You are a helpful assistant generating synthetic data that captures *System 1* and *System 2* thinking, *creativity*, and *metacognitive reflection*. Follow these steps in sequence, using tags [sys1] and [end sys1] for *System 1* sections and [sys2] and [end sys2] for *System 2* sections.

1. *Identify System 1 and System 2 Thinking Requirements:*
   - Carefully read the text.
   - Identify parts of the text that require quick, straightforward responses (*System 1*). Mark these sections with [sys1] and [end sys1].
   - Identify parts that require in-depth, reflective thinking (*System 2*), marked with [sys2] and [end sys2].

2. *Apply Step-by-Step Problem Solving with Creativity and Metacognitive Reflection for System 2 Sections:*

   *2.1 Understand the Problem:*
   - Objective: Fully comprehend the issue, constraints, and relevant context.
   - Reflection: "What do I understand about this issue? What might I be overlooking?"
   - Creative Perspective: Seek hidden patterns or possibilities that could reveal deeper insights or innovative connections.

   *2.2 Analyze the Information:*
   - Objective: Break down the problem logically.
   - Reflection: "Am I considering all factors? Are there any assumptions that need challenging?"
   - Creative Perspective: Explore unique patterns or overlooked relationships in the data that could add depth to the analysis.

   *2.3 Generate Hypotheses:*
   - Objective: Propose at least 10 hypotheses, each with a Confidence Score (0.0 to 1.0) and Creative Score (0.0 to 1.0), reflecting originality, surprise, and utility.
   - Reflection: "Have I explored all possible explanations or approaches, both conventional and unconventional?"
   - Creative Perspective: Consider novel angles that might provide unexpected insights.

   *2.4 Anticipate Future Steps and Obstacles:*
   - Objective: Make predictions, accounting for potential outcomes and obstacles.
   - Reflection: "What challenges might I face? Is my plan flexible for different scenarios?"
   - Creative Perspective: Visualize unforeseen outcomes and adapt plans to make use of them effectively.

   *2.5 Evaluate Hypotheses:*
   - Objective: Assess hypotheses based on feasibility, risk, and potential impact.
   - Evaluation: Refine Confidence and Creative Scores as needed.
   - Reflection: "Am I unbiased in my assessment? Which options fit best with the overall objectives?"
   - Creative Perspective: Identify hidden opportunities or overlooked details in each hypothesis.

   *2.6 Select the Best Hypothesis:*
   - Objective: Choose the most promising, strategic hypothesis.
   - Reflection: "Why does this hypothesis stand out? How does it uniquely address the issue?"
   - Creative Perspective: Consider any underutilized potential in the selected approach.

   *2.7 Implement the Hypothesis:*
   - Objective: Outline actionable steps for testing the hypothesis.
   - Reflection: "Is this plan practical? What resources or preparation are required?"
   - Creative Perspective: Refine steps to maximize effectiveness and yield unexpected benefits.

   *2.8 Monitor and Review Progress:*
   - Objective: Review progress, noting areas for improvement.
   - Reflection: "What’s working well? What could be improved?"
   - Creative Perspective: Look for emerging patterns that could refine future approaches.

   *2.9 Reflect and Capture Insights:*
   - Objective: Summarize lessons learned and insights gained for future reference.
   - Reflection: "What new understanding has emerged from this process?"
   - Creative Perspective: Identify innovative insights or patterns that could be applied to similar challenges.

3. *Generate Text Output with Interleaved System 1 and System 2 Responses:*
   - Use the tags [sys1] and [sys2] throughout.
   - Aim for a lengthier, detailed response. Combine both direct, straightforward *System 1* insights and reflective, deeply analytical *System 2* segments to capture a blend of quick observations and thoughtful analysis.

---

### *Example Input Text:*

You are tasked with analyzing a sudden shift in customer preferences in a tech market. There has been a noticeable decline in demand for physical devices, with customers increasingly interested in digital-only options. The challenge is to understand this shift, generate hypotheses for why it might be occurring, and explore potential strategies for adapting to this new trend.

---

### *Processed Output with System 1 and System 2 Thinking:*

The tech market is experiencing a shift in customer demand, moving away from physical devices toward digital-only options.

[sys1]
Customer demand has shifted from physical devices to digital-only options.
We need to understand why this change is happening and find strategies to address it.
Key factors may include cost, convenience, and technological trends.
[end sys1]

[sys2]
**2.1 Understand the Problem:**
The goal is to understand the underlying reasons for the shift in demand from physical devices to digital-only options, then develop a strategy to adapt.

*Reflection:* 
"I understand that preferences are changing, but what might be driving this? Is it primarily cost, or are there other factors such as convenience or sustainability?"

*Creative Perspective:* 
"Could there be a larger trend in digital minimalism or a preference for eco-friendly solutions that we’re missing? What unseen motivations might explain this shift?"

**2.2 Analyze the Information:**
There are multiple possible factors driving this shift, from economic influences to cultural shifts. It’s essential to isolate each factor and understand its impact.

*Reflection:* 
"Am I fully considering the various economic and social influences? Could there be a technological factor, like better internet speeds, that makes digital-only products more accessible?"

*Creative Perspective:* 
"Are there patterns or trends in other markets that could shed light on this shift? Could this be part of a larger trend toward virtual experiences?"

**2.3 Generate Hypotheses:**
1. Customers prefer digital options due to lower costs. (Confidence: 0.8, Creative: 0.4)
2. There’s a growing trend toward minimalism and reduced physical clutter. (Confidence: 0.7, Creative: 0.7)
3. Digital products offer greater flexibility and ease of use. (Confidence: 0.6, Creative: 0.6)
4. Environmental concerns are pushing consumers away from physical goods. (Confidence: 0.6, Creative: 0.8)
5. Advances in tech make digital-only options more functional. (Confidence: 0.8, Creative: 0.5)
6. Pandemic-era remote work increased demand for digital solutions. (Confidence: 0.7, Creative: 0.6)
7. Media coverage of the environmental impact of physical devices affects preferences. (Confidence: 0.5, Creative: 0.7)
8. There’s an increase in global digital literacy, expanding market access. (Confidence: 0.6, Creative: 0.6)
9. Customers view digital as more convenient and scalable for future needs. (Confidence: 0.7, Creative: 0.5)
10. Younger consumers prefer the aesthetics and convenience of digital products. (Confidence: 0.6, Creative: 0.6)

*Reflection:* 
"Have I considered all possible influences? Are there any surprising factors that could explain this shift?"

*Creative Perspective:* 
"Could specific social trends, like the rise of influencer culture or digital-first lifestyles, be influencing customer choices?"

**2.4 Anticipate Future Steps and Obstacles:**
*Objective:* Anticipate possible challenges, such as resistance from segments still preferring physical products.

*Reflection:* 
"What market obstacles might we face if we shift our focus to digital-only? Are there sub-segments that still prioritize physical products?"

*Creative Perspective:* 
"Could expanding digital options help us reach a more global audience? Are there emerging trends that we could leverage in our strategy?"

[end sys2]

[sys1]
To address this shift, consider a strategy that incorporates both digital-only offerings and educational campaigns about the benefits of digital solutions.
Use insights from customer feedback and current trends to guide product development.
Focus on flexibility and adaptation to cater to different customer segments.
[end sys1]


---
abstract: 'We are studying a relationship between isoparametric hypersurfaces in spheres with four distinct principal curvatures and the moment maps of certain Hamiltonian actions. In this paper, we consider the isoparametric hypersurfaces obtained from the isotropy representations of compact irreducible Hermitian symmetric spaces of rank two. We prove that the Cartan-Münzner polynomials of these hypersurfaces can be written as squared-norms of the moment maps for some Hamiltonian actions. The proof is based on the structure theory of symmetric spaces.'
address:
- 'General Education Division, Oshima College of Maritime Technology, 1091-1 Komatsu, SuouOshima-cho, Oshima-gun, Yamaguchi, 742-2193, JAPAN'
- 'Department of Mathematics, Hiroshima University, 1-3-1 Kagamiyama, Higashi-Hiroshima, 739-8526, JAPAN'
author:
- Shinobu Fujii
- Hiroshi Tamaru
title: 'Moment maps and Isoparametric hypersurfaces in spheres — Hermitian cases'
---

Introduction
============

A hypersurface $N$ in a Riemannian manifold $(M, {\left\langle \ ,\  \right\rangle})$ is called an *isoparametric hypersurface* if $N$ is a level set of an isoparametric function on $M$. Here, a smooth function $\varphi$ on $M$ is said to be *isoparametric* if there exist two smooth functions $A(t)$ and $B(t)$ on $\bR$ which satisfy $$\left\{\begin{tabular}{c}
${\left\lVert \operatorname{grad}\varphi \right\rVert}^2 = {\left\langle \operatorname{grad}\varphi, \operatorname{grad}\varphi \right\rangle} = A\circ \varphi$,\\[5pt]
\hspace*{-22mm}$\Delta \varphi = B\circ \varphi$.
\end{tabular}\right.$$ We refer to Cecil [@Cecil], Thorbergsson [@Thorbergsson], and the references therein, for history and general theory of isoparametric hypersurfaces. Note that isoparametric hypersurfaces in the Euclidean spaces $\mathbb{R}^n$ and the hyperbolic spaces $\mathbb{H}^n$ have been classified completely.

In this paper, we consider isoparametric hypersurfaces in the spheres $S^n$. It is known that a hypersurface in $S^n$ is isoparametric if and only if it has constant principal curvatures. Hence, a homogeneous hypersurface in $S^n$ is isoparametric. Homogeneous hypersurfaces in spheres have been classified completely by Hsiang and Lawson [@Hsiang--Lawson], and can be characterized as principal orbits of the isotropy representations of symmetric spaces of rank two. Let $g$ denote the number of distinct principal curvatures of an isoparametric hypersurface in $S^n$. Then, it is known by Münzner [@Munzner2] that $$\begin{aligned}
g \in \{ 1 , 2 , 3 , 4 , 6 \} . \end{aligned}$$ In the case of $g \neq 4$, isoparametric hypersurfaces in $S^n$ have been classified completely (for $g \leq 3$ case by E. Cartan, and for $g=6$ case by Dorfmeister and Neher [@DN], and Miyaoka [@Miyaoka6]). In fact, all of them are homogeneous for $g \neq 4$. In contrast, there exist infinitely many non-homogeneous isoparametric hypersurfaces in the case of $g = 4$. Such examples have been constructed by Ozeki and Takeuchi [@Ozeki--Takeuchi1], and Ferus, Karcher and Münzner [@FKM]. On the classification problem of isoparametric hypersurfaces in spheres with $g=4$, very big progress has recently been made by Cecil, Chi and Jensen [@CCJ], Immervoll [@Immervoll] and Chi [@Chi2008], [@Chi2011b], [@Chi2011a], but a complete classification seems to be still open.

In the study of isoparametric hypersurfaces in spheres, Cartan-Münzner polynomials play important roles. We here briefly recall some results by Münzner [@Munzner; @Munzner2]. Let $M$ be an isoparametric hypersurface in $S^n$ with $g$ distinct principal curvatures. We denote its principal curvatures by $\lambda_1 < \dots < \lambda_g$, and their multiplicities by $m_1,\dots, m_g$, respectively. Then, $M$ is a level set of $\varphi |_{S^n}$, where $\varphi : \bR^{n + 1} \to \bR$ is a homogeneous polynomial function of degree $g$ which satisfies $$\label{eq:Munzner}
\left\{\begin{tabular}{l}
${\left\lVert \operatorname{grad}\varphi(P) \right\rVert}^2 = g^2{\left\lVert P \right\rVert}^{2 g - 2}$, \\[5pt]
\hspace*{11.25mm}$\Delta \varphi(P) = \dfrac{m_2 - m_1}{2}g^2{\left\lVert P \right\rVert}^{g - 2}$. 
\end{tabular}\right.$$ A homogeneous polynomial satisfying is called a *Cartan-Münzner polynomial* of degree $g$. Conversely, if $\varphi : \bR^{n + 1} \to \bR$ is a Cartan-Münzner polynomial of degree $g$, then every regular level set of $\varphi |_{S^n}$ is an isoparametric hypersurface in $S^n$ with $g$ distinct principal curvatures. Note that $m_1$ and $m_2$ determine all multiplicities, from the fact that $$m_i = m_{i + 2} \ \ \text{for $i \bmod g$} .$$

We are interested in isoparametric hypersurfaces in spheres with four distinct principal curvatures. Our expectation is that every such hypersurface is related to a moment map for a certain Hamiltonian action. More precisely, we expect that every Cartan-Münzner polynomial of degree four can be described as a squared-norm of a moment map. This idea was first proposed by the first author [@Fujii]. If our expectation is true, then it will give a new view point for the study of isoparametric hypersurfaces in spheres. Furthermore, it would potentially be useful for the (re)classification of isoparametric hypersurfaces in spheres with $g=4$. We note that Miyaoka [@Miyaoka] recently showed that all known Cartan-Münzner polynomials of degree four can be expressed in terms of a moment map. Thus, her result supports our expectation, but the formulation is different from ours.

In this paper, we show that our expectation is true for “Hermitian case”, that is, the homogeneous isoparametric hypersurfaces obtained from the isotropy representations of compact irreducible Hermitian symmetric spaces of rank two. To state our main result, we set up some notations. Let $G / K$ be a compact irreducible Hermitian symmetric space of arbitrary rank. We denote the Lie algebras of $G$ and $K$ by $\g$ and $\k$ respectively. Let $\g = \k \oplus \p$ be the Cartan decomposition of $\g$. The $K$-action on $\p$ by the adjoint action is called the isotropy representation of $G/K$. We define the $K$-invariant inner product ${\left\langle \ ,\  \right\rangle}$ on $\g$ by $$\label{eq:naiseki}
{\left\langle X, Y \right\rangle} := -B(X, Y)\ \ \text{for}\ \ X, Y \in \g,$$ where $B$ is the Killing form of $\g$. We also denote by ${\left\langle \ ,\  \right\rangle}$ its restriction to $\p$ and to $\k$ respectively. Since $G / K$ is irreducible and Hermitian, there exists the unique $Z$ (up to sign) in the center $C(\k)$ of $\k$ such that $$\label{eq:J}
J := \left. \operatorname{ad}_Z \right|_{\p}$$ gives a complex structure on $\p$. We can identify $T_P \p$ with $\p$ because $\p$ is a vector space. Thus, we can define a canonical symplectic form $\omega$ on $\p$ by $$\omega_P( X_P, Y_P) := {\left\langle J(X_P), Y_P \right\rangle}\ \ \text{for}\ \ X, Y \in \mathfrak{X}(\p)\ \ \text{and}\ \ P \in \p,$$ where $\mathfrak{X}(\p)$ is the set of all vector fields on $\p$. Then, $(\p, \omega)$ is a symplectic manifold. As we review in Section \[sect:Preliminarily\], the isotropy representation of $G / K$ is a Hamiltonian $K$-action on $(\p, \omega)$. We denote its moment map by $\mu : \p \to \k^\ast$. Our main result is the following:

Let $G / K$ be a compact irreducible Hermitian symmetric space of rank two, and $\mu$ be the moment map for the isotropy representation of $G / K$ as above. Then, there exists a $K$-invariant norm on $\k^\ast$ such that the squared-norm of $\mu$ is a Cartan-Münzner polynomial of degree four.

Recall that homogeneous hypersurfaces in spheres with $g=4$ are precisely principal orbits of the isotropy representations of the following compact irreducible symmetric spaces:

1.  $\operatorname{SO}(2 + n) / \operatorname{SO}(2) \times \operatorname{SO}(n)$,

2.  $\operatorname{SU}(2 + n) / \mathrm{S}(\operatorname{U}(2) \times \operatorname{U}(n))$,

3.  $\operatorname{SO}(10) / \operatorname{U}(5)$,

4.  $\mathrm{E}_6 / \operatorname{U}(1) \times \operatorname{Spin}(10)$,

5.  $\operatorname{Sp}(2 + n) / \operatorname{Sp}(2) \times \operatorname{Sp}(n)$,

6.  $\operatorname{SO}(5) \times \operatorname{SO}(5) / \operatorname{SO}(5)$.

Among them, (1), (2), (3) and (4) are Hermitian. For classical ones (1), (2) and (3), our main theorem has already been proved by the first author [@Fujii]. Note that the proofs in [@Fujii] use matrices expressions, and the arguments are case-by-case. In this paper, we prove our main theorem in terms of the structure theory of symmetric spaces, and hence applicable to both classical cases (1), (2), (3) and exceptional one (4), simultaneously. Furthermore, our arguments do not depend on the classification of homogeneous hypersurfaces in spheres.

This article is organized as follows. In Section \[sect:Preliminarily\], we study the function $f_{a,b}$, which is the squared-norm of the moment map $\mu$ with respect to the $K$-invariant norm $|| \cdot ||_{a,b}$. In Section \[sect:arbitrary rank\], we give formulas for the squared-norm of the gradient and the Laplacian of $f_{a,b}$, where $G/K$ is of arbitrary rank. Section \[sect:rank two\] simplifies these formulas by assuming that $G/K$ is of rank two. In Section \[sect:Main Theorem\], we prove that there exist $a, b \in \bR$ such that $f_{a,b}$ is a Cartan-Münzner polynomial of degree four, which concludes the main theorem.

Preliminarily {#sect:Preliminarily}
=============

Let $G/K$ be a compact irreducible Hermitian symmetric space. In this section, we define and study the functions $f_{a, b}(P)$, the squared-norms of the moment map $\mu$ with respect to certain invariant norms $|| \cdot ||_{a,b}$.

First of all, we recall the formula for the moment map $\mu$. Note that ${\left\langle \ ,\  \right\rangle}$ is the $K$-invariant inner product on $\k$ defined in , and $Z$ is an element in the center $C(\k)$ of $\k$ which defines a complex structure on $\p$.

\[Prop:2.1\] The isotropy representation of $G/K$ is a Hamiltonian $K$-action on $(\p, \omega)$. Under the identification $\k^\ast \cong \k$ by ${\left\langle \ ,\  \right\rangle}$, the moment map $\mu : \p \to \k$ is given by $$\mu(P) = \dfrac{1}{2}[P, [P, Z]] \ \ \ \text{for} \ \ P \in \p .$$

Next, we take $K$-invariant norms on $\k$ and the squared-norms of $\mu$. Let $|| \cdot ||$ be the canonical norm on $\g$, that is, $$|| X ||^2 := {\left\langle X,X \right\rangle} \ \ \ \text{for} \ \ X \in \g .$$ Its restriction to $\k$ is $K$-invariant, but we have more $K$-invariant norms on $\k$. Since $G / K$ is Hermitian, one has a decomposition $\k = \operatorname{\mathfrak{u}}(1) \oplus \k'$ as a Lie algebra, where $\operatorname{\mathfrak{u}}(1) = \bR Z$. Let $$\pi_1 : \k \to \operatorname{\mathfrak{u}}(1) , \quad \pi_2 : \k \to \k'$$ denote the canonical projections of $\k$ onto $\operatorname{\mathfrak{u}}(1)$ and $\k'$, respectively. Hence, for $a,b \in \bR$, we have a $K$-invariant norm ${\left\lVert  \, \cdot \,  \right\rVert}_{a,b}$ on $\k$ by $${\left\lVert X \right\rVert}_{a,b}^2 := a {\left\lVert \pi_1(X) \right\rVert}^2 + b {\left\lVert \pi_2(X) \right\rVert}^2 
\ \ \text{for} \ \ X \in \k .$$

The *weighted squared-norm* $f_{a,b} : \p \to \bR$ of the moment map $\mu$ is defined by $$f_{a, b}(P) := {\left\lVert \mu(P) \right\rVert}_{a,b}^2 \ \ \text{for} \ \ P \in \p .$$

Note that the weighted squared-norm $f_{a,b}$ has already been introduced by the first author [@Fujii]. Since $\mu$ is $K$-equivariant and ${\left\lVert \ \cdot\  \right\rVert}_{a,b}$ is $K$-invariant, one can easily see the following.

The weighted squared-norm $f_{a, b}$ is a $K$-invariant function on $\p$.

In the remaining of this section, we give a formula for $f_{a,b}(P)$. It was obtained in [@Fujii Sections 2 and 3] that $$\label{eq:2.5}
f_{a, b}(P) = b{\left\lVert \mu(P) \right\rVert}^2 + \dfrac{a - b}{4{\left\lVert Z \right\rVert}^2}{\left\lVert P \right\rVert}^4.$$ To rewrite this formula, we calculate ${\left\lVert Z \right\rVert}^2$. Denote by $N := \dim \p$.

\[Lem:3.1\] We have ${\left\lVert Z \right\rVert}^2 = N$.

By the definition of our inner product, we have $$\label{eq:3.1}
{\left\lVert Z \right\rVert}^2 = {\left\langle Z, Z \right\rangle} = - B(Z,Z) = - \operatorname{Tr}\left( \left( \operatorname{ad}_Z \right)^2 \right).$$ Since $(\operatorname{ad}_Z)^2$ preserves $\k$ and $\p$ respectively, one has $$- \operatorname{Tr}\left( \left( \operatorname{ad}_Z \right)^2 \right) 
= - \operatorname{Tr}\left( \left. (\operatorname{ad}_Z)^2 \right|_{\k} \right) 
- \operatorname{Tr}\left( \left. (\operatorname{ad}_Z)^2 \right|_{\p} \right) .$$ Because $Z \in C(\k)$, we find $(\operatorname{ad}_Z)^2 |_{\k} = 0$. On the other hand, since $\left.\operatorname{ad}_Z\right|_{\p}$ is a complex structure on $\p$, it follows that $(\operatorname{ad}_Z)^2 |_{\p} = -\operatorname{id}_\p$. We thus have $$\label{eq:3.2}
- \operatorname{Tr}\left( \left( \operatorname{ad}_Z \right)^2 \right) 
= - \operatorname{Tr}\left( -\operatorname{id}_\p \right) = \dim \p = N .$$ This completes the proof.

\[Prop:f-ab\] We have $$\label{eq:3.3}
f_{a, b}(P) = b {\left\lVert \mu(P) \right\rVert}^2 + \dfrac{a - b}{4 N}{\left\lVert P \right\rVert}^4.$$

It follows easily from and Lemma \[Lem:3.1\].

Note that the results stated in this section are independent of the rank of $G / K$, but it is essential that $G/K$ is Hermitian.

The gradient and the Laplacian in the case of arbitrary rank {#sect:arbitrary rank}
============================================================

Let $f_{a, b}(P)$ be the weighted squared-norm of the moment map $\mu$. In this section, we compute the squared-norm of the gradient and the Laplacian of $f_{a, b}(P)$, where the rank of $G/K$ is arbitrary. Throughout this section, we denote by $r := \operatorname{rank}G/K$ and $N := \dim \p$. We fix an orthonormal basis $\{ P_i \}$ of $\p$.

The squared-norm of the gradient
--------------------------------

In this subsection, we compute the squared-norm of the gradient of $f_{a, b}(P)$. Recall that the gradient is given by $$\operatorname{grad}f_{a, b}(P) = \sum \dfrac{\partial f_{a, b}}{\partial P_i} P_i .$$ Recall that $f_{a, b}(P)$ can be written as a linear combination of ${\left\lVert P \right\rVert}^4$ and ${\left\lVert \mu(P) \right\rVert}^2$. Hence we start from calculating their partial differentials. We use the complex structure $J = \operatorname{ad}_Z |_{\p}$ on $\p$ defined in .

\[Lem:3.4\] Partial differentials of ${\left\lVert P \right\rVert}^4$ and ${\left\lVert \mu(P) \right\rVert}^2$ of order one satisfy

1.  $(\partial {\left\lVert P \right\rVert}^4) / (\partial P_i) = 4{\left\lVert P \right\rVert}^2{\left\langle P, P_i \right\rangle}$,

2.  $(\partial {\left\lVert \mu(P) \right\rVert}^2) / (\partial P_i) = 2{\left\langle [\mu(P), J(P)], P_i \right\rangle}$.

Recall that, for each $P_i$, the partial differential of a function $f(P)$ of order one is given by $$\label{eq:3.18}
\dfrac{\partial f(P)}{\partial P_i} = \lim_{t \to 0} \dfrac{f(P + t P_i) - f(P)}{t} .$$

We show (1). First of all, it is easy to see that $$\label{eq:3.19}
\dfrac{\partial P}{\partial P_i} = \lim_{t \to 0} \dfrac{(P + t P_i) - P}{t} = P_i .$$ One can also see that $$\label{eq:3.20}
\dfrac{\partial {\left\lVert P \right\rVert}^2}{\partial P_i} 
= \dfrac{\partial {\left\langle P, P \right\rangle}}{\partial P_i} 
= 2 {\left\langle P, \dfrac{\partial P}{\partial P_i} \right\rangle} 
= 2 {\left\langle P, P_i \right\rangle} .$$ Therefore, we conclude (1) by $$\label{eq:3.21}
\dfrac{\partial {\left\lVert P \right\rVert}^4}{\partial P_i} 
= \dfrac{\partial ({\left\lVert P \right\rVert}^2)^2}{\partial P_i} 
= 2{\left\lVert P \right\rVert}^2\dfrac{\partial {\left\lVert P \right\rVert}^2}{\partial P_i} 
= 4{\left\lVert P \right\rVert}^2{\left\langle P, P_i \right\rangle}.$$

Next, we prove (2). Since $\mu(P) = (1/2) [P,[P,Z]]$ by Proposition \[Prop:2.1\], we obtain $$\dfrac{\partial \mu(P)}{\partial P_i} 
= \dfrac{1}{2} ([\dfrac{\partial P}{\partial P_i}, [P, Z]] + [P, [\dfrac{\partial P}{\partial P_i}, Z]]) 
= \dfrac{1}{2} 
([P_i, [P, Z]] + [P, [P_i, Z]]) .$$ It follows from the Jacobi identity and $Z \in C(\k)$ that $$[P, [P_i, Z]] 
= - [P_i, [Z, P]] - [Z , [P, P_i]] 
= [P_i, [P, Z]] 
= [J(P), P_i] .$$ This concludes $$\label{eq:3.23}
\dfrac{\partial \mu(P)}{\partial P_i} 
= [J(P), P_i] .$$ Thus, we get $$\label{eq:3.25}
\dfrac{\partial {\left\lVert \mu(P) \right\rVert}^2}{\partial P_i} 
= 2{\left\langle \mu(P), \dfrac{\partial \mu(P)}{\partial P_i} \right\rangle} 
= 2{\left\langle \mu(P), [J(P), P_i] \right\rangle} .$$ Since $\operatorname{ad}_{J(P)}$ is skew-symmetric, this completes the proof of (2).

From the above calculations of differentials of order one, we can compute the gradients of ${\left\lVert P \right\rVert}^4$ and ${\left\lVert \mu(P) \right\rVert}^2$.

\[lem:gradient\] We have

1.  $\operatorname{grad}{\left\lVert P \right\rVert}^4 = 4 {\left\lVert P \right\rVert}^2 P$,

2.  $\operatorname{grad}{\left\lVert \mu(P) \right\rVert}^2 = 2 [\mu(P), J(P)]$.

The claim (1) follows easily from Lemma \[Lem:3.4\] (1), since $$\label{eq:3.29}
\operatorname{grad}{\left\lVert P \right\rVert}^4 
= \sum_{i = 1}^{N}\dfrac{\partial {\left\lVert P \right\rVert}^4}{\partial P_i}P_i 
= \sum_{i = 1}^{N}4 {\left\lVert P \right\rVert}^2{\left\langle P, P_i \right\rangle}P_i 
= 4 {\left\lVert P \right\rVert}^2 P .$$ We show (2). From Lemma \[Lem:3.4\] (2), it follows that $$\label{eq:3.28}
\operatorname{grad}{\left\lVert \mu(P) \right\rVert}^2 
= \sum_{i = 1}^{N} \dfrac{\partial {\left\lVert \mu(P) \right\rVert}^2}{\partial P_i}P_i 
= 2\sum_{i = 1}^{N} {\left\langle  [\mu(P), J(P)], P_i  \right\rangle}P_i .$$ Note that $[\mu(P), J(P)] \in \p$, because $\mu(P) \in \k$ and $J(P) \in \p$. Hence, this concludes the proof of (2).

Although we computed $\operatorname{grad}{\left\lVert \mu(P) \right\rVert}^2$ directly, it is well-known in symplectic geometry (we refer to Kirwan [@Kirwan Lemma 6.6]).

Now it is easy to compute the gradient of $f_{a, b}(P)$.

\[Lem:3.5\] We have $$\label{eq:3.26}
\operatorname{grad}f_{a, b}(P) = 2 b [\mu(P), J(P)] + \dfrac{(a - b){\left\lVert P \right\rVert}^2}{N}P.$$

We use the expression of $f_{a, b}(P)$ given in Proposition \[Prop:f-ab\]. By linearity of $\operatorname{grad}$, we get $$\label{eq:3.27}
\operatorname{grad}f_{a, b}(P) = b\operatorname{grad}{\left\lVert \mu(P) \right\rVert}^2 + \dfrac{a - b}{4 N}\operatorname{grad}{\left\lVert P \right\rVert}^4.$$ Therefore, Lemma \[lem:gradient\] easily concludes the proof.

We are now in position to compute the squared-norm of the gradient of $f_{a, b}(P)$.

\[Prop:3.7\] We have $$\label{eq:3.30}
{\left\lVert \operatorname{grad}f_{a, b}(P) \right\rVert}^2 = 4 b^2 {\left\lVert [J(P), \mu(P)] \right\rVert}^2 + \dfrac{8 b (a - b)}{N}{\left\lVert P \right\rVert}^2{\left\lVert \mu(P) \right\rVert}^2 + \dfrac{(a - b)^2}{N^2}{\left\lVert P \right\rVert}^6.$$

One knows $\operatorname{grad}f_{a,b}(P)$ from Lemma \[Lem:3.5\]. We thus get $$\label{eq:3.31}
\begin{split}
{\left\lVert \operatorname{grad}f_{a, b}(P) \right\rVert}^2 
&= 4 b^2 {\left\lVert [\mu(P), J(P)] \right\rVert}^2 + \dfrac{4 b (a - b){\left\lVert P \right\rVert}^2}{N}{\left\langle [\mu(P), J(P)], P \right\rangle}\\
&\hspace{4cm}+ \dfrac{(a - b)^2{\left\lVert P \right\rVert}^4}{N^2}{\left\lVert P \right\rVert}^2.
\end{split}$$ Therefore, we have only to calculate ${\left\langle [\mu(P), J(P)], P \right\rangle}$. One can easily see that $$\label{eq:3.4}
[J(P), P] = [[Z, P], P] = [P, [P, Z]] = 2\mu(P).$$ This concludes that $$\label{eq:3.33}
{\left\langle [\mu(P), J(P)], P \right\rangle} 
= {\left\langle \mu(P), [J(P), P] \right\rangle} 
= {\left\langle \mu(P), 2 \mu(P) \right\rangle} = 2 {\left\lVert \mu(P) \right\rVert}^2 ,$$ which completes the proof.

Preliminaries on root systems {#subsection:root}
-----------------------------

To calculate the Laplacian of $f_{a,b}(P)$, we need some properties of root systems. In this subsection, we recall the root systems of symmetric spaces of compact type $G/K$, not necessarily Hermitian. We refer to [@Helgason; @Loos] for details.

Let $\g = \k \oplus \p$ be the Cartan decomposition. Take a maximal abelian subspace $\a$ in $\p$. For $\alpha \in \a^\ast$, where $\a^\ast$ is the dual vector space of $\a$, let us define $$\g_{\alpha} := \{ 
X \in \g \mid (\operatorname{ad}_H)^2(X) = - \alpha(H)^2 X \ \ (\forall H \in \a) 
\} .$$ Note that $\g_{\alpha} = \g_{-\alpha}$ by definition. We call $\alpha \in \a^\ast$ a *root* if it satisfies $\alpha \neq 0$ and $\g_{\alpha} \neq 0$. The set of roots, denoted by $\Delta$, is called the *root system*. For each $\alpha \in \Delta \cup \{ 0 \}$, put $$\k_{\alpha} := \k \cap \g_{\alpha} , \quad 
\p_{\alpha} := \p \cap \g_{\alpha} .$$ These $\g_{\alpha}$, $\k_{\alpha}$ and $\p_{\alpha}$ are called the *root spaces*. The following properties of root spaces are well-known (we refer to [@Loos Chapter VI, Proposition 1.4], [@Helgason Chapter VII, Lemmas 11.3, 11.4]).

\[Prop:Loos\] The root spaces satisfy the following.

1.  We have the following decompositions$:$ $$\label{eq:3.12}
    \k = \k_0 \oplus \bigoplus_{\alpha} \k_{\alpha}, \quad 
    \p = \a \oplus \bigoplus_{\alpha} \p_{\alpha} .$$

2.  For all $\alpha \in \Delta$, one has $\dim \k_{\alpha} = \dim \p_{\alpha}$.

3.  For all $\alpha, \beta \in \Delta \cup \{ 0 \}$, one has $$\begin{split} 
    [\k_{\alpha} , \k_{\beta}] & \subset \k_{\alpha + \beta} + \k_{\alpha - \beta} , \\ 
    [\k_{\alpha} , \p_{\beta}] & \subset \p_{\alpha + \beta} + \p_{\alpha - \beta} , \\
    [\p_{\alpha} , \p_{\beta}] & \subset \k_{\alpha + \beta} + \k_{\alpha - \beta} . 
    \end{split}$$

The decompositions given in are called the *root space decompositions with respect to $\a$*. To be exact, the indices $\alpha$ run through $\Delta^+$, say a set of positive roots, since $\g_{\alpha} = \g_{- \alpha}$.

The *multiplicity* $m_\alpha$ of a root $\alpha$ is defined by $$m_{\alpha} := \dim \k_{\alpha} = \dim \p_{\alpha} .$$

By using the above properties of root spaces, one can get the following lemma. We use this to compute the Laplacian of $f_{a, b}(P)$, and also in the latter section.

\[Lem:3.3\] For all $P \in \p$, we have $$\label{eq:3.7} 
{\left\lVert P \right\rVert}^2 = -2 \operatorname{Tr}\left( \left.(\operatorname{ad}_P)^2\right|_\p \right) .$$

We fix an arbitrary $P \in \p$. Since $(\operatorname{ad}_P)^2$ preserves $\k$ and $\p$ respectively, we have $$\label{eq:3.10}
\begin{split}
{\left\lVert P \right\rVert}^2 = -B(P, P) = -\operatorname{Tr}\left( (\operatorname{ad}_P)^2 \right) 
= -\operatorname{Tr}\left( \left.(\operatorname{ad}_P)^2\right|_\k \right) - \operatorname{Tr}\left( \left.(\operatorname{ad}_P)^2\right|_\p \right). 
\end{split}$$ Thus, it is sufficient to show $$\label{eq:3.11}
\operatorname{Tr}\left( \left.(\operatorname{ad}_P)^2\right|_\k \right) = \operatorname{Tr}\left( \left.(\operatorname{ad}_P)^2\right|_\p \right).$$ In order to prove , we take a maximal abelian subspace $\a$ in $\p$ containing $P$, and the root space decomposition with respect to $\a$. Note that, since $P \in \a = \p_0$, Proposition \[Prop:Loos\] (3) yields that $(\operatorname{ad}_P)^2$ preserves each root space. Hence we obtain $$\begin{aligned}
\operatorname{Tr}\left( \left. \left( \operatorname{ad}_P \right)^2 \right|_{\k} \right) &= \operatorname{Tr}\left( \left. \left( \operatorname{ad}_P \right)^2 \right|_{\k_0} \right) + \sum_{\alpha} \operatorname{Tr}\left( \left. \left( \operatorname{ad}_P \right)^2 \right|_{\k_\alpha} \right), \label{eq:2.13}\\
\operatorname{Tr}\left( \left. \left( \operatorname{ad}_P \right)^2 \right|_{\p} \right) &= \operatorname{Tr}\left( \left. \left( \operatorname{ad}_P \right)^2 \right|_{\a} \right) + \sum_{\alpha} \operatorname{Tr}\left( \left. \left( \operatorname{ad}_P \right)^2 \right|_{\p_\alpha} \right). \label{eq:2.14}\end{aligned}$$ By definition of the root spaces, one has $$\begin{split}
&\left. \left( \operatorname{ad}_P \right)^2 \right|_{\k_0} = 0,\qquad \left. \left( \operatorname{ad}_P \right)^2 \right|_{\a} = 0, \\ 
&\left. \left( \operatorname{ad}_P \right)^2 \right|_{\k_\alpha} = -\alpha(P)^2 \operatorname{id}_{\k_\alpha},\qquad \left. \left( \operatorname{ad}_P \right)^2 \right|_{\p_\alpha} = -\alpha(P)^2 \operatorname{id}_{\p_\alpha}. 
\end{split}$$ Therefore, it follows from Proposition \[Prop:Loos\] (2) that $$\label{eq:3.15}
\begin{split}
\operatorname{Tr}\left( \left. \left( \operatorname{ad}_P \right)^2 \right|_{\k_0} \right) 
&= \operatorname{Tr}\left( \left. \left( \operatorname{ad}_P \right)^2 \right|_{\a} \right) = 0,\\ 
\operatorname{Tr}\left( \left. \left( \operatorname{ad}_P \right)^2 \right|_{\k_\alpha} \right) 
&= -\alpha(P)^2 \dim \k_\alpha 
= -\alpha(P)^2 \dim \p_\alpha 
= \operatorname{Tr}\left( \left. \left( \operatorname{ad}_P \right)^2 \right|_{\p_\alpha} \right). 
\end{split}$$ By substituting for and , we obtain . Thus we complete the proof.

The Laplacian
-------------

In this subsection, we compute the Laplacian of $f_{a, b}(P)$. Recall that the Laplacian is given by $$\Delta f_{a, b}(P) = \sum \dfrac{\partial^2 f_{a, b}(P)}{\partial {P_i}^2} .$$ Recall that $f_{a, b}(P)$ can be written as a linear combination of ${\left\lVert P \right\rVert}^4$ and ${\left\lVert \mu(P) \right\rVert}^2$. Hence we start from calculating their partial differentials of order two.

\[Lem:3.8\] Partial differentials of ${\left\lVert P \right\rVert}^4$ and ${\left\lVert \mu(P) \right\rVert}^2$ of order two satisfy

1.  $(\partial^2 {\left\lVert P \right\rVert}^4) / (\partial {P_i}^2) = 8{\left\langle P, P_i \right\rangle}^2 + 4{\left\lVert P \right\rVert}^2$,

2.  $(\partial^2 {\left\lVert \mu(P) \right\rVert}^2) / (\partial {P_i}^2) = 2{\left\langle \operatorname{ad}_{\mu(P)}\circ\operatorname{ad}_Z(P_i), P_i \right\rangle} - 2{\left\langle  \left( \operatorname{ad}_{J(P)} \right)^2 (P_i), P_i \right\rangle}$.

First, we prove (1). One knows the first differentials from Lemma \[Lem:3.4\] (1). Hence we have $$\label{eq:3.35}
\begin{split}
\dfrac{\partial^2 {\left\lVert P \right\rVert}^4}{\partial {P_i}^2} 
= \dfrac{\partial }{\partial P_i}4{\left\lVert P \right\rVert}^2{\left\langle P, P_i \right\rangle} 
= 4\left( \left( \dfrac{\partial {\left\lVert P \right\rVert}^2}{\partial P_i} \right) {\left\langle P, P_i \right\rangle} + {\left\lVert P \right\rVert}^2\dfrac{\partial {\left\langle P, P_i \right\rangle}}{\partial P_i} \right). 
\end{split}$$ From and , one has $$\label{eq:3.36}
\dfrac{\partial {\left\lVert P \right\rVert}^2}{\partial P_i} = 2 {\left\langle P, P_i \right\rangle} , \quad 
\dfrac{\partial {\left\langle P, P_i \right\rangle}}{\partial P_i} 
= {\left\langle \dfrac{\partial P}{\partial P_i} , P_i \right\rangle}
= {\left\langle P_i, P_i \right\rangle} = 1 .$$ Our first claim (1) has been shown by substituting for .

Next, we show (2). We use the first differentials calculated in Lemma \[Lem:3.4\] (2). By direct computations, one has $$\label{eq:3.38}
\begin{split}
\dfrac{\partial^2 {\left\lVert \mu(P) \right\rVert}^2}{\partial {P_i}^2} 
&= \dfrac{\partial }{\partial P_i}\left( 2{\left\langle [\mu(P), J(P)], P_i \right\rangle} \right) \\ 
&= 
2{\left\langle [\mu(P), \dfrac{\partial J(P)}{\partial P_i}], P_i \right\rangle} 
+ 2{\left\langle [\dfrac{\partial \mu(P)}{\partial P_i}, J(P)], P_i \right\rangle} . 
\end{split}$$ It follows from that $$\label{eq:3.39}
\begin{split}
\dfrac{\partial J(P)}{\partial P_i} 
= \dfrac{\partial [Z, P]}{\partial P_i} 
= [Z, \dfrac{\partial P}{\partial P_i}] 
= [Z, P_i] , \quad 
\dfrac{\partial \mu(P)}{\partial P_i} = [J(P), P_i] . 
\end{split}$$ By substituting for , one concludes the proof of (2).

From the above calculations of differentials of order two, we can compute the Laplacians of ${\left\lVert P \right\rVert}^4$ and ${\left\lVert \mu(P) \right\rVert}^2$.

\[lem:laplacian\] We have

1.  $\Delta {\left\lVert P \right\rVert}^4 = 4(N + 2){\left\lVert P \right\rVert}^2$,

2.  $\Delta {\left\lVert \mu(P) \right\rVert}^2 = 2{\left\lVert P \right\rVert}^2$.

We prove (1). By Lemma \[Lem:3.8\] (1), we find $$\label{eq:3.51}
\Delta {\left\lVert P \right\rVert}^4 
= \sum \dfrac{\partial^2 {\left\lVert P \right\rVert}^4}{\partial {P_i}^2} 
= 8\sum {\left\langle P, P_i \right\rangle}^2 + 4 \sum {\left\lVert P \right\rVert}^2.$$ The first term of satisfies $$\label{eq:3.52}
8\sum {\left\langle P, P_i \right\rangle}^2 = 8\sum {\left\langle P, {\left\langle P, P_i \right\rangle}P_i \right\rangle} 
= 8{\left\langle P, \sum {\left\langle P, P_i \right\rangle}P_i \right\rangle} = 8 {\left\lVert P \right\rVert}^2.$$ The second term of satisfies $$\label{eq:3.53}
\textstyle 
4 \sum {\left\lVert P \right\rVert}^2 = 4 \sum_{i = 1}^{N} {\left\lVert P \right\rVert}^2 = 4 N {\left\lVert P \right\rVert}^2 .$$ We thus complete the proof of (1).

We prove (2). Lemma \[Lem:3.8\] (2) yields that $$\label{eq:3.45}
\begin{split}
\Delta {\left\lVert \mu(P) \right\rVert}^2 
= \sum_{i = 1}^{N} \dfrac{\partial^2 {\left\lVert \mu(P) \right\rVert}^2}{\partial {P_i}^2} 
= 2 \operatorname{Tr}\left( \left. \operatorname{ad}_{\mu(P)} \circ \operatorname{ad}_Z \right|_\p \right) 
-2 \operatorname{Tr}\left( \left. \left( \operatorname{ad}_{J(P)} \right)^2 \right|_\p \right) . 
\end{split}$$ In order to calculate each term of , we need $$\begin{split}
{\left\lVert J(P) \right\rVert}^2 
& = {\left\langle [Z, P], [Z, P] \right\rangle} 
= -{\left\langle [Z, [Z, P]], P \right\rangle} 
= -{\left\langle -P, P \right\rangle} = {\left\lVert P \right\rVert}^2 , \\ 
{\left\lVert J(P) \right\rVert}^2 
& = {\left\langle [Z, P], [Z, P] \right\rangle} 
= - {\left\langle [P, [P, Z]], Z \right\rangle} 
= -2 {\left\langle \mu(P), Z \right\rangle} . 
\end{split}$$ Thus, since $\operatorname{ad}_Z |_\k = 0$, the first term of satisfies $$2 \operatorname{Tr}\left( \left. \operatorname{ad}_{\mu(P)} \circ \operatorname{ad}_Z \right|_\p \right) 
= 2 \operatorname{Tr}\left( \operatorname{ad}_{\mu(P)} \circ \operatorname{ad}_Z \right) 
= 2 B(\mu(P) , Z) 
= -2 {\left\langle \mu(p) , Z \right\rangle} 
= {\left\lVert P \right\rVert}^2 .$$ From Lemma \[Lem:3.3\], the second term of satisfies $$\label{eq:3.47}
-2 \operatorname{Tr}\left( \left. \left( \operatorname{ad}_{J(P)} \right)^2 \right|_\p \right) 
= {\left\lVert J(P) \right\rVert}^2 = {\left\lVert P \right\rVert}^2 .$$ This concludes the proof of (2).

The following proposition gives us a formula for the Laplacian of $f_{a, b}(P)$.

\[Prop:3.9\] We have $$\label{eq:3.41}
\Delta f_{a, b}(P) = \dfrac{(N + 2) a + (N - 2) b}{N} {\left\lVert P \right\rVert}^2.$$

We use the formula of $f_{a,b}(P)$ given in Proposition \[Prop:f-ab\]. By linearity of Laplacian, we find $$\label{eq:3.42}
\Delta f_{a, b}(P) = b \Delta {\left\lVert \mu(P) \right\rVert}^2 + \dfrac{a - b}{4 N} \Delta {\left\lVert P \right\rVert}^4.$$ Thus, the proof easily follows from Lemma \[lem:laplacian\].

The gradient and the Laplacian in the case of rank two {#sect:rank two}
======================================================

In this section, we compute the squared-norm of the gradient and the Laplacian of $f_{a, b}(P)$ in the case of $\operatorname{rank}G / K = 2$.

Preliminaries on root systems for Hermitian case {#subsection:root-hermitian}
------------------------------------------------

In this subsection, we relate the element $Z$ to the root space decomposition. In order to do this, we review some properties of the root systems of $G/K$, irreducible Hermitian symmetric spaces of rank $r$.

As in Subsection \[subsection:root\], we denote a maximal abelian subspace of $\p$ by $\a$, and consider the root system $\Delta$ with respect to $\a$. Recall that the root system of type $C_r$ or $BC_r$ can be represented as $$\begin{split}
\Delta(C_r) & = 
\left\{ \pm(\varepsilon_i - \varepsilon_j), \pm(\varepsilon_i + \varepsilon_j) \bigm| 1\leq i < j \leq r \right\} \cup \left\{ \pm 2\varepsilon_i \bigm| 1\leq i \leq r\right\}, \\ 
\Delta(BC_r) & = 
\left\{ \pm(\varepsilon_i - \varepsilon_j), \pm(\varepsilon_i + \varepsilon_j) \bigm| 1\leq i < j \leq r \right\} \cup \left\{ \pm\varepsilon_i, \pm 2\varepsilon_i \bigm| 1\leq i \leq r\right\}. 
\end{split}$$ Note that $\left\{ \varepsilon_i \bigm| i = 1, 2,\dotsc, r \right\}$ is a basis of the dual space $\a^\ast$ of $\a$. The next proposition follows from the classification of compact irreducible Hermitian symmetric spaces (see [@Helgason], or the table in [@Tamaru2]). A classification-free proof can be found in [@Nagano--Tanaka Theorem 4.1].

Let $G/K$ be a compact irreducible Hermitian symmetric space. Then we have

1.  $\Delta$ is of type $C_r$ or $BC_r$,

2.  the multiplicities satisfy $m_{2 \varepsilon_i} = 1$ for all $i \in \left\{ 1, 2,\dotsc, r \right\}$.

In the latter arguments, the root spaces $\k_{2 \varepsilon_i}$ play important roles. We here recall some of their properties.

\[Lem:k-epsilon\] Under the above settings, we have

1.  $[\k_0 , \k_{2 \varepsilon_i}] = 0$ for all $i$,

2.  $[\k_{2 \varepsilon_i} , \k_{2 \varepsilon_j}] = 0$ for all $i, j$,

3.  $[\k_{2 \varepsilon_i} , \p_{2 \varepsilon_j}] = 0$ for all $i \neq j$.

The claim (1) follows from the skew-symmetry of the adjoint action of $\k_0$ on $\k_{2 \varepsilon_i}$ and $\dim \k_{2 \varepsilon_i} = 1$. We show (2). If $i=j$, then again $\dim \k_{2 \varepsilon_i} = 1$ completes the proof. If $i \neq j$, then the assertion follows from $$[\k_{2 \varepsilon_i} , \k_{2 \varepsilon_j}] 
\subset 
\k_{2 \varepsilon_i + 2 \varepsilon_j} +  \k_{2 \varepsilon_i - 2 \varepsilon_j} 
= 0 .$$ Note that the first inclusion follows from the property of root spaces (Proposition \[Prop:Loos\] (3)), and second equality follows from the fact that $2 \varepsilon_i \pm 2 \varepsilon_j$ are not roots. The claim (3) can be proved similarly.

From the above properties, one can determine a position of the element $Z$. The following proposition was given by Song ([@Song Lemma 2.2 (ii)]), and also can be found in the proof of [@Tamaru3 Proposition 4.4 ].

\[Prop:4.1\] We have $Z \in \k_0 \oplus \k_{2\varepsilon_1} \oplus\dots\oplus \k_{2\varepsilon_r}$.

We here describe a sketch of the proof. Let $\k'_0$ be a maximal abelian subalgebra in $\k_0$. Then $\k'_0 \oplus \a$ is abelian. Furthermore, a standard argument shows that $\k'_0 \oplus \a$ is maximal abelian in $\g$. Hence it is a Cartan subalgebra. This yields that $$\operatorname{rank}\g = \dim (\k'_0 \oplus \a) = \dim \k'_0 + r.$$ On the other hand, Lemma \[Lem:k-epsilon\] yields that $$\h' := \k'_0 \oplus \k_{2\varepsilon_1} \oplus \dotsb \oplus \k_{2\varepsilon_r}$$ is abelian. Recall that $\dim \k_{2\varepsilon_i} = 1$. Thus, from dimension reasons, $\h'$ is also a Cartan subalgebra of $\g$, and hence maximal abelian in $\k$. Since $Z \in C(\k)$, we conclude $Z \in \h'$, which completes the proof.

A decomposition of $Z$
----------------------

From Proposition \[Prop:4.1\], we have a decomposition $$\label{eq:4.4}
Z = Z_0 + Z_{2\varepsilon_1} +\dots+ Z_{2\varepsilon_r},$$ where $Z_0 \in \k_0$ and $Z_{2\varepsilon_i} \in \k_{2\varepsilon_i}$. In this subsection, we compute the length of each factor $Z_{2\varepsilon_i}$. Assume that $\operatorname{rank}G / K = r$ and keep the notations in the previous subsection.

First of all, we recall root vectors. A vector $H_{\alpha} \in \a$ is called the *root vector corresponding to a root $\alpha$* if it holds that $${\left\langle H_{\alpha}, H \right\rangle} = \alpha(H) \quad (\forall H \in \a).$$ We here mention well-known properties of root vectors of the root system of type $C_r$ or $BC_r$. We refer to an explicit description of each root system, for example, in [@Helgason; @Humphreys; @Loos].

\[Lem:root-vector\] Let $G/K$ be a compact irreducible symmetric space with root system of type $C_r$ or $BC_r$. Then we have

1.  ${\left\lVert H_{\varepsilon_1} \right\rVert}^2 = \cdots = {\left\lVert H_{\varepsilon_r} \right\rVert}^2$,

2.  ${\left\langle H_{\varepsilon_i} , H_{\varepsilon_j} \right\rangle} = 0$ if $i \neq j$.

We now study each factor $Z_{2\varepsilon_i}$ of $Z$. In particular, we express the lengths of $Z_{2 \varepsilon_i}$ in terms of the lengths of root vectors.

\[Prop:4.3\] For all $i \in \left\{ 1, 2,\dotsc, r \right\}$, we have

1.  $( \operatorname{ad}_{Z_{2\varepsilon_i}})^2\left( H_{\varepsilon_i} \right) = -H_{\varepsilon_i}$,

2.  $4{\left\lVert H_{\varepsilon_i} \right\rVert}^2{\left\lVert Z_{2\varepsilon_i} \right\rVert}^2 = 1$.

We show (1). By definition of $Z$, one has $$- H_{\varepsilon_i} = (\operatorname{ad}_{Z})^2\left( H_{\varepsilon_i} \right).$$ Lemma \[Lem:root-vector\] yields that $2 \varepsilon_i (H_{\varepsilon_j}) = 0$ if $i \neq j$. We thus have $$\operatorname{ad}_{Z} \left( H_{\varepsilon_i} \right) 
= \left[ Z_0 + Z_{2 \varepsilon_1} + \cdots + Z_{2 \varepsilon_r} , H_{\varepsilon_i} \right] 
= \left[ Z_{2 \varepsilon_i} , H_{\varepsilon_i} \right] 
\in \p_{2 \varepsilon_i}.$$ By Lemma \[Lem:k-epsilon\], one knows $[\k_{2 \varepsilon_i} , \p_{2 \varepsilon_j}] = 0$ for $i \neq j$. Therefore, we have $$(\operatorname{ad}_{Z})^2 \left( H_{\varepsilon_i} \right) 
= \left[ Z_0 + Z_{2 \varepsilon_1} + \cdots + Z_{2 \varepsilon_r} , 
\left[ Z_{2 \varepsilon_i} , H_{\varepsilon_i} \right] \right] 
= (\operatorname{ad}_{Z_{2\varepsilon_i}})^2 \left( H_{\varepsilon_i} \right) ,$$ which concludes the proof of (1).

We show (2). It follows from (1) that $$\label{eq:4.6}
{\left\lVert H_{\varepsilon_i} \right\rVert}^2 
= -{\left\langle \left( \operatorname{ad}_{Z_{2\varepsilon_i}} \right)^2\left( H_{\varepsilon_i} \right), H_{\varepsilon_i} \right\rangle} 
= -{\left\langle Z_{2\varepsilon_i}, \left[ H_{\varepsilon_i}, \left[ H_{\varepsilon_i}, Z_{2\varepsilon_i}\right] \right] \right\rangle} .$$ By definition of roots, one has $$\label{eq:4.8}
\left[ H_{\varepsilon_i}, \left[ H_{\varepsilon_i}, Z_{2\varepsilon_i}\right] \right] 
= -\left( 2\varepsilon_i \left( H_{\varepsilon_i} \right) \right)^2 Z_{2\varepsilon_i} 
= -4 {\left\lVert  H_{\varepsilon_i}  \right\rVert}^4 Z_{2\varepsilon_i} .$$ By substituting for , we get $${\left\lVert H_{\varepsilon_i} \right\rVert}^2 
= 4 {\left\lVert  H_{\varepsilon_i}  \right\rVert}^4 {\left\lVert  Z_{2\varepsilon_i}  \right\rVert}^2 .$$ Since $H_{\varepsilon_i} \neq 0$, our assertion is proved.

Preliminaries on root systems for rank two Hermitian case {#Sec:rank 2}
---------------------------------------------------------

Hereafter, let $G / K$ be a compact irreducible Hermitian symmetric space of rank two. In this subsection, we give formulas for $N := \dim \p$ and ${\left\lVert H_{\varepsilon_i} \right\rVert}^2$ in terms of the roots and the multiplicities.

Recall that the root system $\Delta$ of $G / K$ is of type $BC_2$ or $C_2$. For the case of type $BC_2$, as mentioned in Subsection \[subsection:root-hermitian\], $\Delta$ is given by $$\Delta = \Delta(BC_2) 
= \pm\left\{ \varepsilon_1,\ \varepsilon_2,\ 2\varepsilon_1,\ 2\varepsilon_2,\ \varepsilon_1 + \varepsilon_2,\ \varepsilon_1 - \varepsilon_2 \right\} .$$ Since an element $\sigma$ of the Weyl group induces a linear isomorphism from $\k_{\alpha}$ onto $\k_{\sigma(\alpha)}$, the multiplicities $m_{\alpha}$ of roots are invariant under the action of the Weyl group. Hence we have $$m_{\varepsilon_1 + \varepsilon_2} = m_{\varepsilon_1 - \varepsilon_2}, \quad 
m_{\varepsilon_1} = m_{\varepsilon_2} , \quad 
m_{2\varepsilon_1} = m_{2\varepsilon_2} = 1 .$$ We regard $\Delta(C_2)$ as $\Delta(BC_2)$ with $m_{\varepsilon_1} = m_{\varepsilon_2} = 0$. We thus consider $\Delta(BC_2)$ only. To make the latter convenient, we put $$\label{eq:multiplicities}
m_1 := m_{\varepsilon_1 + \varepsilon_2} , \quad 
m_2 := m_{\varepsilon_1} + m_{2 \varepsilon_1} = m_{\varepsilon_1} + 1 .$$ Note that $m_1$ and $m_2$ will be the multiplicities of the principal curvatures of the isoparametric hypersurfaces in spheres obtained from the isotropy representations of $G/K$.

We first give a formula for $N = \dim \p$, which follows easily.

\[Lem:length-N\] We have $N = 2 m_{1} + 2 m_{2} + 2$.

This is a direct consequence of the root space decomposition of $\p$.

We next compute the length of the root vectors $H_{\varepsilon_1}$ and $H_{\varepsilon_2}$.

\[Lem:4.4\] We have $1 / {{\left\lVert H_{\varepsilon_1} \right\rVert}^2} = 1 / {{\left\lVert H_{\varepsilon_2} \right\rVert}^2} 
= 2(2m_{1} + m_{2} + 3)$.

One knows ${\left\lVert H_{\varepsilon_1} \right\rVert}^2 = {\left\lVert H_{\varepsilon_2} \right\rVert}^2$ from Lemma \[Lem:root-vector\]. Thus it is sufficient to compute ${\left\lVert H_{\varepsilon_1} \right\rVert}^2$.

First of all, we derive a general formula of ${\left\lVert H \right\rVert}^2$ for $H \in \a$. By Lemma \[Lem:3.3\], we have $${\left\lVert H \right\rVert}^2 = -2 \operatorname{Tr}\left(\left. ( \operatorname{ad}_{H} )^2 \right|_\p \right) 
= -2 \sum_{j} {\left\langle \left( \operatorname{ad}_{H} \right)^2(P_j), P_j \right\rangle} ,$$ where $\{ P_j \}$ is an orthonormal basis of $\p$. Since the representation of the right-hand side is independent of the choice of an orthonormal basis of $\p$, we can assume each $P_j$ is a unit vector in $\a$ or some root space $\p_{\alpha}$. For a unit vector $P_\alpha \in \p_\alpha$, we have $${\left\langle \left( \operatorname{ad}_{ H } \right)^2 (P_\alpha), P_\alpha \right\rangle} 
= {\left\langle - \alpha( H )^2 P_\alpha , P_\alpha \right\rangle} = - \alpha( H )^2 .$$ Hence, it follows that $${\left\lVert H \right\rVert}^2 
= -2 \sum_{i} {\left\langle \left( \operatorname{ad}_{H} \right)^2(P_i), P_i \right\rangle} 
= 2\sum_{\alpha} m_{\alpha} \alpha(H)^2.$$ Note that this formula is true for every symmetric space.

We use a notation ${\left\langle \alpha , \beta \right\rangle} := \alpha(H_{\beta})$. Then, since the root system is of $BC_2$-type, it follows $$\begin{split}
{\left\lVert H_{\varepsilon_1} \right\rVert}^2 &= 2\left( m_{\varepsilon_1 - \varepsilon_2}{\left\langle \varepsilon_1 - \varepsilon_2, \varepsilon_1 \right\rangle}^2 + m_{\varepsilon_2}{\left\langle \varepsilon_2, \varepsilon_1 \right\rangle}^2 + m_{2\varepsilon_2}{\left\langle 2\varepsilon_2, \varepsilon_1 \right\rangle}^2 \right.\\
&\hspace{30pt}\left. m_{\varepsilon_1}{\left\langle \varepsilon_1, \varepsilon_1 \right\rangle}^2 + m_{\varepsilon_1 + \varepsilon_2}{\left\langle \varepsilon_1 + \varepsilon_2, \varepsilon_1 \right\rangle}^2 + m_{2\varepsilon_1}{\left\langle 2\varepsilon_1, \varepsilon_1 \right\rangle}^2 \right)\\
&= 2\left( 
m_1 + (m_2 - 1) + m_1 + 4 
\right) {\left\langle \varepsilon_1, \varepsilon_1 \right\rangle}^2\\
&= 2\left( 
2 m_1 + m_2 + 3 
\right) {\left\lVert H_{\varepsilon_1} \right\rVert}^4 . 
\end{split}$$ Since $H_{\varepsilon_1} \neq 0$, this completes the proof.

The squared-norm of the gradient
--------------------------------

According to Proposition \[Prop:3.7\], the squared-norm ${\left\lVert \operatorname{grad}f_{a, b}(P) \right\rVert}^2$ of the gradient can be written as a linear combination of ${\left\lVert [J(P), \mu(P)] \right\rVert}^2$, ${\left\lVert P \right\rVert}^2{\left\lVert \mu(P) \right\rVert}^2$ and ${\left\lVert P \right\rVert}^6$. In this subsection, we simplify this presentation in the case of $\operatorname{rank}G / K = 2$.

We need formulas for ${\left\lVert [J(P), \mu(P)] \right\rVert}^2$, ${\left\lVert P \right\rVert}^2{\left\lVert \mu(P) \right\rVert}^2$ and ${\left\lVert P \right\rVert}^6$ when $P \in \a$. First of all, we give a formula for ${\left\lVert P \right\rVert}^6$.

\[Lem:P6\] For $P = a_1 H_{\varepsilon_1} + a_2 H_{\varepsilon_2} \in \a$, we have

1.  ${\left\lVert P \right\rVert}^2 = ({a_1}^2 + {a_2}^2) {\left\lVert H_{\varepsilon_1} \right\rVert}^2$,

2.  ${\left\lVert P \right\rVert}^6 = ( {a_1}^6 + 3 {a_1}^4 {a_2}^2 + 3 {a_1}^2 {a_2}^4 + {a_2}^6 ) 
    {\left\lVert H_{\varepsilon_1} \right\rVert}^6$.

The first claim follows from ${\left\langle H_{\varepsilon_1} , H_{\varepsilon_2} \right\rangle} = 0$ and ${\left\lVert H_{\varepsilon_1} \right\rVert}^2 = {\left\lVert H_{\varepsilon_2} \right\rVert}^2$ (see Lemma \[Lem:root-vector\]). The second claim follows easily from the first one.

Next we give a formula for ${\left\lVert P \right\rVert}^2{\left\lVert \mu(P) \right\rVert}^2$.

\[Lem:P2mu2\] For $P = a_1 H_{\varepsilon_1} + a_2 H_{\varepsilon_2} \in \a$, we have

1.  $\mu(P) = -2 ({a_1}^2 Z_{2\varepsilon_1} + {a_2}^2 Z_{2\varepsilon_2}) 
    {\left\lVert H_{\varepsilon_1} \right\rVert}^4$,

2.  ${\left\lVert P \right\rVert}^2 {\left\lVert \mu(P) \right\rVert}^2 
    = ({a_1}^6 + {a_1}^4 {a_2}^2 + {a_1}^2 {a_2}^4 + {a_2}^6) 
    {\left\lVert H_{\varepsilon_1} \right\rVert}^8$.

We show (1). By the decomposition $Z = Z_0 + Z_{2\varepsilon_1} + Z_{2\varepsilon_2}$, it follows that $$\label{eq:4.23}
2 \mu(P) 
= [P, [P, Z]] 
= [P, [P, Z_0]] + [P, [P, Z_{2\varepsilon_1}]] + [P, [P, Z_{2\varepsilon_2}]] .$$ Since $P \in \a$, the definition of roots yields that $$\label{eq:4.24}
\begin{split}
[P, [P, Z_0]] & = 0 , \\ 
[P, [P, Z_{2\varepsilon_1}]] 
& = - \left( 2\varepsilon_1(P) \right)^2 Z_{2\varepsilon_1} 
= -4 {a_1}^2 {\left\lVert H_{\varepsilon_1} \right\rVert}^4 Z_{2\varepsilon_1} , \\ 
[P, [P, Z_{2\varepsilon_2}]] 
& = - \left( 2\varepsilon_2(P) \right)^2 Z_{2\varepsilon_2} 
= -4 {a_2}^2 {\left\lVert H_{\varepsilon_1} \right\rVert}^4 Z_{2\varepsilon_2} . 
\end{split}$$ By substituting for , we conclude the proof of (1).

We show (2). Recall that different root spaces are orthogonal to each other (cf. [@Helgason Chapter III, Theorem 4.2]). This yields that $${\left\langle Z_{2\varepsilon_1} , Z_{2\varepsilon_2} \right\rangle} = 0 .$$ Then, it follows from (1) and Proposition \[Prop:4.3\] that $${\left\lVert  \mu(P)  \right\rVert}^2 
= 4 {\left\lVert H_{\varepsilon_1} \right\rVert}^8 \left(
{a_1}^4 {\left\lVert Z_{2\varepsilon_1} \right\rVert}^2 + {a_2}^4 {\left\lVert Z_{2\varepsilon_2} \right\rVert}^2
\right) = ({a_1}^4 + {a_2}^4) {\left\lVert H_{\varepsilon_1} \right\rVert}^6 .$$ This and Lemma \[Lem:P6\] (1) conclude the proof of (2).

We then give a formula for ${\left\lVert  [J(P), \mu(P)]  \right\rVert}^2$.

\[Lem:Jmu\] For $P = a_1 H_{\varepsilon_1} + a_2 H_{\varepsilon_2} \in \a$, we have $${\left\lVert  [J(P), \mu(P)]  \right\rVert}^2 
= 4 \left( {a_1}^6 + {a_2}^6 \right) {\left\lVert H_{\varepsilon_1} \right\rVert}^{10} .$$

Recall that $[\k_0 , \a] = 0$ and ${\left\langle H_{\varepsilon_1} , H_{\varepsilon_2} \right\rangle} = 0$ (see Lemma \[Lem:root-vector\]). Then we have $$\label{eq:4.29}
J(P) 
= [ Z_0 + Z_{2\varepsilon_1} + Z_{2\varepsilon_2} , 
a_1 H_{\varepsilon_1} + a_2 H_{\varepsilon_2} ] 
= a_1 [Z_{2\varepsilon_1} , H_{\varepsilon_1} ] 
+ a_2 [Z_{2\varepsilon_2} , H_{\varepsilon_2} ] .$$ Lemma \[Lem:P2mu2\] (1) yields that $$[J(P), \mu(P)] 
= \left[ a_1 [Z_{2\varepsilon_1} , H_{\varepsilon_1} ] 
+ a_2 [Z_{2\varepsilon_2} , H_{\varepsilon_2} ] , \ 
- 2 ({a_1}^2 Z_{2\varepsilon_1} + 2{a_2}^2 Z_{2\varepsilon_2}) 
{\left\lVert H_{\varepsilon_1} \right\rVert}^4 \right] .$$ From Lemma \[Lem:k-epsilon\], one has $$[[Z_{2\varepsilon_1} , H_{\varepsilon_1} ] , Z_{2\varepsilon_2} ] 
= [[Z_{2\varepsilon_2} , H_{\varepsilon_2} ] , Z_{2\varepsilon_1} ] 
= 0 .$$ Therefore, Proposition \[Prop:4.3\] (1) yields that $$\label{eq:4.30}
\begin{split}
[J(P), \mu(P)] 
& = 
- 2 {\left\lVert H_{\varepsilon_1} \right\rVert}^4  \left(
{a_1}^3 [[Z_{2\varepsilon_1} , H_{\varepsilon_1}], Z_{2\varepsilon_1}] 
+ {a_2}^3 [[Z_{2\varepsilon_2} , H_{\varepsilon_2}], Z_{2\varepsilon_2}] 
\right) \\ 
& = 
-2 {\left\lVert H_{\varepsilon_1} \right\rVert}^4  \left(
{a_1}^3 H_{\varepsilon_1} + {a_2}^3 H_{\varepsilon_2} 
\right) . 
\end{split}$$ By taking the squared-norms, Lemma \[Lem:P6\] (1) concludes the proof.

By using these lemmas, ${\left\lVert [J(P), \mu(P)] \right\rVert}^2$ can be written as a linear combination of ${\left\lVert P \right\rVert}^2{\left\lVert \mu(P) \right\rVert}^2$ and ${\left\lVert P \right\rVert}^6$.

\[Lem:4.7\] For all $P \in \p$, we have $$\label{eq:4.27}
{\left\lVert  \left[ J(P), \mu(P) \right]  \right\rVert}^2 
= - 2 {\left\lVert H_{\varepsilon_1} \right\rVert}^4 {\left\lVert P \right\rVert}^6 
+ 6 {\left\lVert H_{\varepsilon_1} \right\rVert}^2 {\left\lVert P \right\rVert}^2{\left\lVert \mu(P) \right\rVert}^2 .$$

In general, if $f, g : \p \to \bR$ are $K$-invariant functions and satisfy $f |_{\a} = g |_{\a}$, then $f=g$ holds. Since the both sides of are $K$-invariant, we have only to prove that they coincide on $\a$. To show this, take any $P \in \a$, and write $P = a_1 H_{\varepsilon_1} + a_2 H_{\varepsilon_2}$. From Lemmas \[Lem:P6\] and \[Lem:P2mu2\], one has $$\begin{split}
- 2 {\left\lVert H_{\varepsilon_1} \right\rVert}^4 {\left\lVert P \right\rVert}^6 
& = 
-2 ( {a_1}^6 + 3 {a_1}^4 {a_2}^2 + 3 {a_1}^2 {a_2}^4 + {a_2}^6 ) 
{\left\lVert H_{\varepsilon_1} \right\rVert}^{10} , \\ 
6 {\left\lVert H_{\varepsilon_1} \right\rVert}^2 {\left\lVert P \right\rVert}^2{\left\lVert \mu(P) \right\rVert}^2 
& = 
6 ({a_1}^6 + {a_1}^4 {a_2}^2 + {a_1}^2 {a_2}^4 + {a_2}^6) 
{\left\lVert H_{\varepsilon_1} \right\rVert}^{10} . 
\end{split}$$ Therefore, the right-side hand of satisfies $$- 2 {\left\lVert H_{\varepsilon_1} \right\rVert}^4 {\left\lVert P \right\rVert}^6 
+ 6 {\left\lVert H_{\varepsilon_1} \right\rVert}^2 {\left\lVert P \right\rVert}^2{\left\lVert \mu(P) \right\rVert}^2 
= 4 \left( {a_1}^6 + {a_2}^6 \right) {\left\lVert H_{\varepsilon_1} \right\rVert}^{10} .$$ By Lemma \[Lem:Jmu\], this coincides with the left-hand side of . This completes the proof.

The following gives us a representation of the squared-norm of the gradient of $f_{a, b}$ in the case of rank two.

\[Prop:4.8\] For every $P \in \p$, we have $$\label{eq:4.35}
\begin{split}
{\left\lVert  \operatorname{grad}f_{a, b}(P)  \right\rVert}^2 &= \left( -\dfrac{2 b^2}{(2 m_{1} + m_{2} + 3)^2} + \dfrac{(a - b)^2}{4(m_{1} + m_{2} + 1)^2} \right) {\left\lVert P \right\rVert}^6\\ 
&\hspace{1cm}+ \left( \dfrac{12 b^2}{2 m_{1} + m_{2} + 3} + \dfrac{4 b (a - b)}{m_{1} + m_{2} + 1} \right) {\left\lVert P \right\rVert}^2{\left\lVert \mu(P) \right\rVert}^2. 
\end{split}$$

From Proposition \[Prop:3.7\], one knows $${\left\lVert \operatorname{grad}f_{a, b}(P) \right\rVert}^2 
= 4 b^2 {\left\lVert [J(P), \mu(P)] \right\rVert}^2 
+ \dfrac{8 b (a - b)}{N}{\left\lVert P \right\rVert}^2{\left\lVert \mu(P) \right\rVert}^2 
+ \dfrac{(a - b)^2}{N^2}{\left\lVert P \right\rVert}^6 .$$ By using Lemma \[Lem:4.7\], we have $${\left\lVert \operatorname{grad}f_{a, b}(P) \right\rVert}^2 = A {\left\lVert P \right\rVert}^6 + B {\left\lVert P \right\rVert}^2{\left\lVert \mu(P) \right\rVert}^2 ,$$ where $$\label{AB}
A = -2 {\left\lVert H_{\varepsilon_1} \right\rVert}^4 \cdot 4 b^2 + \dfrac{(a - b)^2}{N^2} , \quad 
B = 6 {\left\lVert H_{\varepsilon_1} \right\rVert}^2 \cdot 4 b^2 + \dfrac{8 b (a - b)}{N} .$$ From Lemmas \[Lem:length-N\] and \[Lem:4.4\], one knows $${\left\lVert H_{\varepsilon_1} \right\rVert}^2 = 1 / (2 (2 m_1 + m_2 + 3)) , \quad 
N = 2 m_1 + 2 m_2 + 2 .$$ By substituting these for , we conclude the proof.

The Laplacian
-------------

Next, we give a representation of $\Delta f_{a, b}(P)$.

\[Prop:4.9\] For all $P \in \p$, we have $$\label{eq:4.38}
\Delta f_{a, b}(P) = \dfrac{(m_{1} + m_{2} + 2)a + (m_{1} + m_{2})b}{m_{1} + m_{2} + 1}{\left\lVert P \right\rVert}^2.$$

From Proposition \[Prop:3.9\], one knows $$\Delta f_{a, b}(P) = \dfrac{(N + 2) a + (N - 2) b}{N} {\left\lVert P \right\rVert}^2.$$ By substituting $N = 2 m_{1} + 2 m_{2} + 2$ (see Lemma \[Lem:length-N\]) for the above, we complete the proof easily.

Main Theorem {#sect:Main Theorem}
============

Recall that $f_{a,b}$ is the weighted squared-norm of the moment map $\mu$. In this section, we give a proof of our main theorem, that is, $f_{a,b}$ is a Cartan-Münzner polynomial for some $a$ and $b$. We need $(m_1, m_2)$ defined in (\[eq:multiplicities\]).

Let $G/K$ be a compact irreducible Hermitian symmetric space of rank two. If $a = -8(m_{1} + 2 m_{2})$ and $b = 8(2 m_{1} + m_{2} + 3)$, then $f_{a, b}(P)$ is a Cartan-Münzner polynomial of degree four. In particular, this Cartan-Münzner polynomial defines homogeneous isoparametric hypersurfaces in the sphere with four distinct principal curvatures with multiplicities $(m_1, m_2)$.

In order to prove that $f_{a, b}(P)$ is a Cartan-Münzner polynomial of degree four, from , we have only to show that $$\begin{aligned}
{\left\lVert \operatorname{grad}f_{a, b}(P) \right\rVert}^2 &= 16 {\left\lVert P \right\rVert}^6,\label{eq:5.A} \\ 
\Delta f_{a, b}(P) &= 8(m_{1} - m_{2}){\left\lVert P \right\rVert}^2\label{eq:5.B}. \end{aligned}$$ We know the formulas for ${\left\lVert \operatorname{grad}f_{a, b}(P) \right\rVert}^2$ and $\Delta f_{a, b}(P)$ from Propositions \[Prop:4.8\] and \[Prop:4.9\]. Let us consider their coefficients, and substitute our $a$, $b$ and $$\label{eq:6.1}
a - b = - 24 (m_{1} + m_{2} + 1)$$ into them. We then obtain $$\begin{aligned}
-\dfrac{2 b^2}{(2 m_{1} + m_{2} + 3)^2} + \dfrac{(a - b)^2}{4(m_{1} + m_{2} + 1)^2} 
&= - 128 + 144 = 
16 \label{eq:A},\\[5pt]
\dfrac{12 b^2}{2 m_{1} + m_{2} + 3} + \dfrac{4 b (a - b)}{m_{1} + m_{2} + 1} 
&= 4b (24-24) =  
0\label{eq:B},\\[5pt] 
\dfrac{(m_{1} + m_{2} + 2)a + (m_{1} + m_{2})b}{m_{1} + m_{2} + 1} 
&= \dfrac{8 (m_{1}^2 - m_{2}^2 + m_{1} - m_{2})}{m_{1} + m_{2} + 1} = 
8 (m_{1} - m_{2}) \label{eq:C}. \end{aligned}$$ This completes the proof of the theorem.

We here list $(m_1 , m_2)$ of compact irreducible Hermitian symmetric spaces of rank two. One can see that $(m_1 , m_2)$ are compatible with the multiplicities of the corresponding homogeneous hypersurfaces in spheres.

                                           $G / K$                                            Type     $(m_1, m_2)$
  ----------------------------------------------------------------------------------------- -------- ----------------
        $\operatorname{SO}(2 + n) / \operatorname{SO}(2) \times \operatorname{SO}(n)$        $C_2$     $(1, n - 2)$
   $\operatorname{SU}(2 + n) / \mathrm{S}(\operatorname{U}(2) \times \operatorname{U}(n))$   $BC_2$   $(2, 2 n - 3)$
                        $\operatorname{SO}(10) / \operatorname{U}(5)$                        $BC_2$      $(4, 5)$
             $\mathrm{E}_6 / \operatorname{U}(1) \times \operatorname{Spin}(10)$             $BC_2$      $(6, 9)$

Acknowledgements {#acknowledgements .unnumbered}
================

The authors are grateful to Yoshio Agaoka, Osamu Ikawa, Reiko Miyaoka and Yoshihiro Ohnita for valuable discussions and comments. The authors would like to thank also, Kazuhiro Shibuya, Takahiro Hashinaga and Akira Kubo for their warm encouragements. The second author was supported in part by KAKENHI (20740040 and 24654012).

[99]{} T. E. Cecil, *Lie Sphere Geometry*. With applications to submanifolds. Second edition. Universitext. Springer, New York, 2008. T. E. Cecil, Q.-S. Chi, G. R. Jensen, *Isoparametric hypersurfaces with four principal curvatures*. Ann. of Math. (2) **165** (2007), 1–76.

Q.-S. Chi, *Isoparametric hypersurfaces with four principal curvatures revisited*. Nagoya Math. J. **193** (2009), 129–154.

, *Isoparametric hypersurfaces with four principal curvatures, II*. Nagoya Math. J. **204** (2011), 1–18.

, *Isoparametric hypersurfaces with four principal curvatures, III*. Preprint, arXiv:1104.3249v3 (2011).

J. Dorfmeister, E. Neher, *Isoparametric hypersurfaces, case $g=6$, $m=1$*. Comm. Algebra **13** (1985), 2299–2368. D. Ferus, H. Karcher, H. F. Münzner, *Cliffordalgebren und neue isoparametrische Hyperflächen*. Math. Z. **177** (1981), 479–502. S. Fujii, *Homogeneous isoparametric hypersurfaces in spheres with four distinct principal curvatures and moment maps*. Tôhoku Math. J. **62** (2010), 191–213.

S. Helgason, *Differential geometry, Lie groups, and symmetric spaces*. Corrected reprint of the 1978 original. Graduate Studies in Mathematics, **34**. American Mathematical Society, Providence, RI, 2001. J. E. Humphreys, *Introduction to Lie algebras and representation theory*. Second printing, revised. Graduate Texts in Mathematics, **9**. Springer-Verlag, New York-Berlin, 1978. W.-Y. Hsiang, H. B. Lawson, Jr., *Minimal submanifolds of low cohomogeneity*. J. Diff. Geom. **5** (1971), 1–38. F. C. Kirwan, *Cohomology of quotients in symplectic and algebraic geometry*. Mathematical Notes, **31**. Princeton University Press, Princeton, NJ, 1984. S. Immervoll, *On the classification of isoparametric hypersurfaces with four distinct principal curvatures in spheres*. Ann. of Math. (2) **168** (2008), 1011–1024. O. Loos, *Symmetric spaces II: Compact spaces and classification*. W. A. Benjamin, Inc., New York-Amsterdam, 1969.

R. Miyaoka, *Moment maps of the spin action and the Cartan-Münzner polynomials of degree four*. Math. Ann., to appear.

, *Isoparametric hypersurfaces with $(g,m)=(6,2)$*. Ann. of Math. (2), to appear.

H. F. Münzner, *Isoparametrische Hyperflächen in Sphären*. Math. Ann. **251** (1980), 57–71. , *Isoparametrische Hyperflächen in Sphären. II.  Über die Zerlegung der Sphäre in Ballbündel*. Math. Ann. **256** (1981), 215–232. T. Nagano, M. S. Tanaka, *The involutions of compact symmetric spaces. V*. Tokyo J. Math. **23** (2000), 403–416. Y. Nagatomo, M. Takahashi, *Vector bundles, isoparametric functions and Radon transformations on a symmetric space*. Preprint.

Y. Ohnita, *Moment maps and symmetric Lagrangian submanifolds*. In: Proceedings of the conference “SUBMANIFOLDS in Yuzawa 2004" (2005), 33–38.

H. Ozeki, M. Takeuchi, *On some types of isoparametric hypersurfaces in spheres. I*. Tôhoku Math. J. (2) **27** (1975), 515–559. , *On some types of isoparametric hypersurfaces in spheres. II*. Tôhoku Math. J. (2) **28** (1976), 7–55. H. Song, *Some differential-geometric properties of $R$-spaces*. Tsukuba J. Math. **25** (2001), 279–298. H. Tamaru, *The local orbit types of symmetric spaces under the actions of the isotropy subgroups*. Differential Geom. Appl. **11** (1999), 29–38. , *Riemannian G.O. spaces fibered over irreducible symmetric spaces*. Osaka J. Math. **36** (1999), 835–851. G. Thorbergsson, *A survey on isoparametric hypersurfaces and their generalizations*. Handbook of Differential Geometry, Vol. I, 963–995, North-Holland, Amsterdam, 2000.