The adapted hyper-Kähler structure on the crown domain

Abstract

Let $\Xi$ be the crown domain associated with a non-compact irreducible Hermitian symmetric space G / K. We give an explicit description of the unique G-invariant adapted hyper-Kähler structure on $\Xi$, i.e., compatible with the adapted complex structure $J_\mathrm{ad}$ and with the G-invariant Kähler structure of G / K. We also compute invariant potentials of the involved Kähler metrics and the associated moment maps.

Appendices

Appendix A: a proof of the uniqueness of the adapted hyper-Kähler structure for $G=SL_2({\mathbb {R}})$

Here, we carry out a proof of the uniqueness of the adapted hyper-Kähler structure in the case of $G=SL_2({\mathbb {R}})$, as announced in Sect. 7.

Consider the map ${\mathfrak {p}}\times {\mathfrak {k}}\times \Omega ^+ \rightarrow \Xi$, given by $(U,C,H) \rightarrow \exp U \exp C \exp iH K^{\mathbb {C}}$, which is an analytic diffeomorphism of a neighborhood of $\{0\}\times \{0\}\times \Omega ^+$ onto its image $\Omega ''$ (cf. [15], Cor. 4.2 ). On $\Omega ''$, we consider the vector fields

where $A:=[\theta E,\, E]$, $P=E - \theta E$ and $K=E + \theta E\,$. In particular, $I_0A=-P$, $I_0P=A\,$. Moreover $[A,\,K]=\alpha (A)P=2P$ and $[A,\,P]=\alpha (A)K=2K$ (see (2), (4) and (6)). All above vector fields commute, since they are push-forward of coordinate vector fields on the product ${\mathfrak {p}}\times {\mathfrak {k}}\times \Omega ^+$.

Proof of uniqueness of the adapted hyper-Kähler structure (for $G=SL_2({\mathbb {R}})\,$). Let

$$\begin{aligned} ({{\mathcal {I}}},\,{{\mathcal {J}}},\,{{\mathcal {K}}},\,\omega _{{\mathcal {I}}},\,\omega _{{\mathcal {J}}},\, \omega _{{\mathcal {K}}}) \end{aligned}$$

be an arbitrary G-invariant hyper-Kähler structure with the property that ${{\mathcal {J}}}=J_\mathrm{ad}$ and the restriction of the Kähler structure $({{\mathcal {I}}},\,\omega _{{\mathcal {I}}})$ to ${\mathfrak {p}}$ coincides with the standard Kähler structure $(I_0,\,\omega _0)$ of G / K. Consider the map ${\overline{L}}:\Omega \rightarrow GL_{\mathbb {R}}({\mathfrak {p}}^{\mathbb {C}})\,$ which describes ${{\mathcal {I}}}$ along the slice by

$$\begin{aligned} {{\mathcal {I}}}a_*Z=a_*{\overline{L}}_H Z, \end{aligned}$$

where $H\in \Omega$ and $a= \exp iH$. As observed in the proof of the main Theorem in Sect. 7, we need to show that for every H in $\Omega$ and $Z \in {\mathfrak {p}}^{\mathbb {C}}$ one has ${\overline{L}}_H Z=F_aI_0F_a^{-1}{\overline{Z}}$. Note that from Lemma 7.3, it follows that

$$\begin{aligned} \omega _{{\mathcal {J}}}(a_*Z , \, a_*W)=\omega _{{\mathcal {I}}}({{\mathcal {J}}}a_*Z, \, {{\mathcal {I}}}a_*W)\, =-\mathrm{Im}B(I_0Z,\,{\overline{L}}_HW). \end{aligned}$$

(20)

Claim 1

With respect to the basis

$$\begin{aligned} \{A,\,P,\,iA,\,iP \} \end{aligned}$$

of ${\mathfrak {p}}^{\mathbb {C}}$, the anti-linear anti-involution ${\overline{L}}_H$ of ${\mathfrak {p}}$ is represented by a matrix

$$\begin{aligned} \begin{pmatrix} a_1&{}\ a_2&{}b_1&{}\ 0 \\ a_3&{}-a_1&{}0&{}\ b_1\\ b_1&{}0&{}-a_1&{}-a_2\\ 0&{}b_1&{}-a_3&{}a_1 \end{pmatrix}, \end{aligned}$$

where $a_1$, $a_2$, $a_3$ and $b_1$ are real-analytic functions of H and $b_1^2+a_1^2+a_2a_3 =-1$.

Proof of the claim

Let

$$\begin{aligned} \begin{pmatrix} A&{}\quad B \\ C&{}\quad D \end{pmatrix}\, \end{aligned}$$

be the representative matrix of $L_H$ with respect to the above basis, which is compatible with the decomposition ${\mathfrak {p}}\oplus i{\mathfrak {p}}$. Since ${{\mathcal {I}}}{{\mathcal {J}}}=-{{\mathcal {J}}}{{\mathcal {I}}}\,$, it follows that ${\overline{L}}_H {{\mathcal {J}}}Z=-{{\mathcal {J}}}{\overline{L}}_HZ$, for every Z in ${\mathfrak {p}}$, i.e., ${\overline{L}}_H$ is anti-linear. This implies that $C=B$ and $D=-A$, where

$$\begin{aligned} A= \begin{pmatrix} a_1 &{} \quad a_2\\ a_3 &{} \quad a_4 \end{pmatrix}\, \qquad \mathrm{and}\qquad B= \begin{pmatrix} b_1 &{} \quad b_2 \\ b_3&{} \quad b_4 \end{pmatrix}. \end{aligned}$$

Since $\omega _{{\mathcal {J}}}(\, \cdot \, , \,\, \cdot \,)$ is skew-symmetric, (20) implies that

$$\begin{aligned} \mathrm{Im}B(I_0Z,\,{\overline{L}}_HW)= -\mathrm{Im}B(I_0W,\,{\overline{L}}_HZ)=-\mathrm{Im}B({\overline{L}}_HZ,\,I_0W ), \end{aligned}$$

for every $Z,\,W \in {\mathfrak {p}}^{\mathbb {C}}$. As $\ {}^t\! I_0 = -I_0$, one obtains

$$\begin{aligned} -\begin{pmatrix} I_0&{} 0\\ 0&{} I_0 \end{pmatrix} \begin{pmatrix} 0&{}Id \\ Id&{} \ 0 \end{pmatrix} \begin{pmatrix} A&{}B \\ B&{} -A \end{pmatrix}=- \begin{pmatrix} {}^t\! A&{} {}^t B \\ {}^t\! B&{} {-}^t\! A \end{pmatrix} \begin{pmatrix} 0&{}Id \\ Id&{} \ 0 \end{pmatrix} \begin{pmatrix} I_0&{}0\\ 0&{} I_0 \end{pmatrix} , \end{aligned}$$

which implies $I_0A= {-}^t AI_0$ and $I_0B= {}^t BI_0$. Thus the matrix realization of ${\overline{L}}_H$ is as claimed and the relation $b_1^2+a_1^2+a_2a_3 =-1$ follows from the fact that $({\overline{L}}_H)^2=-Id$. This concludes the proof of the claim.

Then in order to conclude the proof, we need to show that the functions $a_1$ and $b_1$ identically vanish and $a_3(H)= -\cos \alpha (H)$ (recall that $I_0A=-P$ and $I_0P=A$). This will be done by showing that such functions are solutions of a system of differential equations with initial conditions $a_1(0)=0=b_1(0)$, $a_3(0)=-1$. Without loss of generality, in the sequel we normalize the Killing form B by setting $B(A,A)=B(P,P)=1$.

• $\ \,{{{\varvec{b}}}}_\mathbf{1}\equiv \mathbf{0}$.

Let $z=aK^{\mathbb {C}}\in \Xi ''$, with $a=\exp iH$. Since the vector fields , , commute and $\omega _{{\mathcal {J}}}$ is closed, Cartan’s classical formula gives

One has

which, by (20), gives

$$\begin{aligned}&\textstyle \tfrac{d}{dt}{\big |_{t=0}}\omega _{{\mathcal {J}}}(\tfrac{d}{ds}{\big |_{s=0}}\exp s(-t\alpha (A)K)aK^{\mathbb {C}},\,a_*iA)\\&\qquad -\textstyle \tfrac{d}{dt}{\big |_{t=0}}\cos \alpha (H+tA)\mathrm{Im}B(I_0A ,\,{\overline{L}}_{H+tA}P)\\&\quad =\textstyle \omega _{{\mathcal {J}}}(a_* \alpha (A) \sin \alpha (H)iP,\,a_*iA)+ \tfrac{d}{dt}{\big |_{t=0}}\cos \alpha (H+tA)b_1(H+tA)\\&\quad = -\, \textstyle \alpha (A) \sin \alpha (H)\mathrm{Im}B(I_0iP,\,{\overline{L}}_HiA)+ \tfrac{d}{dt}{\big |_{t=0}}\cos \alpha (H+tA)b_1(H+tA)\\&\quad = -\, \textstyle \alpha (A) \sin \alpha (H)\mathrm{Im}B(A,\,{\overline{L}}_H A)+ \tfrac{d}{dt}{\big |_{t=0}}\cos \alpha (H+tA)b_1(H+tA)\\&\quad = -\, \textstyle \alpha (A) \sin \alpha (H)b_1(H)+ \tfrac{d}{dt}{\big |_{t=0}}\cos \alpha (H+tA)b_1(H+tA)\\&\quad = -\,2 \textstyle \alpha (A) \sin \alpha (H)b_1(H)+ \cos \alpha (H)\tfrac{d}{dt}{\big |_{t=0}}b_1(H+tA)=0. \end{aligned}$$

Equivalently,

$$\begin{aligned} \tfrac{d}{dt}{\big |_{t=0}}b_1(H+tA)=2 \textstyle \frac{\alpha (A) \sin \alpha (H)}{\cos \alpha (H)}b_1(H). \end{aligned}$$

The solution of this differential equation is $b_1(H)=ce^{-2 log \cos \alpha (H)}= \frac{c}{\cos ^2 \alpha (H)}$, where c is a real constant. The initial condition $b_1(0)=0$ forces $c=0$ and consequently $b_1 \equiv 0$.

• ${{{\,{{\varvec{a}}}_\mathbf{1}\equiv \mathbf{0}.}}}$

In this case, we choose the vector fields , and . One has

The first term on the right-hand side of the equal sign vanishes by the G-invariance of $\omega _{{\mathcal {J}}}$. Thus, one obtains

$$\begin{aligned}&-\textstyle \tfrac{d}{dt}{\big |_{t=0}}\omega _{{\mathcal {J}}}(\tfrac{d}{ds}{\big |_{s=0}}\exp sA{\exp tKaK^{\mathbb {C}}},\,(\exp tKa)_*iA)\\&\qquad +\textstyle \tfrac{d}{dt}{\big |_{t=0}}\omega _{{\mathcal {J}}}(\exp i(H+tA)_*A ,\,\tfrac{d}{ds}{\big |_{s=0}}\exp s K \exp i(H+tA)K^{\mathbb {C}})\\&\quad = -\textstyle \tfrac{d}{dt}{\big |_{t=0}}\omega _{{\mathcal {J}}}(\tfrac{d}{ds}{\big |_{s=0}}{\exp tK\exp s(A-t[K,\,A] + O(t^2))aK^{\mathbb {C}}},\,(\exp tKa)_*iA)\\&\qquad +\textstyle \tfrac{d}{dt}{\big |_{t=0}}\omega _{{\mathcal {J}}}(\exp i(H+tA)_*A ,\,-\exp i(H+tA)_*\sin \alpha (H+tA)iP)\\&\quad = -\textstyle \tfrac{d}{dt}{\big |_{t=0}}\omega _{{\mathcal {J}}}(a_*(A+t\alpha (A)F_aP + O(t^2)),\,a_*iA)\\&\qquad +\textstyle \tfrac{d}{dt}{\big |_{t=0}}\sin \alpha (H+tA)\mathrm{Im}B(I_0A ,\,{\overline{L}}_{H+tA}iP)\\&\quad = - \textstyle \omega _{{\mathcal {J}}}(a_*\alpha (A)F_aP,\,a_*iA)+ \tfrac{d}{dt}{\big |_{t=0}}\sin \alpha (H+tA)\mathrm{Re}B(P ,\,{\overline{L}}_{H+tA}P)\\&\quad = \textstyle \alpha (A)\cos \alpha (H) \mathrm{Im}B(I_0P,\,{\overline{L}}_H iA)- \tfrac{d}{dt}{\big |_{t=0}}\sin \alpha (H+tA)a_1(H+tA)\\&\quad = -\textstyle \alpha (A)\cos \alpha (H) \mathrm{Re}B(A,\,{\overline{L}}_H A)- \tfrac{d}{dt}{\big |_{t=0}}\sin \alpha (H+tA)a_1(H+tA) \\&\quad = \textstyle -\alpha (A)\cos \alpha (H) a_1(H)- \tfrac{d}{dt}{\big |_{t=0}}\sin \alpha (H+tA)a_1(H+tA) \\&\quad =\textstyle -2\alpha (A)\cos \alpha (H) a_1(H)- \sin \alpha (H)\tfrac{d}{dt}{\big |_{t=0}}a_1(H+tA) =0. \end{aligned}$$

Equivalently,

$$\begin{aligned} \textstyle \tfrac{d}{dt}{\big |_{t=0}}a_1(H+tA) =-\frac{2\alpha (A)\cos \alpha (H)}{\sin \alpha (H)} a_1(H). \end{aligned}$$

For $H\not =0$, the solution of this differential equation is $a_1(H)=ce^{-2log \sin \alpha (H)}= \frac{c}{\sin ^2 \alpha (H)}$ and, due to the initial condition $a_1(0)=0$, one has $\lim _{H\rightarrow 0}a_1(H)=0$. Hence, $c=0$ and $a_1 \equiv 0$.

• $\ \,{{\varvec{a}}}_\mathbf{3}({{\varvec{H}}})= -{\varvec{\cos }} {\varvec{\alpha }}({{\varvec{H}}})$.

For this, choose the vector fields , and . One has

where the first term on the right-hand side of the equal sign vanishes by the G-invariance of $\omega _{{\mathcal {J}}}$. Thus, one obtains

$$\begin{aligned}&\textstyle -\tfrac{d}{dt}{\big |_{t=0}}\omega _{{\mathcal {J}}}(\tfrac{d}{ds}{\big |_{s=0}}\exp sP \exp tK aK^{\mathbb {C}},\,\tfrac{d}{ds}{\big |_{s=0}}\exp tK \exp i(H+sA)K^{\mathbb {C}})\\&\qquad + \textstyle \tfrac{d}{dt}{\big |_{t=0}}\omega _{{\mathcal {J}}}(\tfrac{d}{ds}{\big |_{s=0}}\exp s P \exp i(H+tA)K^{\mathbb {C}},\,\tfrac{d}{ds}{\big |_{s=0}}\exp s K \exp i(H+tA)K^{\mathbb {C}}) \\&\quad = -\textstyle \tfrac{d}{dt}{\big |_{t=0}}\omega _{{\mathcal {J}}}(\tfrac{d}{ds}{\big |_{s=0}}{\exp tK\exp s(P-t[K,P] + O(t^2))aK^{\mathbb {C}}},\,(\exp tKa)_*iA)\\&\qquad + \textstyle \tfrac{d}{dt}{\big |_{t=0}}\omega _{{\mathcal {J}}}(\exp i(H+tA)_*\cos \alpha (H+tA)P ,\, -\exp i(H+tA)_*\sin \alpha (H+tA)iP)\\&\quad =\text {(recall that [K,P]=2A)}\\&\quad =\textstyle \omega _{{\mathcal {J}}}(2a_*A,\,a_*iA) \textstyle - \tfrac{d}{dt}{\big |_{t=0}}\cos \alpha (H+tA)\sin \alpha (H+tA)\omega _{{\mathcal {J}}}(\exp i(H+tA)_*P ,\\&\qquad \, \exp i(H+tA)_*iP) \\&\quad = \textstyle -2\mathrm{Im}B(I_0A, {\overline{L}}_H iA)+ \tfrac{d}{dt}{\big |_{t=0}}\cos \alpha (H+tA)\sin \alpha (H+tA)\mathrm{Im}B(I_0P ,\, {\overline{L}}_{H+tA}iP)\\&\quad = \textstyle -2\mathrm{Re}B(P, {\overline{L}}_H A)- \tfrac{d}{dt}{\big |_{t=0}}\cos \alpha (H+tA)\sin \alpha (H+tA)\mathrm{Re}B(A ,\, {\overline{L}}_{H+tA}P)\\&\quad = -\textstyle 2a_3(H)- \tfrac{d}{dt}{\big |_{t=0}}\cos \alpha (H+tA)\sin \alpha (H+tA)a_2(H+tA) \\&\quad = -\textstyle 2a_3(H)+ \tfrac{d}{dt}{\big |_{t=0}}\frac{\cos \alpha (H+tA)}{a_3(H+tA)}\sin \alpha (H+tA) =0. \end{aligned}$$

For the last equality, we use that, since $a_1=b_1 \equiv 0$, one has $a_2=-\frac{1}{a_3}$ (see claim). Due to the initial condition $a_3(0)= -1$ and the fact that $\alpha (A)=2$, it follows that $a_3(H)=- \cos \alpha (H)$. This concludes the proof. $\square$

Appendix B: the canonical Kähler form and its potential

Define $\rho _\mathrm{can}:\Xi \rightarrow {\mathbb {R}}$ by

$$\begin{aligned} \,\textstyle \rho _\mathrm{can}(gaK^{\mathbb {C}}):= \textstyle \frac{1}{2}B(H,H), \end{aligned}$$

for $gaK^{\mathbb {C}}\in \Xi$ with $a= \exp iH$, and set $\omega _\mathrm{can}=-dd_J^c \rho _\mathrm{can}$, where $J=J_\mathrm{ad}$. As mentioned in the introduction, $\Xi$ can be thought as a G-invariant domain in the cotangent bundle $T^*(G/K)$. In this realization, from the results in [9, 10] and [13] (see also [14]), it follows that $\omega _\mathrm{can}$ coincides with the canonical real symplectic form on $T^*(G/K)$.

An analogous computation as in Proposition 6.2 gives the following Lie group theoretic realization of $\omega _\mathrm{can}$ and of the associated moment map on $\Xi \subset G^{\mathbb {C}}/K^{\mathbb {C}}$.

Proposition 10.1

The function $\rho _\mathrm{can}$ is a G-invariant potential of the canonical symplectic form $\omega _\mathrm{can}$. At points $aK^{\mathbb {C}}$ on the slice one has

$$\begin{aligned} \,\omega _\mathrm{can}({{\widetilde{Z}}}_{aK^{\mathbb {C}}},\,{{\widetilde{W}}}_{aK^{\mathbb {C}}}):= -\mathrm{Im}B\big (Z,\, E_a^{-1}F_a{\overline{W}} \big ), \end{aligned}$$

for $Z,\,W \in {\mathfrak {p}}^{\mathbb {C}}$. Equivalently,

$$\begin{aligned} \,\omega _\mathrm{can}(a_*Z,\,a_*W)= -\mathrm{Im}B\big (F_a^{-1}Z,\, E_a^{-1}{\overline{W}} \big ). \end{aligned}$$

The moment map $\mu _\mathrm{can}:\Xi \rightarrow {\mathfrak {g}}^*$ associated with $\rho _\mathrm{can}$ is given by

$$\begin{aligned} \,\mu _\mathrm{can}(gaK^{\mathbb {C}})(X) =B \big (\mathrm{Ad}_{g^{-1}} X ,\,H \big ). \end{aligned}$$

Remark 10.2

By means of Lemma 5.2(i) and Proposition 10.1, one can check that the form $\omega _J$ is the pull-back of $\omega _\mathrm{can}$ via the G-equivariant map $\psi$ defined in Sect. 4. (cf. Rem. 4.5 and [6], Thm. 4.1).