\mathbf {b}}}(w)_0\), which is the diagonal part of \({{\mathbf {b}}}(w)\). Writing
$$\begin{aligned} {{\mathbf {b}}}(w)_0 = \exp (\phi (w)), \end{aligned}$$
(A.25)
we get the ordinary moment map \(w \mapsto \phi (w) \in {\mathfrak {b}}(n)_0\), where \({\mathfrak {b}}(n)_0\) (the space of real diagonal matrices) is identified with the linear dual of \({\mathfrak {u}}(n)_0\).
The following proposition represents one of the side results of the paper.
Proposition A.6
The Poisson bracket (A.3) is symplectic and, with \({{\mathcal {G}}}_j= 1 + \sum _{k=j}^n \vert w_j\vert ^2\), the corresponding symplectic form on \({\mathbb {C}}^n\) is given by
$$\begin{aligned} \Omega _{{\mathbb {C}}^n} = \frac{{\mathrm{i}}}{2}\sum _{k=1}^n\frac{1}{{{\mathcal {G}}}_k} dw_k\wedge d{\overline{w}}_k + \frac{{\mathrm{i}}}{4}\sum _{k=1}^{n-1}\frac{1}{{{\mathcal {G}}}_k{{\mathcal {G}}}_{k+1}} d{{\mathcal {G}}}_{k+1} \wedge \left( {\overline{w}}_k dw_k - w_k d{\overline{w}}_k\right) .\nonumber \\ \end{aligned}$$
(A.26)
Proof
We start from the coordinate form of the Poisson bracket, copied here for convenience:
$$\begin{aligned} \begin{aligned} \{w_i,w_k\}&= {\mathrm{i}}\,\text {sgn}(i-k)w_iw_k\\ \{w_i,{\overline{w}}_l\}&= {\mathrm{i}}\delta _{il}(2+|w|^2) + {\mathrm{i}}w_i{\overline{w}}_l + {\mathrm{i}}\delta _{il}\sum _{r=1}^n\text {sgn}(r-l) |w_r|^2. \end{aligned} \end{aligned}$$
(A.27)
We shall first invert the Poisson tensor on the dense open submanifold on which all \(\vert w_j \vert >0\), where we can use the parametrization \(w_j = e^{{\mathrm{i}}\varphi _j} \vert w_j\vert \).
Let us consider
$$\begin{aligned} \{w_i, |w_k|^2\}= & {} {\mathrm{i}}\,\text {sgn}(i-k)|w_k|^2w_i +{\mathrm{i}}|w_k|^2w_i +{\mathrm{i}}\delta _{ik}(2+|w|^2)w_i\nonumber \\&+{\mathrm{i}}\delta _{ik}w_i\sum _{r=1}^n\text {sgn}(r-k)|w_r|^2, \end{aligned}$$
(A.28)
from which we easily obtain
$$\begin{aligned} \{|w_i|^2,|w_k|^2\}=0. \end{aligned}$$
(A.29)
Using this, and restricting now to our submanifold, the relation (A.28) implies
$$\begin{aligned} \{e^{{\mathrm{i}}\varphi _i},|w_k|^2\} = \{ \frac{w_i}{|w_i|} , |w_k|^2 \} = {\mathrm{i}}[1-\delta _{ik}+\text {sgn}(i-k)]|w_k|^2 e^{{\mathrm{i}}\varphi _i} + 2{\mathrm{i}}\delta _{ik} {{\mathcal {G}}}_k e^{{\mathrm{i}}\varphi _i}. \nonumber \\ \end{aligned}$$
(A.30)
Plainly, we have the identity
$$\begin{aligned} \{ w_j , w_k \} + e^{2{\mathrm{i}}\varphi _j} e^{2{\mathrm{i}}\varphi _k} \{ {\overline{w}}_j , {\overline{w}}_k \} = 2 \vert w_j w_k \vert \{ e^{{\mathrm{i}}\varphi _j} , e^{{\mathrm{i}}\varphi _k} \}. \end{aligned}$$
(A.31)
The left-hand side can be checked to vanish, and thus, we get
$$\begin{aligned} \{ e^{{\mathrm{i}}\varphi _j} , e^{{\mathrm{i}}\varphi _k} \}=0. \end{aligned}$$
(A.32)
It is convenient to change variables, noting that the map \((|w_1|^2,\dots ,|w_n|^2) \mapsto ({{\mathcal {G}}}_1,\dots ,{{\mathcal {G}}}_n)\) is invertible. Then, it is elementary to derive from (A.30) the relation
$$\begin{aligned} \{e^{{\mathrm{i}}\varphi _i}, {{\mathcal {G}}}_k\} = \left\{ \begin{aligned}&2{\mathrm{i}}{{\mathcal {G}}}_k e^{{\mathrm{i}}\varphi _i},\quad k\le i\\&0, \ \ \ \quad \qquad k>i\end{aligned}\right. \end{aligned}$$
(A.33)
that can be also written as
$$\begin{aligned} \{ \varphi _i, \ln {{\mathcal {G}}}_k\} = \left\{ \begin{aligned}&2,\quad k\le i\\&0,\quad k>i \end{aligned} \right. \end{aligned}$$
(A.34)
This means that the matrix of Poisson brackets, in the variables \( \varphi _i, \ln {{\mathcal {G}}}_k \) has the form
$$\begin{aligned} P = 2\left( \begin{array}{cc}0 &{}\quad A \\ -A^T &{}\quad 0\end{array}\right) \end{aligned}$$
(A.35)
with
$$\begin{aligned} A = {{\varvec{1}}}_n + B + B^2 + \dots +B^{n-1}, \end{aligned}$$
(A.36)
where B is the nilpotent matrix having the entries \(B_{ik} = \delta _{i,k+1}\). Both A and P are invertible, and their inverses are
$$\begin{aligned} A^{-1}={{\varvec{1}}}_n-B \quad \hbox {and}\quad P^{-1} = \frac{1}{2} \left( \begin{array}{cc}0 &{} -(A^{-1})^T \\ A^{-1} &{} 0\end{array}\right) . \end{aligned}$$
(A.37)
Consequently, we obtain the symplectic formFootnote 7 (\(x^\alpha \) represent the local variables \(\varphi _i\) and \(\ln {{\mathcal {G}}}_k\))
$$\begin{aligned} \Omega= & {} \frac{1}{2} \sum _{\alpha ,\beta =1}^{2n}(P^{-1})_{\alpha \beta } dx^\alpha \wedge dx^\beta \nonumber \\= & {} {\textstyle {\frac{1}{2}}}\sum _{k=1}^{n-1} [d\ln {{\mathcal {G}}}_k-d\ln {{\mathcal {G}}}_{k+1}]\wedge d\varphi _k + {\textstyle {\frac{1}{2}}}d\ln {{\mathcal {G}}}_n\wedge d\varphi _n. \end{aligned}$$
(A.38)
If we now substitute the identities
$$\begin{aligned} d \ln {{\mathcal {G}}}_k - d \ln {{\mathcal {G}}}_{k+1}= \frac{ {{\mathcal {G}}}_{k+1} d \vert w_k \vert ^2 - \vert w_k\vert ^2 d {{\mathcal {G}}}_{k+1}}{{{\mathcal {G}}}_k {{\mathcal {G}}}_{k+1}} \end{aligned}$$
(A.39)
and
$$\begin{aligned} d\varphi _k = (2{\mathrm{i}}|w_k|^2)^{-1}({\overline{w}}_k dw_k - w_kd{\overline{w}}_k), \end{aligned}$$
(A.40)
then \(\Omega \) (A.38) takes the form
$$\begin{aligned} \Omega = \frac{{\mathrm{i}}}{2}\sum _{k=1}^n\frac{1}{{{\mathcal {G}}}_k} dw_k\wedge d{\overline{w}}_k + \frac{{\mathrm{i}}}{4}\sum _{k=1}^{n-1}\frac{1}{{{\mathcal {G}}}_k{{\mathcal {G}}}_{k+1}} d {{\mathcal {G}}}_{k+1} \wedge \left( {\overline{w}}_k dw_k - w_k d{\overline{w}}_k\right) . \nonumber \\ \end{aligned}$$
(A.41)
It is clear that both the original Poisson tensor corresponding to (A.27) and \(\Omega \) (A.41) are regular over the whole of \({\mathbb {C}}^n\). As a result, their inverse relationship extends from the dense open submanifold (where \(|w_j|>0\) for all j) to the full phase space. \(\square \)
Remark A.7
The image of the map \(w\mapsto {{\varvec{1}}}_n + w w^\dagger \) is the union of the \({\mathrm{U}}(n)\) orbits in \({{\mathfrak {P}}}(n)\) passing through the degenerate diagonal matrices
$$\begin{aligned} \text {diag}(1 + R^2, 1\dots , 1), \qquad R\ge 0. \end{aligned}$$
(A.42)
For any fixed \(R>0\), the orbit is a symplectic leaf in \({{\mathfrak {P}}}(n)\) of dimension \(2(n-1)\); \(R=0\) corresponds to a trivial symplectic leaf. The union of the orbits consisting of the conjugates of the matrices
$$\begin{aligned} \text {diag}(1 - r^2, 1,\dots , 1), \qquad 0 \le r < 1 \end{aligned}$$
(A.43)
is the image of the map
$$\begin{aligned} z \mapsto {{\varvec{1}}}_n - z z^\dagger \end{aligned}$$
(A.44)
from
$$\begin{aligned} {{\mathcal {B}}}(1):= \{ z \in {\mathbb {C}}^n \mid |z|^2 <1 \} \end{aligned}$$
(A.45)
to \({{\mathfrak {P}}}(n)\). In fact, the open ball \({{\mathcal {B}}}(1)\), identified as a subset of \({\mathbb {R}}^{2n}\), can be equipped with the Poisson bracket
$$\begin{aligned} \begin{aligned} \{z_i,z_k\}&= {\mathrm{i}}\,\text {sgn}(i-k)z_iz_k\\ \{z_i,{\overline{z}}_l\}&= {\mathrm{i}}(|z|^2 -2)\delta _{il} + {\mathrm{i}}z_i{\overline{z}}_l + {\mathrm{i}}\delta _{il}\sum _{r=1}^n\text {sgn}(r-l)|z_r|^2 \end{aligned} \end{aligned}$$
(A.46)
with respect to which the map (A.44) is Poisson. This is also a special case of the Poisson structures found in [56]. The analogue of Proposition A.3 holds for the map \({{\mathbf {b}}}_-: {{\mathcal {B}}}(1) \rightarrow \text {B}(n)\) defined by
$$\begin{aligned} {{\varvec{1}}}_n - z z^\dagger = {{\mathbf {b}}}_-(z) {{\mathbf {b}}}_-(z)^\dagger . \end{aligned}$$
(A.47)
The Poisson map \({{\mathbf {b}}}_-\) can be used to introduce variants of our reduction. Concretely, one may replace one or more of the \({{\mathbf {b}}}\) factors in (3.1) by \({{\mathbf {b}}}_-\) and study the reduced system. The restriction \(\gamma >0\) in the moment map constraint (3.17) then might not be necessary. Let us also note that one obtains a Poisson pencil on \({\mathbb {C}}^n\) if one replaces the last term of (A.3) by \(-\lambda \mathfrak {I}\left( \xi (w)^\dagger \eta (w)\right) \) for any real parameter \(\lambda \), and the formula (A.46) corresponds to \(\lambda =-1\).
Proof of Lemma 5.1
In this section, we work over \(({\mathbb {C}}^{n\times d}, \{\ ,\ \}_{{\mathcal {W}}})\) with the primary spins \((w^\alpha )\), see Sect. 2.2. We set \(\{ \ , \ \}:=\{\ ,\ \}_{{\mathcal {W}}}\) to simplify notations.
As noted in Sect. 3.1, the half-dressed spins \(v^\alpha \) can be defined in \({\mathbb {C}}^{n\times d}\) in terms of the primary spins. It is convenient to introduce the matrices \(b_\alpha ={{\mathbf {b}}}(w^\alpha )\) and \(B^\alpha =b_1\cdots b_\alpha \), so that
$$\begin{aligned} v^\alpha = B^{\alpha -1} w^\alpha \,. \end{aligned}$$
(B.1)
Remark that \(B^\alpha \) is related to the matrix \(B_\alpha \) introduced in (3.2) by \(B_\alpha = b_R B^\alpha \). We also note the following lemma, which follows from Proposition 2.1 by straightforward computations.
Lemma B.1
For any \(1\le \alpha \le d\), \(1\le i , j , l \le n\),
$$\begin{aligned}&\{ w^\alpha _i,(b_\alpha )_{jl} \}=\,{\mathrm{i}}\,[\delta _{ij}+2 \delta _{(i>j)}]\,w^\alpha _j (b_\alpha )_{il} \,, \end{aligned}$$
(B.2)
$$\begin{aligned}&\{ \overline{w}^\alpha _i,(b_\alpha )_{jl} \}=-{\mathrm{i}}\delta _{ij} \overline{w}^\alpha _i (b_\alpha )_{il} -2{\mathrm{i}}\delta _{ij} \sum _{k=j+1}^l \overline{w}^\alpha _k (b_\alpha )_{kl} \,. \end{aligned}$$
(B.3)
Furthermore, the Poisson bracket evaluated on \(((b_\alpha )_{ij},(\overline{b}_\alpha )_{ij})\) is given by (2.7)–(2.8).
Next, we need to describe the Poisson brackets between the matrix entries of \((B^\alpha ,w^\alpha )\), which appear in the decomposition (B.1). To write them down, we introduce the matrices
$$\begin{aligned} B^{\alpha ;\gamma }=b_\alpha \cdots b_\gamma \,, \quad 1\le \alpha \le \gamma \le d\,, \end{aligned}$$
(B.4)
which are such that \(B^{1;\alpha }=B^\alpha \) and \(B^{\alpha ;\alpha }=b_\alpha \). We also set \(B^{\alpha +1;\alpha }:={{\varvec{1}}}_n\) and \(B^0:={{\varvec{1}}}_n\).
Lemma B.2
For any \(1\le \alpha ,\beta \le d\), \(1\le i ,k,l \le n\),
$$\begin{aligned}&\{ w^\alpha _i,B^\beta _{kl} \}= - {\mathrm{i}}\delta _{(\alpha \le \beta )} B^{\alpha -1}_{ki} w_i^\alpha B^{\alpha ;\beta }_{il} +2{\mathrm{i}}\delta _{(\alpha \le \beta )} \sum _{k' \le i} B^{\alpha -1}_{kk'} w_{k'}^\alpha B^{\alpha ;\beta }_{il} \,, \end{aligned}$$
(B.5)
$$\begin{aligned}&\{ \overline{w}^\alpha _i,B^\beta _{kl} \}= +{\mathrm{i}}\delta _{(\alpha \le \beta )} B^{\alpha -1}_{ki} \overline{w}_i^\alpha B^{\alpha ;\beta }_{il} - 2{\mathrm{i}}\delta _{(\alpha \le \beta )} B^{\alpha -1}_{ki} \sum _{i\le u} \overline{w}_u^\alpha B^{\alpha ;\beta }_{ul} \,. \end{aligned}$$
(B.6)
Proof
By construction, for \(\beta \ne \alpha \) we have \(\{ w^\alpha _i,w^\beta _k \}=0\), hence \(\{ w^\alpha _i,(b_\beta )_{kl} \}=0\). We get that
$$\begin{aligned} \{ w^\alpha _i,B^\beta _{kl} \}=0,\,\, \alpha <\beta \,; \quad \{ w^\alpha _i,B^\beta _{kl} \}=\sum _{k \le l' \le l} \{ w^\alpha _i,B^\alpha _{kl'} \}B^{\alpha +1;\beta }_{l'l}\,, \,\, \beta > \alpha \,. \end{aligned}$$
(B.7)
When \(\beta =\alpha \), \(\{ w^\alpha _i,(b_\alpha )_{kl} \}\) is given by (B.2) and we get
$$\begin{aligned} \{ w^\alpha _i,B^\alpha _{kl} \}= -{\mathrm{i}}B^{\alpha -1}_{ki} w_i^\alpha (b_\alpha )_{il} +2{\mathrm{i}}\sum _{k' \le i} B^{\alpha -1}_{kk'} w_{k'}^\alpha (b_\alpha )_{il}\,, \end{aligned}$$
(B.8)
from which the first identity can be obtained. The second case is proved in the same way. \(\square \)
Lemma B.3
For any \(1\le \alpha ,\beta \le d\), \(1\le i,j,k,l\le n\),
$$\begin{aligned} \begin{aligned} \{ B^\alpha _{ij},B^\beta _{kl} \}{\mathop {=}\limits ^{\alpha \leqslant \beta }}&- 2 {\mathrm{i}}B^{\alpha }_{kj}\sum _{r> j}B^{\alpha }_{ir} B^{\alpha +1;\beta }_{rl} - {\mathrm{i}}B^{\alpha }_{kj} B^{\alpha }_{ij} B^{\alpha +1;\beta }_{jl} + 2 {\mathrm{i}}\delta _{(i>k)} B^\alpha _{kj} B^\beta _{il}+ {\mathrm{i}}\delta _{ik} B^\alpha _{kj} B^\beta _{il}\,, \\ \{ B^\alpha _{ij},\overline{B}^\beta _{kl} \}{\mathop {=}\limits ^{\alpha \leqslant \beta }}&-{\mathrm{i}}B_{ij}^\alpha \overline{B}_{kj}^\alpha \overline{B}_{jl}^{\alpha +1;\beta } - 2 {\mathrm{i}}\sum _{s<j} B_{is}^\alpha \overline{B}_{ks}^\alpha \overline{B}_{jl}^{\alpha +1;\beta } +{\mathrm{i}}\delta _{ik} B_{ij}^\alpha \overline{B}_{kl}^\beta + 2 {\mathrm{i}}\delta _{ik} \sum _{r>k} B_{rj}^\alpha \overline{B}_{rl}^\beta \,, \\ \{ B^\alpha _{ij},\overline{B}^\beta _{kl} \}{\mathop {=}\limits ^{\alpha \geqslant \beta }}&-{\mathrm{i}}B_{il}^\beta \overline{B}_{kl}^\beta B_{lj}^{\beta +1;\alpha } - 2 {\mathrm{i}}\sum _{s<l} B_{is}^\beta \overline{B}_{ks}^\beta B_{lj}^{\beta +1;\alpha } +{\mathrm{i}}\delta _{ik} B_{ij}^\alpha \overline{B}_{kl}^\beta + 2 {\mathrm{i}}\delta _{ik} \sum _{r>k} B_{rj}^\alpha \overline{B}_{rl}^\beta \,. \end{aligned} \end{aligned}$$
Proof
For the first equality, we have for \(\alpha \le \beta \) that
$$\begin{aligned} \{ B^\alpha _{ij},B^\beta _{kl} \} =\sum _{1\le \gamma \le \alpha } \sum _{i',j',k',l'}(B^{\gamma -1})_{ii'} (B^{\gamma -1})_{kk'} \{ (b_\gamma )_{i'j'},(b_\gamma )_{k'l'} \} B^{\gamma +1;\alpha }_{j'j} B^{\gamma +1;\beta }_{l'l}\,.\nonumber \\ \end{aligned}$$
(B.9)
A similar expansion holds for \(\{ B^\alpha _{ij},\overline{B}^\beta _{kl} \}\). It then suffices to use Lemma B.1.
\(\square \)
Note that in the case \(\beta =\alpha \), the Poisson brackets from Lemma B.3 take the usual form (2.7)–(2.8) on \(\text {B}(n)\). We can also see that we can use \(\beta =0\) in Lemma B.2 and \(\alpha ,\beta =0\) in Lemma B.3, since in such cases the Poisson bracket vanishes on \(B^0={{\varvec{1}}}_n\).
We can now prove Lemma 5.1 using Lemmae B.2 and B.3 . We will use the definition of the half-dressed spins given by (B.1). To show (5.1) we need to write
$$\begin{aligned} \begin{aligned} \{ v^\alpha _i,v^\beta _k \}=&\sum _{j,l}\{ B^{\alpha -1}_{ij} w^\alpha _j, B^{\beta -1}_{kl} w^\beta _l \} \\ =&\sum _{j,l}\{ B^{\alpha -1}_{ij} , B^{\beta -1}_{kl} \} w^\alpha _j w^\beta _l +\sum _{j,l}\{ B^{\alpha -1}_{ij} , w^\beta _l \} w^\alpha _j B^{\beta -1}_{kl} \\&+\sum _{j,l}\{ w^\alpha _j, B^{\beta -1}_{kl} \} B^{\alpha -1}_{ij} w^\beta _l +\sum _{j,l}\{ w^\alpha _j, w^\beta _l \} B^{\alpha -1}_{ij} B^{\beta -1}_{kl} \end{aligned} \end{aligned}$$
(B.10)
where we assume \(\alpha \le \beta \) without loss of generality. We can then use Lemmae B.2 and B.3 to show that
$$\begin{aligned} \{ v^\alpha _i,v^\alpha _k \}=-{\mathrm{i}}\, \text {sgn}(k-i) v_{k}^\alpha v_i^\alpha \,; \quad \{ v^\alpha _i,v^\beta _k \} =-{\mathrm{i}}\, \text {sgn}(k-i) v_k^\alpha v_i^\beta +{\mathrm{i}}v_k^\alpha v_i^\beta \,,\,\, \alpha < \beta \,.\nonumber \\ \end{aligned}$$
(B.11)
By anti-symmetry, (B.11) implies that (5.1) holds.
The Poisson bracket (5.2) is computed in the same way and requires to remark in the case \(\alpha =\beta \) that
$$\begin{aligned} \sum _{s} B^{\gamma }_{is} \overline{B}^{\gamma }_{ks} =\sum _{\mu =1}^\gamma v^\mu _i \bar{v}^\mu _k + \delta _{ik}\,. \end{aligned}$$
(B.12)
This identity is equivalent to \(B^\gamma (B^\gamma )^\dagger =\sum _{\mu =1}^\gamma v^\mu (v^\mu )^\dagger + {{\varvec{1}}}_n\), which is obtained by induction on \(\gamma \) using (2.33); it becomes (3.7) when \(\gamma =d\).
Proof of Theorem 5.8
Recall that we work over the gauge slice \(\check{{\mathcal {M}}}_{0,+}^\text {reg}\) (5.23) and wish to compute the reduced Poisson brackets \(\{\ ,\ \}_\text {red}\) of the basic evaluation functions \(Q_j=e^{{\mathrm{i}}q_j} \in {\mathrm{U}}(1)\) and \(v(\alpha )_j\in {\mathbb {C}}\), where the latter obey the relations (5.32). Our fundamental tool will be the identity (5.26), which concerns \({\mathrm{U}}(n)\) invariant functions on \({{\mathcal {M}}}\) and their pull-backs on \(\check{{\mathcal {M}}}_{0,+}^\text {reg}\). Knowing the left-hand side of (5.26), we will be able to determine the reduced Poisson brackets. In the particular case at hand, we consider the invariant functions \(f_m,f_m^{\alpha \beta }\in C^\infty ({{\mathcal {M}}})\) defined by (5.27). Their Poisson brackets on \({{\mathcal {M}}}\) are given by Lemma 5.7, and their restrictions (pull-backs) to \(\check{{\mathcal {M}}}_{0,+}^\text {reg}\) are displayed in (5.33). The point is that the right-hand side of (5.26) can be also expressed through the reduced Poisson brackets of the basic variables on \(\check{{\mathcal {M}}}_{0,+}^\text {reg}\), which enables us to derive the explicit formulae of Theorem 5.8.
We begin by giving an auxiliary lemma, which will be used below.
Lemma C.1
The \(n\times n\) matrices \({{\mathcal {E}}},\tilde{{{\mathcal {E}}}}\) given by
$$\begin{aligned} {{\mathcal {E}}}_{kl}=Q_l^k \quad \text {and} \quad \tilde{{{\mathcal {E}}}}_{kl}=Q_l^k {{\mathcal {U}}}_l \end{aligned}$$
(C.1)
are invertible on \(\check{{\mathcal {M}}}_{0,+}^\mathrm{reg}\).
Proof
We can write that \({{\mathcal {E}}}=V Q\) with \(Q=\text {diag}(Q_1,\ldots ,Q_n)\) and \(V=(V_{kl})\), \(V_{kl}=Q_k^{l-1}\), which is a Vandermonde matrix. Since \(Q\in {\mathbb {T}}^n_{\text {reg}}\) on \(\check{{\mathcal {M}}}_{0,+}^\text {reg}\), both V and Q are invertible. We also have that \(\tilde{{{\mathcal {E}}}}={{\mathcal {E}}}D\) where \(D=\text {diag}({{\mathcal {U}}}_1,\ldots ,{{\mathcal {U}}}_n)\). As \({{\mathcal {U}}}_j>0\) on \(\check{{\mathcal {M}}}_{0,+}^\text {reg}\), \(\tilde{{{\mathcal {E}}}}\) is also invertible. \(\square \)
Deriving (5.37)
Lemma C.2
For any \(i,j=1,\ldots ,n\), \(\{ q_i,q_j \}_\mathrm{red}=0\).
Proof
From (5.28) and (5.33), we get for any \(M,N \in {\mathbb {N}}\),
$$\begin{aligned} 0=\xi ^*\{ f_M,f_N \} = \{ \xi ^*f_M,\xi ^*f_N \}_{\text {red}}=-MN\,\sum _{i,j=1}^n e^{{\mathrm{i}}M q_i}e^{{\mathrm{i}}N q_j} \{ q_i,q_j \}_{\text {red}}\,. \end{aligned}$$
Considering this equality for \(M,N=1,\ldots ,n\), this is equivalent to
$$\begin{aligned} {{\mathcal {E}}}\, \hat{U}^{(0)} \, {{\mathcal {E}}}^T=0_{n \times n}\,, \end{aligned}$$
where \(\hat{U}^{(0)} \in {\text {Mat}}_{n \times n}({\mathbb {C}})\) is given by \(\hat{U}^{(0)}_{kl}=\{ q_k,q_l \}_{\text {red}}\). By Lemma C.1, \({{\mathcal {E}}}\) is invertible on \(\check{{\mathcal {M}}}_{0,+}^{\text {reg}}\) so that \(\hat{U}^{(0)}\) is the zero matrix. \(\square \)
Lemma C.3
For any \(i,j=1,\ldots ,n\),
$$\begin{aligned} \{ {{\mathcal {U}}}_i,q_j \}_\mathrm{red}=-\delta _{ij}{{\mathcal {U}}}_i\,, \quad \{ v(\alpha )_i,q_j \}_\mathrm{red}=-\delta _{ij} v(\alpha )_i\,, \quad \{ \overline{v}(\alpha )_i,q_j \}_\mathrm{red}=-\delta _{ij} \overline{v}(\alpha )_i\,.\nonumber \\ \end{aligned}$$
(C.2)
Proof
From (5.29), after summing over all \(\alpha ,\beta \) we get for any \(M,N \in {\mathbb {N}}\)
$$\begin{aligned}&\sum _{i,j}\{ {{\mathcal {U}}}_i^2 e^{{\mathrm{i}}M q_i}, e^{{\mathrm{i}}N q_j} \}_{\text {red}} = \sum _{\alpha ,\beta }\{ \xi ^*f_M^{\alpha \beta },\xi ^*f_N \}_{\text {red}} = -2{\mathrm{i}}N \sum _{\alpha ,\beta } \xi ^*f_{M+N}^{\alpha \beta } \\&\quad = -2{\mathrm{i}}N \sum _i {{\mathcal {U}}}_i^2 e^{{\mathrm{i}}(M+N) q_i}\,. \end{aligned}$$
Using Lemma C.2, we obtain
$$\begin{aligned} \sum _{i,j} e^{{\mathrm{i}}M q_i}{{\mathcal {U}}}_i e^{{\mathrm{i}}N q_j}\{ {{\mathcal {U}}}_i , q_j \}_{\text {red}} = -\sum _i {{\mathcal {U}}}_i^2 e^{{\mathrm{i}}(M+N) q_i}\,. \end{aligned}$$
We can rewrite this for \(N,M=1,\ldots ,n\) as
$$\begin{aligned} \tilde{{{\mathcal {E}}}} \, \hat{U}^{(1)} \, {{\mathcal {E}}}^T=\tilde{{{\mathcal {E}}}} \, {U}^{(1)} \, {{\mathcal {E}}}^T\,, \end{aligned}$$
where the \(n\times n\) matrices are given by \(\hat{U}^{(1)}_{kl}=\{ {{\mathcal {U}}}_k,q_l \}_{\text {red}}\), \({U}^{(1)}_{kl}=-\delta _{kl}{{\mathcal {U}}}_k\). By Lemma C.1, both \({{\mathcal {E}}}\) and \(\tilde{{{\mathcal {E}}}}\) are invertible. Hence, \(\hat{U}^{(1)}={U}^{(1)}\).
For the second identity, we use (5.29) with summation over all \(\beta \), and we get for any \(M,N \in {\mathbb {N}}\)
$$\begin{aligned} \sum _{i,j}\{ {{\mathcal {U}}}_i v(\alpha )_i e^{{\mathrm{i}}M q_i}, e^{{\mathrm{i}}N q_j} \}_{\text {red}} = -2{\mathrm{i}}N \sum _i v(\alpha )_i{{\mathcal {U}}}_i e^{{\mathrm{i}}(M+N) q_i}\,. \end{aligned}$$
Now that the first identity is proved, we can use it to get
$$\begin{aligned} \sum _{i,j}e^{{\mathrm{i}}M q_i}{{\mathcal {U}}}_i e^{{\mathrm{i}}N q_j} \{ v(\alpha )_i , q_j \}_{\text {red}} = -\sum _i v(\alpha )_i {{\mathcal {U}}}_i e^{{\mathrm{i}}(M+N) q_i}\,. \end{aligned}$$
As before, we write this for \(N,M=1,\ldots ,n\) as
$$\begin{aligned} \tilde{{{\mathcal {E}}}} \, \hat{U}^{(2)} \, {{\mathcal {E}}}^T=\tilde{{{\mathcal {E}}}} \, {U}^{(2)} \, {{\mathcal {E}}}^T\,, \end{aligned}$$
where the \(n\times n\) matrices are given by \(\hat{U}^{(2)}_{kl}=\{ v(\alpha )_k,q_l \}_{\text {red}}\), \({U}^{(2)}_{kl}=-\delta _{kl}v(\alpha )_k\). Again by invertibility of \({{\mathcal {E}}}\) and \(\tilde{{{\mathcal {E}}}}\), we get \(\hat{U}^{(2)}={U}^{(2)}\).
The last identity follows from the second one by complex conjugation. \(\square \)
From now on, we do not provide complete proofs of the different results that are stated. They can be successively obtained by direct computations in the same way as we got Lemmas C.2 and C.3 .
Deriving (5.38)
We first need two preliminary lemmae.
Lemma C.4
For any \(i,j=1,\ldots ,n\),
$$\begin{aligned} \begin{aligned} \{ {{\mathcal {U}}}_i,{{\mathcal {U}}}_j \}_\mathrm{red}=&\,\,\frac{1}{2} {\mathrm{i}}\delta _{(i\ne j)} \frac{Q_i+Q_j}{Q_i-Q_j}{{\mathcal {U}}}_i{{\mathcal {U}}}_j +\frac{1}{4} {\mathrm{i}}\sum _{\mu ,\nu } \mathrm{sgn}(\nu -\mu ) \left[ v(\nu )_i v(\mu )_j - \overline{v}(\nu )_i \overline{v}(\mu )_j \right] \\&+\frac{1}{4} {\mathrm{i}}\sum _{\mu }\left[ v(\mu )_i \overline{v}(\mu )_j - v(\mu )_j \overline{v}(\mu )_i \right] + \frac{d}{2} {\mathrm{i}}(L_{ij}-L_{ji}) \\&+\frac{1}{2} {\mathrm{i}}\sum _{\nu }\sum _{\mu <\nu }\left[ v(\mu )_i \overline{v}(\mu )_j - v(\mu )_j \overline{v}(\mu )_i \right] \,. \end{aligned} \end{aligned}$$
(C.3)
Proof
It suffices to use (5.30) where we sum over all \(\alpha ,\beta ,\gamma ,\epsilon \). After elementary manipulations, we arrive at
$$\begin{aligned} \begin{aligned} \sum _{i,j} Q_i^M{{\mathcal {U}}}_i Q_j^N {{\mathcal {U}}}_j \{ {{\mathcal {U}}}_i,{{\mathcal {U}}}_j \}_{\text {red}} =&\sum _{i,j}Q_i^M{{\mathcal {U}}}_i Q_j^N {{\mathcal {U}}}_j \, {U}^{(3)}_{ij} \,, \end{aligned} \end{aligned}$$
(C.4)
where \({U}^{(3)}_{ij}\) is the right-hand side of (C.3). We can then write the equalities with \(N,M=1,\ldots ,n\) as
$$\begin{aligned} \tilde{{{\mathcal {E}}}} \, \hat{U}^{(3)} \, \tilde{{{\mathcal {E}}}}^T=\tilde{{{\mathcal {E}}}} \, {U}^{(3)} \, \tilde{{{\mathcal {E}}}}^T\,, \end{aligned}$$
(C.5)
where the \(n\times n\) matrix \(\hat{U}^{(3)}\) is given by \(\hat{U}^{(3)}_{kl}=\{ {{\mathcal {U}}}_k,{{\mathcal {U}}}_l \}_{\text {red}}\). By invertibility of \({\tilde{{{\mathcal {E}}}}}\), this proves the claim (C.3). \(\square \)
Lemma C.5
For any \(i,j=1,\ldots ,n\),
$$\begin{aligned} \begin{aligned} \{ v(\alpha )_i,{{\mathcal {U}}}_j \}_\mathrm{red}=&\,\,\frac{1}{2} {\mathrm{i}}\delta _{(i\ne j)} \frac{Q_i+Q_j}{Q_i-Q_j}v(\alpha )_j {{\mathcal {U}}}_i +\frac{1}{2} {\mathrm{i}}\sum _\kappa \mathrm{sgn}(\kappa -\alpha ) v(\alpha )_j v(\kappa )_i \\&-\frac{1}{4} {\mathrm{i}}\frac{v(\alpha )_i}{{{\mathcal {U}}}_i} \sum _{\mu ,\nu } \mathrm{sgn}(\nu -\mu ) \left[ v(\nu )_i v(\mu )_j + \overline{v}(\nu )_i \overline{v}(\mu )_j \right] \\&+\frac{1}{2} {\mathrm{i}}v(\alpha )_i \overline{v}(\alpha )_{j} -\frac{1}{4} {\mathrm{i}}\frac{v(\alpha )_i}{{{\mathcal {U}}}_i}\sum _{\mu }\left[ v(\mu )_i \overline{v}(\mu )_j + v(\mu )_j \overline{v}(\mu )_i \right] \\&+ {\mathrm{i}}\sum _{\kappa< \alpha } v(\kappa )_i \overline{v}(\kappa )_j -\frac{1}{2} {\mathrm{i}}\frac{v(\alpha )_i}{{{\mathcal {U}}}_i}\sum _{\nu }\sum _{\mu <\nu } \left[ v(\mu )_i \overline{v}(\mu )_j + v(\mu )_j \overline{v}(\mu )_i \right] \\&+\frac{1}{2} {\mathrm{i}}\left[ 2L_{ij} - d \frac{v(\alpha )_i}{{{\mathcal {U}}}_i} (L_{ij}+L_{ji})\right] \,. \end{aligned} \end{aligned}$$
(C.6)
Proof
It suffices to use (5.30) after summing over \(\beta ,\gamma ,\epsilon \). We arrive at
$$\begin{aligned} \begin{aligned} \sum _{i,j} Q_i^M {{\mathcal {U}}}_i Q_j^N {{\mathcal {U}}}_j \{ v(\alpha )_i , {{\mathcal {U}}}_j \}_{\text {red}} =\sum _{i,j}Q_i^M{{\mathcal {U}}}_i Q_j^N {{\mathcal {U}}}_j \, {U}^{(4)}_{ij}\,, \end{aligned} \end{aligned}$$
(C.7)
where \({U}^{(4)}_{ij}\) is the right-hand side of (C.6). We can then write the equalities with \(N,M=1,\ldots ,n\) as
$$\begin{aligned} \tilde{{{\mathcal {E}}}} \, \hat{U}^{(4)} \, \tilde{{{\mathcal {E}}}}^T= \tilde{{{\mathcal {E}}}} \, {U}^{(4)} \, \tilde{{{\mathcal {E}}}}^T\,, \end{aligned}$$
where the \(n\times n\) matrix \(\hat{U}^{(4)}\) is given by \(\hat{U}^{(4)}_{kl}=\{ v(\alpha )_k,{{\mathcal {U}}}_l \}_{\text {red}}\). By invertibility of \({\tilde{{{\mathcal {E}}}}}\), we obtain the equality (C.5). \(\square \)
Summing over \(\beta ,\epsilon \) in (5.30) and using the previous results, we can get
$$\begin{aligned} \sum _{i,j} Q_i^M {{\mathcal {U}}}_i Q_j^N {{\mathcal {U}}}_j \{ v(\alpha )_i , v(\gamma )_j \}_{\text {red}} =\sum _{i,j}Q_i^M{{\mathcal {U}}}_i Q_j^N {{\mathcal {U}}}_j \,{U}^{(5)}_{ij}\,, \end{aligned}$$
(C.8)
where \({U}^{(5)}_{ij}\) is the right-hand side of (5.38). We can then write the equalities (C.8) with \(N,M=1,\ldots ,n\) as
$$\begin{aligned} \tilde{{{\mathcal {E}}}} \, \hat{U}^{(5)} \, \tilde{{{\mathcal {E}}}}^T=\tilde{{{\mathcal {E}}}} \, {U}^{(5)} \, \tilde{{{\mathcal {E}}}}^T\,, \end{aligned}$$
where the \(n\times n\) matrix \(\hat{U}^{(5)}\) is given by \(\hat{U}^{(5)}_{kl}=\{ v(\alpha )_k,v(\gamma )_l \}_{\text {red}}\). By invertibility of \({\tilde{{{\mathcal {E}}}}}\), this implies that (5.38) holds.
Deriving (5.39)
By anti-symmetry and complex conjugation, we get \(\{ {{\mathcal {U}}}_i,\overline{v}(\epsilon )_j \}_{\text {red}}\) from Lemma C.5. We can then use the previous results as well as (5.30) after summing over \(\beta ,\gamma \) in order to get
$$\begin{aligned} \sum _{i,j} Q_i^M {{\mathcal {U}}}_i Q_j^N {{\mathcal {U}}}_j \{ v(\alpha )_i , \overline{v}(\epsilon )_j \}_{\text {red}} =\sum _{i,j}Q_i^M{{\mathcal {U}}}_i Q_j^N {{\mathcal {U}}}_j {U}^{(6)}_{ij}\,, \end{aligned}$$
(C.9)
where \({U}^{(6)}_{ij}\) is the right-hand side of (5.39). We can then write the equalities (C.9) with \(N,M=1,\ldots ,n\) as
$$\begin{aligned} \tilde{{{\mathcal {E}}}} \, \hat{U}^{(6)} \, \tilde{{{\mathcal {E}}}}^T=\tilde{{{\mathcal {E}}}} \, {U}^{(6)} \, \tilde{{{\mathcal {E}}}}^T\,, \end{aligned}$$
where the \(n\times n\) matrix \(\hat{U}^{(6)}\) is given by \(\hat{U}^{(6)}_{kl}=\{ v(\alpha )_k,\overline{v}(\epsilon )_l \}_{\text {red}}\). By invertibility of \(\tilde{{{\mathcal {E}}}}\), we can conclude that (5.39) holds.
Poisson Brackets of Collective Spins
Recall the matrix \((S_{ij})\) defined before Theorem 5.8. The reduced Poisson brackets of the so-called collective spins F (3.29) can be computed in the following form.
Lemma D.1
Denoting \(q_{ab}:=q_a-q_b\), the following identity holds on \(\check{{\mathcal {M}}}_{0,+}^\mathrm{reg}\)
$$\begin{aligned} \begin{aligned} \{&F_{ij},F_{kl} \}_\mathrm{red} = \,\,{\mathrm{i}}\left( \frac{S_{ik}}{{{\mathcal {U}}}_i {{\mathcal {U}}}_k}-\frac{S_{lj}}{{{\mathcal {U}}}_l {{\mathcal {U}}}_j} + \frac{S_{kj}}{{{\mathcal {U}}}_k {{\mathcal {U}}}_j} - \frac{S_{il}}{{{\mathcal {U}}}_i {{\mathcal {U}}}_l} \right) F_{ij} F_{kl} \\&+\frac{1}{2} \left[ \delta _{(i \ne k)} \cot (\frac{q_{ik}}{2}) + \delta _{(j\ne l)} \cot \left( \frac{q_{jl}}{2}\right) + \delta _{(k \ne j)} \cot \left( \frac{q_{kj}}{2}\right) + \delta _{(i \ne l)} \cot \left( \frac{q_{li}}{2}\right) \right] \, F_{ij}F_{kl}\\&+\left[ \delta _{(i \ne k)} \cot \left( \frac{q_{ik}}{2}\right) + \delta _{(j \ne l)} \cot (\frac{q_{jl}}{2}) - \cot \left( \frac{q_{jk}}{2}-{\mathrm{i}}\gamma \right) + \cot \left( \frac{q_{li}}{2} - {\mathrm{i}}\gamma \right) \right] \, F_{il}F_{kj} \\&+\frac{1}{2} \left[ \delta _{(k \ne i)} \cot \left( \frac{q_{ki}}{2}\right) - \cot \left( \frac{q_{li}}{2}-{\mathrm{i}}\gamma \right) \right] \frac{{{\mathcal {U}}}_k}{{{\mathcal {U}}}_i} F_{ij}F_{il}\\&+\frac{1}{2} \left[ \delta _{(j \ne k)} \cot \left( \frac{q_{jk}}{2} \right) + \cot \left( \frac{q_{lj}}{2}-{\mathrm{i}}\gamma \right) \right] \frac{{{\mathcal {U}}}_k}{{{\mathcal {U}}}_j} F_{ij} F_{jl} \\&+\frac{1}{2} \left[ \delta _{(i\ne k)}\cot \left( \frac{q_{ki}}{2}\right) + \cot \left( \frac{q_{jk}}{2}-{\mathrm{i}}\gamma \right) \right] \frac{{{\mathcal {U}}}_i}{{{\mathcal {U}}}_k} F_{kj} F_{kl}\\&+\frac{1}{2} \left[ \delta _{(i \ne l)} \cot \left( \frac{q_{il}}{2} \right) - \cot \left( \frac{q_{jl}}{2} - {\mathrm{i}}\gamma \right) \right] \frac{{{\mathcal {U}}}_i}{{{\mathcal {U}}}_l} F_{lj}F_{kl} \\&+\frac{1}{2} \left[ \delta _{(i \ne l)} \cot \left( \frac{q_{il}}{2} \right) - \cot \left( \frac{q_{ik}}{2} - {\mathrm{i}}\gamma \right) \right] \frac{{{\mathcal {U}}}_l}{{{\mathcal {U}}}_i} F_{ij} F_{ki}\\&+\frac{1}{2} \left[ \delta _{(l \ne j)} \cot \left( \frac{q_{lj}}{2} \right) + \cot \left( \frac{q_{jk}}{2} - {\mathrm{i}}\gamma \right) \right] \frac{{{\mathcal {U}}}_l}{{{\mathcal {U}}}_j} F_{ij} F_{kj} \\&+\frac{1}{2} \left[ \delta _{(j \ne k)} \cot ( \frac{q_{jk}}{2} ) + \cot \left( \frac{q_{ki}}{2} - {\mathrm{i}}\gamma \right) \right] \frac{{{\mathcal {U}}}_j}{{{\mathcal {U}}}_k} F_{ik} F_{kl}\\&+\frac{1}{2} \left[ \delta _{(j \ne l)} \cot \left( \frac{q_{lj}}{2} \right) - \cot \left( \frac{q_{li}}{2} - {\mathrm{i}}\gamma \right) \right] \frac{{{\mathcal {U}}}_j}{{{\mathcal {U}}}_l} F_{il} F_{kl} \end{aligned} \end{aligned}$$
This follows from Theorem 5.8 by direct calculation. The reader can easily check the reality condition \(\{ F_{ji},F_{lk} \}_{\text {red}}=\{ \overline{F}_{ij},\overline{F}_{kl} \}_{\text {red}}=\overline{ \{ F_{ij},F_{kl} \} }_{\text {red}} \). Taking \(i=j\) and \(k=l\) in Lemma D.1, everything cancels out except for the third line, which can be rewritten as follows:
$$\begin{aligned} \{ F_{jj},F_{kk} \}_{\text {red}} = F_{jk} F_{kj} \frac{2\cot (\frac{q_{jk}}{2})}{1+\sinh ^{-2}(\gamma )\,\sin ^2(\frac{q_{jk}}{2})}\,, \quad \text {for }j\ne k\,. \end{aligned}$$
(D.1)
Let us now assume that \(d=1\), so that \(F_{jk} F_{kj} = F_{jj} F_{kk}\). Note that the formula of L (3.29) shows that \(F_{jj} >0\). Motivated by the form of the equations of motion (1.8) and the spinless Hamiltonian (1.11), we make the change of variables
$$\begin{aligned} F_{jj} = e^{2\theta _j}\prod _{i\ne j} \left[ 1+ \frac{\sinh ^2\gamma }{1 + \sin ^2\frac{q_i - q_j}{2}}\right] ^{\frac{1}{2}}. \end{aligned}$$
(D.2)
Using (5.37) and (D.1), it turns out that \((q_j, \theta _j)\) are Darboux variables, and we recover the standard chiral RS Hamiltonian (1.11) for \({{\mathcal {H}}}= \sum _j F_{jj}\).