1. Introduction
In 1970, Lawson [Reference Lawson11] constructed minimal surfaces in
$S^3$
of arbitrary genus. This work initiated a new direction in differential geometry by moving the study of minimal surfaces from Euclidean space to positively curved ambient spaces. Later, Hsiang and Lawson [Reference Hsiang and Lawson7] extended related techniques to higher-dimensional spheres. On the other hand, Lawson’s work also clarified topological restrictions for minimal surfaces in positively curved manifolds and influenced the study of incompressible surfaces in three-manifolds by Schoen and Yau [Reference Schoen and Yau18].
Besides the existence problem, an equally important question is to understand the geometric and topological properties of minimal hypersurfaces. Among these, the Morse index is a central analytical invariant: it measures stability and reflects deep information about the geometry of the hypersurface. It has been conjectured that the index controls the basic topology of minimal hypersurfaces in an effective way.
Conjecture 1 (Schoen-Marques-Neves, see [Reference Marques and Neves12]). Let
$(N^{n+1},g)$
be a closed Riemannian manifold with positive Ricci curvature. Then there exists a positive constant
$C$
such that, for every closed embedded minimal hypersurface
$M^n\subset N^{n+1}$
,
\begin{equation*} \operatorname {index}\!(M)\ge C\, b_1(M). \end{equation*}
A particular ambient manifold for which this conjecture is known is the round sphere. This was proved by Savo [Reference Savo17], building on earlier work of Ros [Reference Ros16]. For flat three-dimensional tori, Ros proved that the index of a closed minimal surface is bounded from below by an affine function of the genus. Since, for a closed orientable surface, the first Betti number is twice the genus, this is a close analogue of the conjectural picture above.
Ambrozio, Carlotto, and Sharp [Reference Ambrozio, Carlotto and Sharp1] developed a general framework to study this conjecture in other positively curved ambient manifolds. Their method gives a lower bound for the Morse index in terms of the first Betti number under a suitable extrinsic inequality.
Theorem 1.1 (Ambrozio-Carlotto-Sharp [Reference Ambrozio, Carlotto and Sharp1]). Let
$(N^{n+1},g)$
be a Riemannian manifold isometrically embedded in some Euclidean space
${\mathbb{R}}^d$
. Let
$M^n$
be a closed embedded minimal hypersurface of
$N^{n+1}$
. Assume that for every nonzero vector field
$X$
on
$M$
,
\begin{equation} \begin{aligned} &\int _M \Bigg [\sum _{k=1}^n R^N(e_k,X,e_k,X)+\operatorname {Ric}_N\!(\nu ,\nu )|X|^2\Bigg ] \, dM \\ \gt &\int _M \Bigg [\sum _{k=1}^n |\Pi (e_k,X)|^2 +\sum _{k=1}^n |\Pi (e_k,\nu )|^2|X|^2\Bigg ] \, dM, \end{aligned} \end{equation}
where
$R^N$
is the curvature tensor of
$N$
,
$\Pi$
is the second fundamental form of the embedding
$N\hookrightarrow {\mathbb{R}}^d$
,
$\nu$
is a local unit normal vector field along
$M$
, and
$\{e_k\}_{k=1}^n$
is a local orthonormal frame on
$M$
. Then,
\begin{equation*} \operatorname {index}(M)\ge \frac {2}{d(d-1)}\, b_1(M). \end{equation*}
For convenience, we refer to (1) as the ACS condition. The same strategy has since been extended to free boundary and weighted settings; see [Reference Ambrozio, Carlotto and Sharp2, Reference Chen, Ge and Zhang4, Reference Impera and Rimoldi8, Reference Impera, Rimoldi and Savo9].
In this paper, we apply the ACS framework to minimal isoparametric hypersurfaces with positive Ricci curvature in the unit sphere. We hope to provide further evidence for the Schoen-Marques-Neves conjecture by adding new examples in this class.
For an isoparametric hypersurface
$N^{n+1}\subset S^{n+2}$
, the number
$g$
of distinct principal curvatures can only be
$1,2,3,4$
or
$6$
, and the multiplicities satisfy
$m_i=m_{i+2}$
(indices modulo
$g$
). Hence, there are at most two distinct multiplicities, say
$m_1$
and
$m_2$
; see [Reference Cecil and Ryan3, Reference Ge, Qian, Tang and Yan10]. For a minimal isoparametric hypersurface
$N^{n+1}\subset S^{n+2}$
, the cases with positive Ricci curvature are the following [Reference Qian, Tang and Yan15]:
-
(1) when
$g=1$
, the Ricci curvature is obviously positive; -
(2) when
$g=2$
, the Ricci curvature is positive unless
$N$
is
$S^1(r_1)\times S^n(r_2)$
; -
(3) when
$g=3$
, the multiplicities are equal and the Ricci curvature is positive for
$m_1=m_2\gt 1$
; -
(4) when
$g=4$
, the Ricci curvature is positive provided
$m_1,m_2\ge 2$
.
Our main result is the following.
Theorem 1.2.
Let
$N^{n+1}\subset S^{n+2}$
be a minimal isoparametric hypersurface. Then the following statements hold.
-
(1) If
$g=3$
and
$m_1=m_2\in \{4,8\}$
, then
$N$
satisfies the ACS condition.
-
(2) If
$g=4$
and
$ \min \{m_1,m_2\}\ge 4,$
then
$N$
satisfies the ACS condition.
Consequently, if
$M^n$
is a closed minimal hypersurface of any of the above ambient manifolds, then
\begin{equation*} \operatorname {index}\!(M)\ge \frac {2}{d(d-1)}\, b_1(M), \end{equation*}
where
$d=n+3$
for the natural embedding
$N\subset S^{n+2}\subset {\mathbb{R}}^{n+3}$
.
Gorodski, Mendes, and Radeschi [Reference Gorodski, Mendes and Radeschi6] also produced many examples supporting the Schoen-Marques-Neves conjecture. They obtained robust index lower bounds for minimal hypersurfaces in several isoparametric and symmetric ambient spaces by a different method. Our contribution is instead to verify a sufficient ACS inequality for the ambient manifold itself. In the
$g=4$
case, we work with the pointwise quantity
$\Delta (X,\nu )$
and obtain progressively tighter bounds. A direct estimate of
$\Delta$
using only the largest absolute curvature yields the range
$\min \{m_1,m_2\}\ge 5$
. By retaining a quadratic term in Proposition 3.1 and completing the square, we lower the coefficient to
$\tfrac {17}{4}$
, which extends the result to
$\min \{m_1,m_2\}=4$
with
$m_1+m_2\le 19$
. Finally, a moment-relaxation technique (Proposition 3.2) that exploits the asymmetry between the largest and smallest principal curvatures produces the sharp coefficient
$4$
, establishing the ACS condition for all pairs with
$\min \{m_1,m_2\}\ge 4$
without restriction on
$m_1+m_2$
. We observe in Remark 3.4 that
$\min \{m_1,m_2\}\ge 4$
is also a necessary condition for any argument based solely on pointwise positivity of
$\Delta$
. Thus, our result is sharp with respect to this method.
The cases not covered by Theorem1.2 remain open from the point of view of the present computation. More precisely:
-
(1) for
$g=3$
, the case
$m_1=m_2=2$
is not decided by our argument, while for
$m_1=m_2=1$
the Ricci curvature is not positive; -
(2) for
$g=4$
, our estimate does not decide the ACS condition when at least one multiplicity lies in
$\{2,3\}$
; -
(3) for
$g=6$
, the Ricci curvature is not positive.
2. Preliminaries
An isoparametric hypersurface is a hypersurface that arises as a regular level set of an isoparametric function. More precisely, a hypersurface
$N^{n+1}$
in a Riemannian manifold
$\widetilde N^{n+2}$
is called isoparametric if there exists a nonconstant smooth function
$f$
on
$\widetilde N$
and smooth functions
$a,b$
such that
-
(1)
$N=f^{-1}(c)$
for some regular value
$c$
of
$f$
; -
(2)
$|\nabla f|^2=b(f)$
; -
(3)
$\Delta f=a(f)$
.
In a space form, this is equivalent to requiring that all principal curvatures are constant.
For isoparametric hypersurfaces in the unit sphere
$S^{n+2}$
, the possible numbers of distinct principal curvatures are
$1,2,3,4$
and
$6$
. The cases may be summarized as follows (see [Reference Cecil and Ryan3, Reference Chi5, Reference Ge, Qian, Tang and Yan10, Reference Münzner13, Reference Münzner14]):
-
(1) if
$g=1$
, then
$N$
is totally umbilic; -
(2) if
$g=2$
, then
$N$
is a Clifford hypersurface
$S^k(r)\times S^{n+1-k}(\sqrt {1-r^2})$
; -
(3) if
$g=3$
, then all multiplicities are equal and can be
$1,2,4$
or
$8$
; -
(4) if
$g=6$
, then all multiplicities are equal and can be
$1$
or
$2$
; -
(5) if
$g=4$
, then the examples are either homogeneous or of OT-FKM type; among the multiplicity pairs, only
$(2,2)$
and
$(4,5)$
are not of OT-FKM type.
For OT-FKM hypersurfaces, the multiplicities are of the form
$(m_1,m_2)=(m_1,\ell -m_1-1)$
with
$\ell \in \mathbb N^*$
.
We will also use the following result of Münzner.
Theorem 2.1 (Münzner, see [Reference Chi5, Reference Münzner13, Reference Münzner14]). Let
$M^n$
be an isoparametric hypersurface with
$g$
distinct principal curvatures in
$S^{n+1}$
. Let
\begin{equation*} \lambda _1=\cot \theta _1\gt \cdots \gt \lambda _g=\cot \theta _g, \qquad \theta _i\in (0,\pi ), \end{equation*}
be the principal curvatures, with multiplicities
$m_1,\ldots ,m_g$
. Then:
-
(1) there exists a homogeneous polynomial
$F$
of degree
$g$
on
${\mathbb{R}}^{n+2}$
, called the Cartan-Münzner polynomial, such that
where
\begin{equation*} |\nabla F|^2=g^2 r^{2g-2}, \qquad \Delta F=\frac {m_2-m_1}{2}\, g^2 r^{g-2}, \end{equation*}
$r$
is the radial function on
${\mathbb{R}}^{n+2}$
;
-
(2) if
$f=F|_{S^{n+1}}$
, then the range of
$f$
is
$[{-}1,1]$
, the only critical values are
$\pm 1$
, and the critical sets
$M_\pm =f^{-1}({\pm} 1)$
are connected focal submanifolds;
-
(3) for each
$c\in ({-}1,1)$
, the level set
$M_c=f^{-1}(c)$
is an isoparametric hypersurface, the multiplicities satisfy
$m_i=m_{i+2}$
(indices modulo
$g$
), and
in particular, if
\begin{equation*} \theta _i=\theta _1+\frac {i-1}{g}\pi ; \end{equation*}
$g$
is odd then all multiplicities are equal;
-
(4) the unique minimal member in the isoparametric family is
$M_{c_0}=f^{-1}(c_0)$
, where
\begin{equation*} c_0=\frac {m_2-m_1}{m_1+m_2}. \end{equation*}
3. The computations
Let
$N^{n+1}\subset S^{n+2}$
be a minimal isoparametric hypersurface. To check the ACS condition, it is enough to study the pointwise integrand in (1).
Fix a closed minimal hypersurface
$M^n\subset N^{n+1}$
. Let
$\nu$
be a local unit normal vector field of
$M$
in
$N$
, and let
$\{e_k\}_{k=1}^n$
be a local orthonormal frame on
$M$
. We define
\begin{equation} \begin{aligned} \Delta (X,\nu )\;:\!=\;{}&\sum _{k=1}^n R^N(e_k,X,e_k,X)+\operatorname {Ric}_N\!(\nu ,\nu )|X|^2 \\ &-\sum _{k=1}^n |\Pi (e_k,X)|^2-\sum _{k=1}^n |\Pi (e_k,\nu )|^2|X|^2. \end{aligned} \end{equation}
If
$\Delta (X,\nu )\gt 0$
for every nonzero tangent vector
$X$
, then
$N$
satisfies the ACS condition.
Let
$\{\varepsilon _i\}_{i=1}^{n+1}$
be a local orthonormal frame of principal directions of
$N\subset S^{n+2}$
, with corresponding principal curvatures
$\{\lambda _i\}_{i=1}^{n+1}$
. Write
\begin{equation*} X=\sum _{i=1}^{n+1} x_i\varepsilon _i, \qquad \nu =\sum _{i=1}^{n+1} \nu _i\varepsilon _i. \end{equation*}
Since (2) is homogeneous in
$X$
, we may assume
$|X|=1$
. We also have
$|\nu |=1$
and
$\langle X,\nu \rangle =0$
.
Proposition 3.1. With the notation above,
\begin{align*} \Delta (X,\nu ) ={}&(2n-2)-2\sum _{i=1}^{n+1} x_i^2\lambda _i^2-2\sum _{i=1}^{n+1} \nu _i^2\lambda _i^2 +2\bigg (\sum _{i=1}^{n+1} x_i\nu _i\lambda _i\bigg )^2 \\ &\qquad +\bigg (\sum _{i=1}^{n+1} \nu _i^2\lambda _i\bigg )^2 -\bigg (\sum _{i=1}^{n+1} \nu _i^2\lambda _i\bigg ) \bigg (\sum _{i=1}^{n+1} x_i^2\lambda _i\bigg ). \end{align*}
Proof.
Let
$\eta$
be a local unit normal vector field of
$N$
in
$S^{n+2}$
, and let
$\xi$
be the unit normal vector field of
$S^{n+2}$
in
${\mathbb{R}}^{n+3}$
. Then, for tangent vectors
$U,V\in TN$
,
\begin{equation*} \Pi (U,V)=\Pi ^{\eta }(U,V)+\Pi ^{\xi }(U,V)=\langle AU,V\rangle \eta +\langle U,V\rangle \xi , \end{equation*}
where
$A$
is the shape operator of
$N\subset S^{n+2}$
. In the principal frame,
\begin{equation*} A(\varepsilon _i)=\lambda _i\varepsilon _i, \qquad \Pi (\varepsilon _i,\varepsilon _j)=\lambda _i\delta _{ij}\eta +\delta _{ij}\xi . \end{equation*}
We first compute the curvature term. Since
$\{e_1,\ldots ,e_n,\nu \}$
is an orthonormal frame of
$TN$
,
\begin{equation*} \sum _{k=1}^n R^N(e_k,X,e_k,X)=\operatorname {Ric}_N(X,X)-R^N(\nu ,X,\nu ,X). \end{equation*}
By the Gauss equation for
$N\subset S^{n+2}$
,
\begin{equation*} K(\varepsilon _i,\varepsilon _j)=1+\lambda _i\lambda _j \qquad (i\neq j), \end{equation*}
and the minimality of
$N$
implies
$\sum _i \lambda _i=0$
. Therefore,
\begin{align*} \operatorname {Ric}_N(\varepsilon _i,\varepsilon _i) &=\sum _{j\ne i} K(\varepsilon _j,\varepsilon _i) =\sum _{j\ne i}(1+\lambda _j\lambda _i) =n+\lambda _i\sum _{j\ne i}\lambda _j =n-\lambda _i^2. \end{align*}
Hence
\begin{equation} \operatorname {Ric}_N(X,X)=\sum _{i=1}^{n+1}(n-\lambda _i^2)x_i^2. \end{equation}
On the other hand,
\begin{align*} \Pi (\nu ,\nu ) &=\bigg (\sum _{i=1}^{n+1}\nu _i^2\lambda _i\bigg )\eta +\xi , \\ \Pi (X,X) &=\bigg (\sum _{i=1}^{n+1}x_i^2\lambda _i\bigg )\eta +\xi , \\ \Pi (\nu ,X) &=\bigg (\sum _{i=1}^{n+1}x_i\nu _i\lambda _i\bigg )\eta , \end{align*}
since
$\langle \nu ,X\rangle =0$
. Applying the Gauss equation again, we obtain
\begin{align} R^N(\nu ,X,\nu ,X) &=\langle \Pi (\nu ,\nu ),\Pi (X,X)\rangle -|\Pi (\nu ,X)|^2 \notag \\ &=\bigg (\sum _{i=1}^{n+1}\nu _i^2\lambda _i\bigg ) \bigg (\sum _{i=1}^{n+1}x_i^2\lambda _i\bigg ) +1 -\bigg (\sum _{i=1}^{n+1}x_i\nu _i\lambda _i\bigg )^2. \end{align}
\begin{align} \sum _{k=1}^n R^N(e_k,X,e_k,X) ={}&\sum _{i=1}^{n+1}(n-\lambda _i^2)x_i^2 -\Big (\sum _{i=1}^{n+1}\nu _i^2\lambda _i\Big ) \bigg (\sum _{i=1}^{n+1}x_i^2\lambda _i\bigg ) \notag \\ &\qquad -1 +\bigg (\sum _{i=1}^{n+1}x_i\nu _i\lambda _i\bigg )^2. \end{align}
Moreover,
\begin{equation} \operatorname {Ric}_N(\nu ,\nu )|X|^2= \sum _{i=1}^{n+1}(n-\lambda _i^2)\nu _i^2. \end{equation}
We next compute the second fundamental form terms. Since
$\{e_1,\ldots ,e_n,\nu \}$
is an orthonormal basis of
$TN$
,
\begin{align*} \sum _{k=1}^n |\Pi (e_k,X)|^2 &=\sum _{i=1}^{n+1}|\Pi (\varepsilon _i,X)|^2-|\Pi (\nu ,X)|^2, \\ \sum _{k=1}^n |\Pi (e_k,\nu )|^2 &=\sum _{i=1}^{n+1}|\Pi (\varepsilon _i,\nu )|^2-|\Pi (\nu ,\nu )|^2. \end{align*}
Now,
\begin{equation*} \Pi (\varepsilon _i,X)=x_i\lambda _i\eta +x_i\xi , \qquad \Pi (\varepsilon _i,\nu )=\nu _i\lambda _i\eta +\nu _i\xi , \end{equation*}
so that
\begin{equation*} |\Pi (\varepsilon _i,X)|^2=x_i^2(\lambda _i^2+1), \qquad |\Pi (\varepsilon _i,\nu )|^2=\nu _i^2(\lambda _i^2+1). \end{equation*}
Therefore,
\begin{align} \sum _{k=1}^n |\Pi (e_k,X)|^2 +\sum _{k=1}^n |\Pi (e_k,\nu )|^2|X|^2 ={}&\sum _{i=1}^{n+1}x_i^2(\lambda _i^2+1) +\sum _{i=1}^{n+1}\nu _i^2(\lambda _i^2+1) \notag \\ &-\bigg (\sum _{i=1}^{n+1}x_i\nu _i\lambda _i\bigg )^2 \notag \\ &-\Bigg [\bigg (\sum _{i=1}^{n+1}\nu _i^2\lambda _i\bigg )^2+1\Bigg ]. \end{align}
Finally, substituting (5), (6), and (7) into (2), and using
$\sum _i x_i^2=\sum _i \nu _i^2=1$
, we obtain the claimed formula.
Proposition 3.2.
Assume
$g=4$
, set
\begin{equation*} q\;:\!=\;\min \{m_1,m_2\}=4, \qquad s\;:\!=\;m_1+m_2, \end{equation*}
and let
\begin{equation*} a^2=\frac {1+\sqrt {1-4/s}}{1-\sqrt {1-4/s}}. \end{equation*}
If
$s\ge 15$
, then for every orthogonal unit vectors
$X,\nu \in T_pN$
one has
\begin{equation*} \Delta (X,\nu )\ge 4s-4-4a^2\gt 0. \end{equation*}
In particular,
$N$
satisfies the ACS condition.
Proof.
After interchanging
$m_1$
and
$m_2$
if necessary, we may assume
\begin{equation*} m_1=4\le m_2. \end{equation*}
For the minimal leaf with
$g=4$
, we have
\begin{equation*} \cos 2\theta _1=\sqrt {\frac {m_2}{s}}, \qquad \sin 2\theta _1=\sqrt {\frac {m_1}{s}}=\sqrt {\frac {4}{s}}, \end{equation*}
and therefore
\begin{equation*} \lambda _1^2 =\frac {1+\sqrt {m_2/s}}{1-\sqrt {m_2/s}}, \qquad \lambda _4^2 =\frac {1+\sqrt {m_1/s}}{1-\sqrt {m_1/s}}. \end{equation*}
Since
$m_2\ge m_1$
and the function
\begin{equation*} t\longmapsto \frac {1+\sqrt t}{1-\sqrt t} \end{equation*}
is strictly increasing on
$(0,1)$
, it follows that
$\lambda _1^2\ge \lambda _4^2$
. By definition of
$a$
, we therefore have
\begin{equation*} a=\lambda _1=\cot \theta _1\gt 0. \end{equation*}
Moreover,
$\lambda _1$
has multiplicity
$m_1=4$
, so the
$a$
-eigenspace has dimension exactly
$4$
.
Now write
$\lambda _4=-b\lt 0$
. Then,
\begin{equation*} b\;:\!=\;-\lambda _4 =-\cot \Bigl (\theta _1+\frac {3\pi }{4}\Bigr ) =\tan \Bigl (\theta _1+\frac {\pi }{4}\Bigr ) =\frac {a+1}{a-1}. \end{equation*}
Hence, all principal curvatures lie in the interval
$[{-}b,a]$
.
Next we explain why
$b\le a/2$
when
$s\ge 15$
. Set
\begin{equation*} x(s)\;:\!=\;\sqrt {1-\frac {4}{s}}\in (0,1). \end{equation*}
Then,
\begin{equation*} a^2=\frac {1+x(s)}{1-x(s)}. \end{equation*}
Since
$x(s)$
is increasing in
$s$
and the function
$x\mapsto \frac {1+x}{1-x}$
is strictly increasing on
$(0,1)$
, the quantity
$a^2$
is increasing in
$s$
. Hence, the assumption
$s\ge 15$
gives
\begin{equation*} a^2\ge \frac {1+\sqrt {11/15}}{1-\sqrt {11/15}} =\frac {13+\sqrt {165}}{2} \gt \frac {13+3\sqrt {17}}{2} =\left (\frac {3+\sqrt {17}}{2}\right )^2. \end{equation*}
Therefore,
\begin{equation*} a\gt \frac {3+\sqrt {17}}{2}. \end{equation*}
On the other hand, since
$a\gt 1$
and
$b=\frac {a+1}{a-1}$
, we have
\begin{equation*} b\le \frac {a}{2} \iff \frac {a+1}{a-1}\le \frac {a}{2} \iff 2(a+1)\le a(a-1) \iff a^2-3a-2\ge 0 \iff a\ge \frac {3+\sqrt {17}}{2}. \end{equation*}
Thus, the previous inequality for
$a$
implies
\begin{equation*} b\le \frac {a}{2}. \end{equation*}
Now define
\begin{equation*} A\;:\!=\;\sum _{i=1}^{n+1}x_i^2\lambda _i, \qquad B\;:\!=\;\sum _{i=1}^{n+1}\nu _i^2\lambda _i, \qquad C\;:\!=\;\sum _{i=1}^{n+1}x_i\nu _i\lambda _i, \end{equation*}
and
\begin{equation*} S_x\;:\!=\;\sum _{i=1}^{n+1}x_i^2\lambda _i^2, \qquad S_\nu \;:\!=\;\sum _{i=1}^{n+1}\nu _i^2\lambda _i^2. \end{equation*}
By Proposition 3.1,
\begin{equation*} \Delta =(2n-2)-2S_x-2S_\nu +2C^2+B^2-AB. \end{equation*}
Equivalently,
\begin{equation*} \Delta =(2n-2)-F+2C^2, \qquad F\;:\!=\;2S_x+2S_\nu -B^2+AB. \end{equation*}
Since each principal curvature
$\lambda _i$
lies in
$[{-}b,a]$
, we have
\begin{equation*} (\lambda _i-a)(\lambda _i+b)\le 0, \end{equation*}
hence
\begin{equation*} \lambda _i^2\le (a-b)\lambda _i+ab. \end{equation*}
Averaging against the weights
$x_i^2$
and
$\nu _i^2$
, we obtain
\begin{equation*} S_x\le (a-b)A+ab, \qquad S_\nu \le (a-b)B+ab. \end{equation*}
Therefore,
\begin{equation*} F\le G(A,B), \end{equation*}
where
\begin{equation*} G(A,B)\;:\!=\;2(a-b)A+2(a-b)B+4ab-B^2+AB. \end{equation*}
Since
$A,B\in [{-}b,a]$
, we have
\begin{equation*} \frac {\partial G}{\partial A}=2(a-b)+B\ge 2(a-b)-b=2a-3b\ge \frac {a}{2}\gt 0, \end{equation*}
because
$b\le a/2$
. Thus
$G$
is strictly increasing in
$A$
throughout the square
$[{-}b,a]\times [{-}b,a]$
, and so its maximum is attained at
$A=a$
. Hence
\begin{equation*} G(A,B)\le G(a,B) =-B^2+(3a-2b)B+2a^2+2ab. \end{equation*}
This is a concave quadratic in
$B$
, with axis
\begin{equation*} B_0=\frac {3a-2b}{2}=\frac {3a}{2}-b\ge a, \end{equation*}
since
$b\le a/2$
. It follows that
$G(a,B)$
is increasing on
$[{-}b,a]$
, and therefore
\begin{equation*} F\le G(A,B)\le G(a,a)=4a^2. \end{equation*}
Consequently,
\begin{equation*} \Delta (X,\nu )\ge (2n-2)-4a^2. \end{equation*}
Since
$n=2s-1$
, this becomes
\begin{equation*} \Delta (X,\nu )\ge 4s-4-4a^2. \end{equation*}
Finally, let
\begin{equation*} x\;:\!=\;\sqrt {1-\frac {4}{s}}\in (0,1). \end{equation*}
Then,
\begin{equation*} a^2=\frac {1+x}{1-x}, \qquad s=\frac {4}{1-x^2}, \end{equation*}
so
\begin{equation*} s-1-a^2 = \frac {3+x^2}{1-x^2}-\frac {1+x}{1-x} = \frac {2-2x}{1-x^2} = \frac {2}{1+x}\gt 0. \end{equation*}
Hence,
\begin{equation*} 4s-4-4a^2=4(s-1-a^2)\gt 0, \end{equation*}
and therefore
$\Delta (X,\nu )\gt 0$
for every orthogonal unit pair
$(X,\nu )$
.
Moreover, this lower bound is sharp. Indeed, the
$a$
-eigenspace has dimension
$q=4$
, so we may choose orthogonal unit vectors
$X,\nu$
in that eigenspace. Then,
\begin{equation*} A=B=a,\qquad S_x=S_\nu =a^2,\qquad C=0, \end{equation*}
and Proposition 3.1 gives
\begin{equation*} \Delta (X,\nu )=4s-4-4a^2. \end{equation*}
Proof of Theorem 1.2. For simplicity, write
\begin{equation*} A\;:\!=\;\sum _{i=1}^{n+1} x_i^2\lambda _i, \qquad B\;:\!=\;\sum _{i=1}^{n+1} \nu _i^2\lambda _i, \qquad C\;:\!=\;\sum _{i=1}^{n+1} x_i\nu _i\lambda _i, \end{equation*}
and
\begin{equation*} S_x\;:\!=\;\sum _{i=1}^{n+1} x_i^2\lambda _i^2, \qquad S_\nu \;:\!=\;\sum _{i=1}^{n+1} \nu _i^2\lambda _i^2. \end{equation*}
Then, Proposition 3.1 gives
\begin{equation*} \Delta =(2n-2)-2S_x-2S_\nu +2C^2+B^2-AB. \end{equation*}
It suffices to show that
$\Delta \gt 0$
.
(1) The case
$g=3$
. For a minimal isoparametric hypersurface with
$g=3$
, the principal curvatures are
\begin{equation*} \lambda _1=\sqrt {3},\qquad \lambda _2=0,\qquad \lambda _3=-\sqrt {3}. \end{equation*}
If the common multiplicity is
$m\in \{4,8\}$
, then
\begin{equation*} n=3m-1\ge 11, \qquad 2n-2\ge 20. \end{equation*}
Since
$|X|=|\nu |=1$
and
$\langle X,\nu \rangle =0$
, we have
\begin{equation*} \sum _i x_i^2=\sum _i \nu _i^2=1, \qquad \sum _i x_i\nu _i=0. \end{equation*}
Hence,
\begin{equation*} \sum _i x_i^2\lambda _i^2\le 3, \qquad \sum _i \nu _i^2\lambda _i^2\le 3, \qquad \left |\bigg (\sum _i \nu _i^2\lambda _i\bigg )\bigg (\sum _i x_i^2\lambda _i\bigg )\right |\le 3. \end{equation*}
Discarding the nonnegative terms in the expression for
$\Delta$
, we obtain
\begin{equation*} \Delta \ge (2n-2)-6-6-3\ge 20-6-6-3\gt 0. \end{equation*}
Therefore,
$N$
satisfies the ACS condition when
$g=3$
and
$m\in \{4,8\}$
.
(2) The case
$g=4$
. Set
\begin{equation*} s\;:\!=\;m_1+m_2, \qquad n=2s-1. \end{equation*}
By Theorem2.1, the principal curvatures are
\begin{equation*} \lambda _i=\cot \bigg (\theta _1+\frac {i-1}{4}\pi \bigg ), \qquad i=1,2,3,4, \end{equation*}
where, for the minimal leaf,
\begin{equation*} \cos 4\theta _1=c_0=\frac {m_2-m_1}{m_1+m_2}. \end{equation*}
Let
\begin{equation*} a^2\;:\!=\;\max _{1\le i\le 4}\lambda _i^2=\max \{\lambda _1^2,\lambda _4^2\}. \end{equation*}
Since
\begin{equation*} S_x\le a^2, \qquad S_\nu \le a^2, \qquad |A|\le a, \qquad |B|\le a, \end{equation*}
we first record the rough estimate obtained by discarding the nonnegative terms
$2C^2$
and
$B^2$
and bounding
$|AB|\le a^2$
:
\begin{equation} \Delta \ge (2n-2)-5a^2. \end{equation}
Thus, it is enough to require
\begin{equation} a^2\lt \frac {2n-2}{5}=\frac {4}{5}(s-1). \end{equation}
We do not assume any ordering between
$m_1$
and
$m_2$
. Since
$0\lt 2\theta _1\lt \pi /2$
, we have
\begin{equation*} \cos 2\theta _1=\sqrt {\frac {1+\cos 4\theta _1}{2}}=\sqrt {\frac {m_2}{s}}, \qquad \sin 2\theta _1=\sqrt {\frac {1-\cos 4\theta _1}{2}}=\sqrt {\frac {m_1}{s}}. \end{equation*}
Therefore,
\begin{align*} \lambda _1^2 &=\cot ^2\theta _1 =\frac {1+\cos 2\theta _1}{1-\cos 2\theta _1} =\frac {1+\sqrt {m_2/s}}{1-\sqrt {m_2/s}}, \\ \lambda _4^2 &=\cot ^2\bigg (\theta _1+\frac {3\pi }{4}\bigg ) =\frac {1+\sin 2\theta _1}{1-\sin 2\theta _1} =\frac {1+\sqrt {m_1/s}}{1-\sqrt {m_1/s}}. \end{align*}
If we set
\begin{equation*} p\;:\!=\;\max \{m_1,m_2\}, \qquad q\;:\!=\;\min \{m_1,m_2\}=s-p, \end{equation*}
then
\begin{equation*} a^2=\frac {1+\sqrt {p/s}}{1-\sqrt {p/s}}. \end{equation*}
The rough estimate (8) gives the range
$q\ge 5$
as follows. Condition (9) is equivalent to
\begin{equation*} \sqrt {\frac {p}{s}}\lt \frac {\frac 45(s-1)-1}{\frac 45(s-1)+1}, \end{equation*}
and hence to
\begin{equation} q=s-p\gt s\left [1-\left (\frac {\frac 45(s-1)-1}{\frac 45(s-1)+1}\right )^2\right ] =\frac {80s(s-1)}{(4s+1)^2}. \end{equation}
Now,
\begin{equation*} 5(4s+1)^2-80s(s-1)=120s+5\gt 0, \end{equation*}
so
\begin{equation*} \frac {80s(s-1)}{(4s+1)^2}\lt 5. \end{equation*}
Thus, (10) is certainly satisfied if
$q\ge 5$
, i.e. if
\begin{equation*} \min \{m_1,m_2\}\ge 5. \end{equation*}
To handle the borderline case
$q=4$
, we keep the term
$B^2$
instead of discarding it. Since
$2C^2\ge 0$
,
\begin{equation*} \Delta \ge (2n-2)-2S_x-2S_\nu +B^2-AB. \end{equation*}
Completing the square yields
\begin{equation*} B^2-AB=\left (B-\frac {A}{2}\right )^2-\frac {A^2}{4}\ge -\frac {A^2}{4}. \end{equation*}
Moreover, since
$\sum _i x_i^2=1$
, the Cauchy–Schwarz inequality gives
\begin{equation*} A^2=\left (\sum _i x_i^2\lambda _i\right )^2\le \sum _i x_i^2\lambda _i^2=S_x. \end{equation*}
Therefore,
\begin{equation} \Delta \ge (2n-2)-\tfrac {9}{4}\,S_x-2S_\nu \ge (2n-2)-\tfrac {17}{4}\,a^2. \end{equation}
Hence, it is enough to require
\begin{equation} a^2\lt \frac {4}{17}(2n-2)=\frac {16}{17}(s-1). \end{equation}
If
$q=4$
, then
$p=s-4$
, so
\begin{equation*} a^2=\frac {1+\sqrt {1-4/s}}{1-\sqrt {1-4/s}}. \end{equation*}
Substituting this into (12) and simplifying, we obtain
\begin{equation*} 16s^2-304s-1\lt 0. \end{equation*}
For integers
$s$
, this holds exactly when
$s\le 19$
: the left-hand side is increasing for
$s\ge 10$
, equals
$-1$
at
$s=19$
, and equals
$319$
at
$s=20$
. Therefore, (11) yields
$\Delta \gt 0$
whenever
$q=4$
and
$s\le 19$
. On the other hand, for
$q=4$
and
$s\ge 15$
, Proposition 3.2 gives
\begin{equation*} \Delta (X,\nu )\ge 4s-4-4a^2\gt 0. \end{equation*}
Since the ranges
$s\le 19$
and
$s\ge 15$
cover all
$s\ge 8$
, the ACS condition holds for every multiplicity pair with
$q=4$
.
This proves the ACS condition in all cases listed in Theorem1.2. The index estimate then follows immediately from Theorem1.1.
Remark 3.3.
The argument above leaves the following cases unresolved by this method: the case
$g=3$
with multiplicity
$m=2$
, and the case
$g=4$
when at least one multiplicity belongs to
$\{2,3\}$
. The cases
$g=3$
with
$m=1$
and
$g=6$
are excluded because the ambient Ricci curvature is not positive.
Remark 3.4.
The bound
$\min \{m_1,m_2\}\ge 4$
in Theorem
1.2
(2) is sharp with respect to the pointwise method. Indeed, the principal curvature whose absolute value equals
$a$
has multiplicity
$q\;:\!=\;\min \{m_1,m_2\}$
. Since positive Ricci curvature forces
$q\ge 2$
in the case
$g=4$
, the corresponding eigenspace has dimension
$\ge 2$
, and we may choose orthogonal unit vectors
$X,\nu$
supported entirely in that eigenspace. For such a choice,
\begin{equation*} C=0, \qquad S_x=S_\nu =a^2, \qquad A=B=\lambda \quad ({where } |\lambda |=a), \end{equation*}
and the terms
$B^2-AB$
and
$2C^2$
all vanish. Proposition 3.1 therefore gives
\begin{equation*} \Delta (X,\nu )=(2n-2)-4a^2. \end{equation*}
Since pointwise positivity requires
$\Delta (X,\nu )\gt 0$
for
$\mathrm{every}$
orthogonal unit pair
$(X,\nu )$
, and the configuration above attains the value
$(2n-2)-4a^2$
, such positivity forces
$(2n-2)-4a^2\gt 0$
. Substituting
\begin{equation*} a^2=\frac {1+\sqrt {1-q/s}}{1-\sqrt {1-q/s}}, \qquad s=m_1+m_2, \end{equation*}
and simplifying, this condition is equivalent to
$q\gt 4-4/s$
and hence to
$q\ge 4$
for integer
$q$
and
$s\ge 4$
. Thus,
$\min \{m_1,m_2\}\ge 4$
is a
$\mathrm{necessary}$
condition for any argument based only on pointwise positivity of
$\Delta$
.
In the present paper, this necessary condition is also shown to be sufficient. Indeed, the completed-square estimate (
11
) covers the range
$8\le s\le 19$
, while Proposition 3.2 proves that for
$s\ge 15$
the moment-relaxation bound
$F\le 4a^2$
is achieved precisely at the extremal configuration above. Since
\begin{equation*} 4(s-1)-4a^2=\frac {8}{1+\sqrt {1-4/s}}\gt 0, \end{equation*}
the argument closes for every
$s\ge 8$
, i.e. for every multiplicity pair with
$q=4$
.
