Moment maps of the spin action and the Cartan–Münzner polynomials of degree four

Abstract

We express all the known Cartan–Münzner polynomials of degree four in terms of the moment map of certain group actions.

Appendix

For hypersurfaces of OT-FKM type which are at the same time isotropy orbits of Hermitian symmetric spaces, we give a relation between the matrix expression in [11] and the argument in Sect. 6.

In the case of Hermitian symmetric space $SO(2+n)/SO(2)\times SO(n)$, we have

$$\begin{aligned} \begin{array}{ll} {\mathfrak k }=\left\{ \begin{pmatrix}A&0\\ 0&A^{\prime }\end{pmatrix}, \,A\in {\mathfrak o }(2),\,A^{\prime }\in {\mathfrak o }(n)\right\} ,\\ {\mathfrak p }=\left\{ \begin{pmatrix}0&V\\ -{}^tV&0\end{pmatrix}, \,V\in M_{2,n}({\mathbb R })\right\} , \end{array} \end{aligned}$$

and from $\dim {\mathfrak p }=2n$, $n=l$ holds in Sect. 5. We may consider an element $H(\xi _1,\xi _2)\in {\mathfrak a }$ which corresponds to

$$\begin{aligned} V=\begin{pmatrix}\xi _1&0&\ldots&0\\ 0&\xi _2&\ldots&0\end{pmatrix},\quad \xi _1,\xi _2\in {\mathbb R }. \end{aligned}$$

Namely, with respect to an orthonormal basis $e_1,\ldots , e_l$ of ${\mathbb R }^l$, take a double copy ${\mathbb R }^l\oplus {\mathbb R }^l$ consisting of $\pm 1$ eigenspaces of $P_0$. We may consider $H=(\xi _1e_1,\xi _2e_2)\in {\mathfrak a }\subset {\mathbb R }^l\oplus {\mathbb R }^l$. Since we can use $P_0,P_1$ with respect to this decomposition (ark 6.2), $P_0H=(\xi _1e_1,-\xi _2e_2)$ and $P_1H=(\xi _2e_2,\xi _1e_1)$ follow, and hence $\langle P_0H,H\rangle =\xi _1^2-\xi _2^2$ and $\langle P_1H,H\rangle =0$ hold, which implies $Y_H=P_1H$. Because $m=1$, it follows that

$$\begin{aligned} \Vert H\Vert ^2=\Vert P_1H\Vert ^2=\xi _1^2+\xi _2^2,\quad \Vert \mu (H,Y_H)\Vert ^2=\langle P_0H,H\rangle ^2=(\xi _1^2-\xi _2^2)^2, \end{aligned}$$

and from our result (3), we obtain

$$\begin{aligned} \begin{array}{ll} F(H)&=\Vert H\Vert ^4-2\Vert \mu (H,Y_H)\Vert ^2\\&=(\xi _1^2+\xi _2^2)^2-2(\xi _1^2-\xi _2^2)^2\\&=3(\xi _1^2+\xi _2^2)^2-4(\xi _1^4+\xi _2^4). \end{array} \end{aligned}$$

This is nothing but (4.4) in [22], and (4.9) in [11] multiplied by $\frac{8}{n}$.

The case of $SU(2+n)/S(U(2)\times U(n))$ is similarly discussed, where from $m=2$, $P_0P_1,P_1P_2,P_2P_0$ generate ${\mathfrak s }{\mathfrak u }(2)$, and we may put $P_2(u,v)=(\sqrt{-1}v,-\sqrt{-1}u)$, and $\langle P_2(u,v),(u,v)\rangle =0$ follows. Thus we obtain the same $F(x)$ as above, which coincides with (4.4) in [22].

In the case of $E_6/U(1)\times Spin(10)$, we know $(m_1,m_2)=(9,6)$, and the spin action on ${\mathfrak p }\cong {\mathbb R }^{32}$ in Sect. 6 coincides with the isotropy action of $Spin(10)$. In fact, by the argument in 6.3 of [10], $P_iP_jx$, $(i,j)=(0,1), (2,3), (1,2i), (1,2i+1), (0,2i),(0,2i+1)$ for $1\le i\le 4$, and $(2i,2j)$ $(2i,2j+1)$ for $1\le i< j\le 4$ span the 30 linearly independent Killing fields on the hypersurface, and hence the action is transitive. Up to now, [12] is not yet at hand, and so Theorem 1.2 gives the unique expression of this case in terms of the moment map $\mu$.