Lichnerowicz-Obata Estimate, Almost Parallel p-form and Almost Product Manifolds

Abstract

We show a Lichnerowicz-Obata type estimate for the first eigenvalue of the Laplacian of n-dimensional closed Riemannian manifolds with an almost parallel p-form ($2\le p \le n/2$) in $L^2$-sense, and give a Gromov-Hausdorff approximation to a product $S^{n-p}\times X$ under some pinching conditions when $2\le p<n/2$.

1 Introduction

In this paper we give an estimate for the first eigenvalue of the Laplacian of closed Riemannian manifolds with positive Ricci curvature and an almost parallel form, and show the Gromov-Hausdorff closeness to a product space for the almost equality case.

One of the most famous theorem about the estimate of the first eigenvalue of the Laplacian is the Lichnerowicz-Obata theorem. Lichnerowicz showed the optimal comparison result for the first eigenvalue when the Riemannian manifold has positive Ricci curvature, and Obata showed that the equality of the Lichnerowicz estimate implies that the Riemannian manifold is isometric to the standard sphere. In the following, $\lambda _k(g)$ denotes the k-th eigenvalue of the Laplacian $\Delta :=-{\mathrm {tr}}_g {\mathrm {Hess}}$ acting on functions.

Theorem 1.1

(Lichnerowicz-Obata theorem) Take an integer $n\ge 2$. Let (M, g) be an n-dimensional closed Riemannian manifold. If ${\mathrm {Ric}}\ge (n-1) g$, then $\lambda _1(g)\ge n$. The equality holds if and only if (M, g) is isometric to the standard sphere of radius 1.

Petersen [19], Aubry [3] and Honda [13] showed the stability result of the Lichnerowicz-Obata theorem. In the following, $d_{GH}$ denotes the Gromov-Hausdorff distance and $S^n$ denotes the n-dimensional standard sphere of radius 1. (see Definition 2.2 for the definition of the Gromov-Hausdorff distance).

Theorem 1.2

([3, 13, 19]) For given an integer $n\ge 2$ and a positive real number $\epsilon >0$, there exists $\delta (n,\epsilon )>0$ such that if (M, g) is an n-dimensional closed Riemannian manifold with ${\mathrm {Ric}}\ge (n-1) g$ and $\lambda _n(g)\le n+\delta$, then $d_{GH}(M,S^n)\le \epsilon$.

Note that Petersen considered the pinching condition on $\lambda _{n+1}(g)$, and Aubry and Honda improved it independently.

We mention some improvements of the Lichnerowicz estimate when the Riemannian manifold has a special structure. If (M, g) is a real n-dimensional Kähler manifold with ${\mathrm {Ric}}\ge (n-1)g$, then the Lichnerowicz estimate is improved as follows:

$$\begin{aligned} \lambda _1(g)\ge 2(n-1). \end{aligned}$$

(1)

See [4, Theorem 11.49] for the proof. If (M, g) is a real n-dimensional quaternionic Kähler manifold with ${\mathrm {Ric}}\ge (n-1)g$, then we have

$$\begin{aligned} \lambda _1(g)\ge \frac{2n+8}{n+8}(n-1). \end{aligned}$$

(2)

See [2] for the proof. In these cases, the Riemannian manifold (M, g) has a non-trivial parallel 2 and 4-form, respectively. When (M, g) is an n-dimensional product Riemannian manifold $(N_1\times N_2,g_1+g_2)$ with ${\mathrm {Ric}}\ge (n-1)g$, then we have

$$\begin{aligned} \lambda _1(g)\ge \min _{i\in \{1,2\}}\left\{ \frac{\dim N_i}{\dim N_i-1}\right\} (n-1), \end{aligned}$$

and M has a non-trivial parallel form if either $N_1$ or $N_2$ is orientable.

Grosjean [12] gave a unified proof of the improvements of the Lichnerowicz estimate when the Riemannian manifold has a non-trivial parallel form.

Theorem 1.3

([12]) Let (M, g) be an n-dimensional closed Riemannian manifold. Assume that ${\mathrm {Ric}}\ge (n-p-1)g$ and that there exists a nontrivial parallel p-form on M $(2\le p\le n/2)$. Then, we have

$$\begin{aligned} \lambda _1(g)\ge n-p. \end{aligned}$$

(3)

Moreover, if $p<n/2$ and if in addition M is simply connected, then the equality in (3) implies that (M, g) is isometric to a product $S^{n-p}\times (X,g')$, where $(X,g')$ is some p-dimensional closed Riemannian manifold.

Remark 1.1

We give several remarks on this theorem.

• When ${\mathrm {Ric}}\ge (n-p-1)g$, the Lichnerowicz estimate is $\lambda _1(g)\ge n(n-p-1)/(n-1)$. Since $n-p>n(n-p-1)/(n-1)$ for $2\le p\le n/2$, the estimate (3) improves the Lichnerowicz estimate.

• Grosjean also showed this type theorem when M has a convex smooth boundary.

• Though Grosjean originally assumed the manifold is orientable, the assumption can be easily removed by taking the orientable double covering.

• If (M, g) is either a Kähler manifold with $n\ge 6$ or a quaternionic Kähler manifold, then the estimate (1) or (2) (with scaling) is stronger than (3).

• There exists no non-trivial parallel 1-form on any closed Riemannian manifold with positive Ricci curvature.

• The assumption $2\le p\le n/2$ (resp. $2\le p< n/2$) implies $n\ge 4$ (resp. $n\ge 5$). For the case $n=4$ and $p=n/2=2$, the complex projective space ${\mathbb {C}}P^2$ also satisfies the equality in (3).

• If there exists a non-trivial parallel p-form $\omega$ ($1\le p\le n-1$) on an n-dimensional Riemannian manifold (M, g), then $\omega (x)\in \bigwedge ^p T^*_x M$ ($x\in M$) is invariant under the Holonomy action, and so the Holonomy group coincides with neither $\mathrm {SO}(n)$ nor $\mathrm {O}(n)$.

The main aim of this paper is to show the almost version of Grosjean’s result. We also give the almost version of the estimate (1) in Appendix B.

We first note that, for a closed Riemannian manifold (M, g), there exists a non zero p-form $\omega$ with $\Vert \nabla \omega \Vert _2^2\le \delta \Vert \omega \Vert _2^2$ for some $\delta >0$ if and only if $\lambda _1(\Delta _{C,p})\le \delta$ holds, where $\lambda _1(\Delta _{C,p})$ is defined by

$$\begin{aligned} \lambda _1(\Delta _{C,p}):=\inf \left\{ \frac{\Vert \nabla \omega \Vert _2^2}{\Vert \omega \Vert _2^2}: \omega \in \Gamma (\bigwedge ^p T^*M) \text { with }\omega \ne 0\right\} . \end{aligned}$$

Let us state our eigenvalue estimate.

Main Theorem 1

For given integers $n\ge 4$ and $2\le p \le n/2$, there exists a constant $C(n,p)>0$ such that if (M, g) is an n-dimensional closed Riemannian manifold with ${\mathrm {Ric}}_g\ge (n-p-1)g$, then we have

$$\begin{aligned} \lambda _1(g)\ge n-p-C(n,p)\lambda _1(\Delta _{C,p})^{1/2}. \end{aligned}$$

We immediately have the following corollary:

Corollary 1.4

For given integers $n\ge 4$ and $2\le p \le n/2$, there exists a constant $C(n,p)>0$ such that if (M, g) is an n-dimensional closed Riemannian manifold with ${\mathrm {Ric}}_g\ge (n-p-1)g$ and

$$\begin{aligned} \frac{n(n-p-1)}{n-1}\le \lambda _1(g)\le n-p, \end{aligned}$$

then we have

$$\begin{aligned} \lambda _1(\Delta _{C,p})\ge \left( \frac{n-p-\lambda _1(g)}{C(n,p)}\right) ^2. \end{aligned}$$

Note that we always have the lower bound on the eigenvalue of the Laplacian $\lambda _1(g)\ge n(n-p-1)/(n-1)$ if ${\mathrm {Ric}}_g\ge (n-p-1)g$ by the Lichnerowicz estimate. An upper bound on C(n, p) is computable. However, we do not know the optimal value of it.

We next state the eigenvalue pinching result.

Main Theorem 2

For given integers $n\ge 5$ and $2\le p < n/2$ and a positive real number $\epsilon >0$, there exists $\delta =\delta (n,p,\epsilon )>0$ such that if (M, g) is an n-dimensional closed Riemannian manifold with ${\mathrm {Ric}}_g\ge (n-p-1)g$,

$$\begin{aligned} \lambda _{n-p+1}(g)\le n-p+\delta \end{aligned}$$

and

$$\begin{aligned} \lambda _1(\Delta _{C,p})\le \delta , \end{aligned}$$

then M is orientable and

$$\begin{aligned} d_{GH}(M,S^{n-p}\times X)\le \epsilon , \end{aligned}$$

where X is some compact metric space.

Remark 1.2

In fact, we prove that there exist constants $C(n,p)>0$ and $\alpha (n)>0$ such that

$$\begin{aligned} d_{GH}(M,S^{n-p}\times X)\le C(n,p)\delta ^{\alpha (n)} \end{aligned}$$

under the assumption of Main Theorem 2. One can easily find the explicit value of $\alpha (n)$ (see Notation 4.35 and Theorem 4.47). However, it might be far from the optimal value. By the Gromov’s pre-compactness theorem, we can take X to be a geodesic space. However, we lose the information about the convergence rate in that case.

Based on Theorem 1.2, one might expect that we can replace the assumption “$\lambda _{n-p+1}(g)\le n-p+\delta$” in Main Theorem 2 to the weaker assumption “$\lambda _{n-p}(g)\le n-p+\delta$”. However, an example shows that we cannot do it even if $\delta =0$ (see Proposition 3.3). Instead of that, replacing $\lambda _1(\Delta _{C,p})$ to $\lambda _1(\Delta _{C,n-p})$, we have the following theorems:

Main Theorem 3

For given integers $n\ge 4$ and $2\le p \le n/2$, there exists a constant $C(n,p)>0$ such that if (M, g) is an n-dimensional closed Riemannian manifold with ${\mathrm {Ric}}_g\ge (n-p-1)g$, then we have

$$\begin{aligned} \lambda _1(g)\ge n-p-C(n,p)\lambda _1(\Delta _{C,n-p})^{1/2}. \end{aligned}$$

Main Theorem 4

For given integers $n\ge 5$ and $2\le p < n/2$ and a positive real number $\epsilon >0$, there exists $\delta =\delta (n,p,\epsilon )>0$ such that if (M, g) is an n-dimensional closed Riemannian manifold with ${\mathrm {Ric}}_g\ge (n-p-1)g$,

$$\begin{aligned} \lambda _{n-p}(g)\le n-p+\delta \end{aligned}$$

and

$$\begin{aligned} \lambda _1(\Delta _{C,n-p})\le \delta , \end{aligned}$$

then we have

$$\begin{aligned} d_{GH}(M,S^{n-p}\times X)\le \epsilon , \end{aligned}$$

where X is some compact metric space.

Note that the assumption “$\lambda _1(\Delta _{C,n-p})\le \delta$” is equivalent to the assumption “$\lambda _1(\Delta _{C,p})\le \delta$” if the manifold is orientable.

We would like to point out that our work was motivated by Honda’s spectral convergence theorem [17], which asserts the continuity of the eigenvalues of the connection Laplacian $\Delta _{C,p}$ acting on p-forms with respect to the non-collapsing Gromov-Hausdorff convergence assuming the two-sided bound on the Ricci curvature. By virtue of his theorem, we can generalize our main theorems to Ricci limit spaces under such assumptions. Note that we show our main theorems without the non-collapsing assumption, i.e., without assuming the lower bound on the volume of the Riemannian manifold.

Our work was also motivated by the Cheeger-Colding almost splitting theorem (see [6, Theorem 9.25]), whose conclusion is the Gromov-Hausdorff approximation to a product ${\mathbb {R}}\times X$. As the almost splitting theorem, we need to show the almost Pythagorean theorem under the assumption of Main Theorem 2.

The structure of this paper is as follows.

In Sect. 2, we recall some basic definitions and facts, and give calculations of differential forms.

In Sect. 3, we estimate the error terms of the Grosjean’s formula when the Riemannian manifold has a non-trivial almost parallel p-form. As a consequence, we prove Main Theorem 1 and Main Theorem 3.

In Sect. 4, we prove Main Theorem 2 and Main Theorem 4. In Sect. 4.1, we list some useful techniques for our pinching problem. In Sect. 4.2, we show some pinching conditions on the eigenfunctions along geodesics under the assumption $\lambda _k(g)\le n-p+\delta$ and $\lambda _1(\Delta _{C,p})\le \delta$. In Sect. 4.3, we show that similar results hold under the assumption $\lambda _k(g)\le n-p+\delta$ and $\lambda _1(\Delta _{C,n-p})\le \delta$. In Sect. 4.4, we show that the eigenfunctions are almost cosine functions in some sense under our pinching condition. In Sect. 4.5, we construct an approximation map and show Main Theorem 2 except for the orientability. In Sect. 4.6, we give some lemmas to prove the remaining part of main theorems. In Sect. 4.7, we show the orientability of the manifold under the assumption of Main Theorem 2, and complete the proof of it. In Sect. 4.8, we show that the assumption of Main Theorem 4 implies that $\lambda _{n-p+1}(g)$ is close to $n-p$, and complete the proof of Main Theorem 4.

In Appendix A, we discuss Ricci limit spaces. Using the technique of Sect. 4.7, we show the stability of unorientability under the non-collapsing Gromov-Hausdorff convergence assuming the two-sided bound on the Ricci curvature and the upper bound on the diameter.

In Appendix B, we give the almost version of the estimate (1) assuming that there exists a 2-form $\omega$ which satisfies that $\Vert \nabla \omega \Vert _2$ and $\Vert J_\omega ^2+{\mathrm {Id}}\Vert _1$ are small, where $J_\omega \in \Gamma (T^*M\otimes T M)$ is defined so that $\omega =g(J_\omega \cdot ,\cdot )$.

2 Preliminaries

2.1 Basic Definitions

We first recall some basic definitions and fix our convention.

Definition 2.1

(Hausdorff distance) Let (X, d) be a metric space. For each point $x_0\in X$, subsets $A,B\subset X$ and $r>0$, define

$$\begin{aligned} d(x_0,A)&:=\inf \{d(x_0,a):a\in A\},\\ B_{r}(x_0)&:=\{x\in X: d(x,x_0)<r\},\\ B_{r}(A)&:=\{x\in X:d(x,A)<r\},\\ d_{H,d}(A,B)&:=\inf \{\epsilon >0:A\subset B_{\epsilon }(B) \text { and } B\subset B_{\epsilon }(A)\} \end{aligned}$$

We call $d_{H,d}$ the Hausdorff distance.

The Hausdorff distance defines a metric on the collection of compact subsets of X.

Definition 2.2

(Gromov-Hausdorff distance) Let $(X,d_X),(Y,d_Y)$ be metric spaces. Define

$$\begin{aligned}&d_{GH}(X,Y):=\inf \Big \{d_{H,d}(X,Y): d\text { is a metric on }X\coprod Y\text { such that}\\&\qquad \qquad \quad d|_X=d_X\text { and }d|_Y=d_Y\Big \}. \end{aligned}$$

The Gromov-Hausdorff distance defines a metric on the set of isometry classes of compact metric spaces (see [20, Proposition 11.1.3]).

Definition 2.3

($\epsilon$-Hausdorff approximation map) Let $(X,d_X),(Y,d_Y)$ be metric spaces. We say that a map $f:X\rightarrow Y$ is an $\epsilon$-Hausdorff approximation map for $\epsilon >0$ if the following two conditions hold.

• (i) For all $a,b\in X$, we have $|d_X(a,b)-d_Y(f(a),f(b))|< \epsilon$,

• (ii) f(X) is $\epsilon$-dense in Y, i.e., for all $y\in Y$, there exists $x\in X$ with $d_Y(f(x),y)< \epsilon$.

If there exists an $\epsilon$-Hausdorff approximation map $f:X\rightarrow Y$, then we can show that $d_{GH}(X,Y)\le 3\epsilon /2$ by considering the following metric d on $X\coprod Y$:

$$\begin{aligned} d(a,b)=\left\{ \begin{array}{cl} d_X(a,b)&{}\quad (a,b\in X),\\ \frac{\epsilon }{2} +\inf _{x\in X}(d_X(a,x)+d_Y(f(x),b))&{}\quad (a\in X,\,b\in Y),\\ d_Y(a,b)&{}\quad (a,b\in Y). \end{array}\right. \end{aligned}$$

If $d_{GH}(X,Y)< \epsilon$, then there exists a $2\epsilon$-Hausdorff approximation map from X to Y.

Let $C(u_1,\ldots ,u_l)>0$ denotes a positive function depending only on the numbers $u_1,\ldots ,u_l$. For a set X, ${\mathrm {Card}}X$ denotes a cardinal number of X.

Let (M, g) be a closed Riemannian manifold. For any $p\ge 1$, we use the normalized $L^p$-norm:

$$\begin{aligned} \Vert f\Vert _p^p:=\frac{1}{{\mathrm {Vol}}(M)}\int _M |f|^p\,d\mu _g, \end{aligned}$$

and $\Vert f\Vert _{\infty }:=\mathop {\mathrm {ess~sup}}\limits _{x\in M}|f(x)|$ for a measurable function f on M. We also use these notation for tensors. We have $\Vert f\Vert _p\le \Vert f\Vert _q$ for any $p\le q \le \infty$.

Let $\nabla$ denotes the Levi-Civita connection. Throughout in this paper, $0=\lambda _0(g)< \lambda _1(g) \le \lambda _2(g) \le \cdots \rightarrow \infty$ denotes the eigenvalues of the Laplacian $\Delta =-{\mathrm {tr}}{\mathrm {Hess}}$ acting on functions. We sometimes identify TM and $T^*M$ using the metric g. Given points $x,y\in M$, let $\gamma _{x,y}$ denotes one of minimal geodesics with unit speed such that $\gamma _{x,y}(0)=x$ and $\gamma _{x,y}(d(x,y))=y$. For given $x\in M$ and $u\in T_x M$ with $|u|=1$, let $\gamma _{u}:{\mathbb {R}}\rightarrow M$ denotes the geodesic with unit speed such that $\gamma _u(0)=x$ and ${\dot{\gamma }}_u(0)=u$.

For any $x\in M$ and $u\in T_x M$ with $|u|=1$, put

$$\begin{aligned} t(u):=\sup \{t\in {\mathbb {R}}_{>0}: d(x,\gamma _u(t))=t\}, \end{aligned}$$

and define $I_x\subset M$ to be the complement of the cut locus at x (see also [21, p.104]), i.e.,

$$\begin{aligned} I_x:=\{\gamma _u (t): u\in T_x M \text { with }|u|=1\text { and } 0\le t< t(u)\}. \end{aligned}$$

Then, $I_x$ is open and ${\mathrm {Vol}}(M\setminus I_x)=0$ [21, III Lemma 4.4]. For any $y\in I_x\setminus \{x\}$, the minimal geodesic $\gamma _{x,y}$ is uniquely determined. The function $d(x,\cdot ):M\rightarrow {\mathbb {R}}$ is differentiable in $I_x\setminus \{x\}$ and $\nabla d(x,\cdot )(y)={\dot{\gamma }}_{x,y}(d(x,y))$ holds for any $y\in I_x\setminus \{x\}$ [21, III Proposition 4.8].

Let V be an n-dimensional real vector space with an inner product $\langle ,\rangle$. We define inner products on $\bigwedge ^k V$ and $V\otimes \bigwedge ^k V$ as follows:

$$\begin{aligned} \begin{aligned}&\langle v_1\wedge \ldots \wedge v_k,w_1\wedge \ldots \wedge w_k\rangle =\det \{\langle v_i,w_j\rangle \}_{i,j},\\&\langle v_0\otimes v_1\wedge \ldots \wedge v_k,w_0\otimes w_1\wedge \ldots \wedge w_k\rangle =\langle v_0,w_0\rangle \det \{\langle v_i,w_j\rangle \}_{i,j}, \end{aligned} \end{aligned}$$

for $v_0,\ldots ,v_k,w_0,\ldots ,w_k\in V$. For $\alpha \in V$ and $\omega \in \bigwedge ^k V$, there exists a unique $\iota (\alpha )\omega \in \bigwedge ^{k-1} V$ such that $\langle \iota (\alpha )\omega ,\eta \rangle =\langle \omega ,\alpha \wedge \eta \rangle$ holds for any $\eta \in \bigwedge ^{k-1} V$. If $k=0$, we define $\iota (\alpha )\omega =0$ and $\bigwedge ^{-1}V=\{0\}$. Then, $\iota$ defines a bi-linear map:

$$\begin{aligned} \iota :V\times \bigwedge ^k V\rightarrow \bigwedge ^{k-1} V. \end{aligned}$$

By identifying V and $V^*$ using $\langle ,\rangle$, we also use the notation $\iota$ for the bi-linear map:

$$\begin{aligned} \iota :V^*\times \bigwedge ^k V\rightarrow \bigwedge ^{k-1} V. \end{aligned}$$

For any Riemannian manifold (M, g), we define operators $\nabla ^*:\Gamma (T^*M\otimes \bigwedge ^k T^*M)\rightarrow \Gamma (\bigwedge ^k T^*M)$ and $d^*:\Gamma (\bigwedge ^k T^*M)\rightarrow \Gamma (\bigwedge ^{k-1}T^*M)$ by

$$\begin{aligned} \nabla ^*(\alpha \otimes \beta ):&=-{\mathrm {tr}}_{T^*M} \nabla (\alpha \otimes \beta ) =-\sum _{i=1}^n \left( \nabla _{e_i}\alpha \right) (e_i)\cdot \beta -\sum _{i=1}^n\alpha (e_i)\cdot \nabla _{e_i}\beta .\\ d^*\omega :&=-\sum _{i=1}^n\iota (e_i)\nabla _{e_i}\omega \end{aligned}$$

for all $\alpha \otimes \beta \in \Gamma (T^*M\otimes \bigwedge ^k T^*M)$ and $\omega \in \Gamma (\bigwedge ^k T^*M)$, where $n=\dim M$ and $\{e_1,\ldots ,e_n\}$ is an orthonormal basis of TM. If M is closed, then we have

$$\begin{aligned} \int _M \langle T,\nabla \alpha \rangle \,d\mu _g&=\int _M \langle \nabla ^*T, \alpha \rangle \,d\mu _g,\\ \int _M \langle \omega ,d\eta \rangle \,d\mu _g&=\int _M \langle d^*\omega , \eta \rangle \,d\mu _g \end{aligned}$$

for all $T\in \Gamma (T^*M\otimes \bigwedge ^k T^*M)$, $\alpha \in \Gamma (\bigwedge ^k T^*M)$, $\omega \in \Gamma (\bigwedge ^k T^*M)$ and $\eta \in \Gamma (\bigwedge ^{k-1} T^*M)$ by the divergence theorem. The Hodge Laplacian $\Delta :\Gamma (\bigwedge ^k T^*M)\rightarrow \Gamma (\bigwedge ^k T^*M)$ is defined by

$$\begin{aligned} \Delta :=d d^*+d^*d. \end{aligned}$$

Notation 2.4

For an n-dimensional Riemannian manifold (M, g), we can take orthonormal basis of TM only locally in general. However, for example, the tensor

$$\begin{aligned} \sum _{i=1}^n e^i\otimes \iota (\nabla _{e_i} \nabla f)\omega \in \Gamma (T^*M\otimes \bigwedge ^{k-1} T^*M)\quad (f\in C^\infty (M),\,\omega \in \Gamma (\bigwedge ^k T^*M)) \end{aligned}$$

is defined independently of the choice of the orthonormal basis $\{e_1,\ldots ,e_n\}$ of TM, where $\{e^1,\ldots ,e^n\}$ denotes its dual. Thus, we sometimes use such notation without taking a particular orthonormal basis.

Finally, we list some important notation. Let (M, g) be a closed Riemannian manifold.

• d denotes the Riemannian distance function.

• ${\mathrm {Ric}}$ denotes the Ricci curvature tensor.

• ${\mathrm {diam}}$ denotes the diameter.

• ${\mathrm {Vol}}$ or $\mu _g$ denotes the Riemannian volume measure.

• $\Vert \cdot \Vert _p$ denotes the normalized $L^p$-norm for each $p\ge 1$, which is defined by

  $$\begin{aligned} \Vert f\Vert _p^p:=\frac{1}{{\mathrm {Vol}}(M)}\int _M |f|^p\,d\mu _g \end{aligned}$$

  for any measurable function f on M.

• $\Vert f\Vert _{\infty }$ denotes the essential sup of |f| for any measurable function f on M.

• $\nabla$ denotes the Levi-Civita connection.

• $\nabla ^2$ denotes the Hessian for functions.

• $\Delta :\Gamma (\bigwedge ^k T^*M)\rightarrow \Gamma (\bigwedge ^k T^*M)$ denotes the Hodge Laplacian defined by $\Delta :=d d^*+d^*d$. We frequently use the Laplacian acting on functions. Note that $\Delta =-{\mathrm {tr}}_g \nabla ^2$ holds for functions under our sign convention.

• $0=\lambda _0(g)< \lambda _1(g) \le \lambda _2(g) \le \cdots \rightarrow \infty$ denotes the eigenvalues of the Laplacian acting on functions.

• $\gamma _{x,y}:[0,d(x,y)]\rightarrow M$ denotes one of minimal geodesics with unit speed such that $\gamma _{x,y}(0)=x$ and $\gamma _{x,y}(d(x,y))=y$ for any $x,y\in M$.

• $\gamma _{u}:{\mathbb {R}}\rightarrow M$ denotes the geodesic with unit speed such that $\gamma _u(0)=x$ and ${\dot{\gamma }}(0)=u$ for any $x\in M$ and $u\in T_x M$ with $|u|=1$.

• $I_x\subset M$ denotes the complement of the cut locus at $x\in M$. We have ${\mathrm {Vol}}(M\setminus I_x)=0$. We have that $\gamma _{x,y}$ is uniquely determined and $\nabla d(x,\cdot )={\dot{\gamma }}_{x,y}(d(x,y))$ holds for any $y\in I_x\setminus \{x\}$.

• $\Delta _{C,k}=\nabla ^*\nabla :\Gamma (\bigwedge ^k T^*M)\rightarrow \Gamma (\bigwedge ^k T^*M)$ denotes the connection Laplacian acting on k-forms.

• $0\le \lambda _1(\Delta _{C,k}) \le \lambda _2(\Delta _{C,k}) \le \cdots \rightarrow \infty$ denotes the eigenvalues of the connection Laplacian $\Delta _{C,k}$ acting on k-forms.

• $S^n(r)$ denotes the n-dimensional standard sphere of radius r.

• $S^n:=S^n(1)$.

Note that the lowest eigenvalue of the Laplacian $\Delta$ acting on function is always equal to 0, and so we start counting the eigenvalues of it from $i=0$. This is not the case with the connection Laplacian $\Delta _{C,k}$ acting on k-forms, and so we start counting the eigenvalues of it from $i=1$. For any $i\in {\mathbb {Z}}_{>0}$, we have

$$\begin{aligned} \lambda _i(\Delta _{C,0})=\lambda _{i-1}(g). \end{aligned}$$

2.2 Calculus of Differential Forms

In this subsection, we recall some facts about differential forms, and do some calculations.

We first recall the decomposition:

$$\begin{aligned} T^*M\otimes \bigwedge ^k T^*M=T^{k,1}M\oplus \bigwedge ^{k+1} T^*M\oplus \bigwedge ^{k-1} T^*M. \end{aligned}$$

See also [23, Section 2].

Let V be an n-dimensional real vector space with an inner product $\langle ,\rangle$. We put

$$\begin{aligned}&P_1:V\otimes \bigwedge ^k V\rightarrow \bigwedge ^{k+1} V,\quad P_1(\alpha \otimes \omega ):=\left( \frac{1}{k+1}\right) ^\frac{1}{2}\alpha \wedge \omega ,\\&P_2:V\otimes \bigwedge ^k V\rightarrow \bigwedge ^{k-1} V,\quad P_2(\alpha \otimes \omega ):=\left( \frac{1}{n-k+1}\right) ^\frac{1}{2}\iota (\alpha )\omega ,\\&Q_1:\bigwedge ^{k+1} V\rightarrow V\otimes \bigwedge ^k V,\quad Q_1(\zeta ):=\left( \frac{1}{k+1}\right) ^\frac{1}{2}\sum _{i=1}^n e^i\otimes \iota (e^i)\zeta ,\\&Q_2:\bigwedge ^{k-1} V\rightarrow V\otimes \bigwedge ^k V,\quad Q_2(\eta ):=\left( \frac{1}{n-k+1}\right) ^\frac{1}{2}\sum _{i=1}^n e^i\otimes e^i\wedge \eta , \end{aligned}$$

where $\{e^1,\ldots ,e^n\}$ is orthonormal basis of V. Then, we have

• ${\mathrm {Im}}Q_1\bot {\mathrm {Im}}Q_2$,

• $P_i\circ Q_i={\mathrm {Id}}$ for each $i=1,2$,

• $Q_1$ and $Q_2$ preserve the norms,

• $Q_i\circ P_i:V\otimes \bigwedge ^k V\rightarrow V\otimes \bigwedge ^k V$ is symmetric and $(Q_i\circ P_i)^2=Q_i\circ P_i$ for each $i=1,2$.

Therefore, $Q_i\circ P_i$ is the orthogonal projection $V\otimes \bigwedge ^k V\rightarrow {\mathrm {Im}}Q_i$. Since $\bigwedge ^{k+1} V\cong {\mathrm {Im}}Q_1$ and $\bigwedge ^{k-1} V \cong {\mathrm {Im}}Q_2$, we can regard $\bigwedge ^{k+1} V$ and $\bigwedge ^{k-1} V$ as subspaces of $V\otimes \bigwedge ^k V$.

Take an n-dimensional Riemannian manifold (M, g) and consider the case when $V=T^*_x M$ ($x\in M$). We can take a sub-bundle $T^{k,1}M$ of $T^*M\otimes \bigwedge ^k T^*M$ such that

$$\begin{aligned} T^*M\otimes \bigwedge ^k T^*M=T^{k,1}M\oplus \bigwedge ^{k+1} T^*M\oplus \bigwedge ^{k-1} T^*M \end{aligned}$$

is an orthogonal decomposition. Then, for $\omega \in \Gamma (\bigwedge ^k T^*M)$, we can decompose $\nabla \omega \in \Gamma (T^*M\otimes \bigwedge ^k T^*M)$, the $\bigwedge ^{k+1} T^*M$-component is equal to $\left( 1/(k+1)\right) ^{1/2}d\omega$ and the $\bigwedge ^{k-1} T^*M$-component is equal to $-\left( 1/(n-k+1)\right) ^{1/2} d^*\omega$. Let $T(\omega )$ denotes the remaining part ($T:\Gamma (\bigwedge ^k T^*M)\rightarrow \Gamma (T^{k,1}M)$). Then, we have

$$\begin{aligned} \nabla \omega =T(\omega )+ \left( \frac{1}{k+1}\right) ^\frac{1}{2} Q_1(d\omega )-\left( \frac{1}{n-k+1}\right) ^\frac{1}{2}Q_2(d^*w). \end{aligned}$$

Therefore, we get

$$\begin{aligned} |\nabla \omega |^2=|T(\omega )|^2+\frac{1}{k+1} |d\omega |^2+\frac{1}{n-k+1}|d^*\omega |^2. \end{aligned}$$

(4)

If $d^*\omega =0$ and $T(\omega )=0$, then $\omega$ is called a Killing k-form (see also [23, Definition 2.1]).

We next recall the Bochner-Weitzenböck formula.

Definition 2.5

Let (M, g) be an n-dimensional Riemannian manifold. We define a homomorphism ${\mathcal {R}}_k:\bigwedge ^k T^*M\rightarrow \bigwedge ^k T^*M$ as

$$\begin{aligned} {\mathcal {R}}_k \omega =-\sum _{i,j}e^i\wedge \iota (e_j)\left( R(e_i,e_j)\omega \right) \end{aligned}$$

for any $\omega \in \bigwedge ^k T^*M$, where $\{e_1,\ldots ,e_n\}$ is an orthonormal basis of TM, $\{e^1,\ldots ,e^n \}$ is its dual and $R(e_i,e_j)\omega$ is defined by

$$\begin{aligned} R(e_i,e_j)\omega =\nabla _{e_i}\nabla _{e_j}\omega -\nabla _{e_j}\nabla _{e_i}\omega -\nabla _{[e_i,e_j]}\omega \in \Gamma (\bigwedge ^k T^*M). \end{aligned}$$

Note that if $k=1$, then we have ${\mathcal {R}}_1 \omega ={\mathrm {Ric}}(\omega ,\cdot )$ for any $\omega \in \Gamma (T^*M)$.

The Bochner-Weitzenböck formula is stated as follows:

Theorem 2.6

(Bochner-Weitzenböck formula) For any $\omega \in \Gamma (\bigwedge ^k T^*M)$, we have

$$\begin{aligned} \Delta \omega =\nabla ^*\nabla \omega +{\mathcal {R}}_k \omega . \end{aligned}$$

In particular, we have the following theorem when $k=1$:

Theorem 2.7

(Bochner-Weitzenböck formula for 1-forms) For any $\omega \in \Gamma (T^*M)$, we have

$$\begin{aligned} \Delta \omega =\nabla ^*\nabla \omega + {\mathrm {Ric}}(\omega ,\cdot ). \end{aligned}$$

Let us do some calculations of differential forms.

Lemma 2.8

Let (M, g) be an n-dimensional Riemannian manifold. Take a vector field $X\in \Gamma (TM)$, a p-form $\omega \in \Gamma (\bigwedge ^p T^*M)$ $(p\ge 1)$ and a local orthonormal bases $\{e_1,\ldots ,e_n\}$ of TM.

• (i) We have

  $$\begin{aligned} {\mathcal {R}}_{p-1}(\iota (X)\omega )=\iota (X) {\mathcal {R}}_p \omega +\iota ({\mathrm {Ric}}(X))\omega +2\sum _{i=1}^n\iota (e_i)(R(X,e_i)\omega ). \end{aligned}$$

• (ii) We have

  $$\begin{aligned} \Delta (\iota (X)\omega )= & {} \iota (\Delta X)\omega +\iota (X)\Delta \omega +2\sum _{i=1}^n\iota (e_i) (R(X,e_i)\omega )\\&-2\sum _{i=1}^n\iota (\nabla _{e_i}X) (\nabla _{e_i}\omega ). \end{aligned}$$

• (iii) We have

  $$\begin{aligned} \sum _{i=1}^n\iota (e_i) (R(X,e_i)\omega ) =-\nabla _X d^*\omega +d^*\nabla _X \omega +\sum _{i,j=1}^n \langle \nabla _{e_j} X, e_i\rangle \iota (e_j)\nabla _{e_i}\omega . \end{aligned}$$

Proof

Let $\{e^1,\ldots ,e^n\}$ be the dual basis of $\{e_1,\ldots ,e_n\}$.

We first show (i). If $p=1$, both sides are equal to 0. Let us assume $p\ge 2$. We have

$$\begin{aligned} \begin{aligned}&\iota ({\mathrm {Ric}}(X))\omega \\&\quad =\frac{1}{(p-1)!}\sum _{i,i_1,\ldots ,i_{p-1}}\omega (R(X,e_i)e_i,e_{i_1},\cdots ,e_{i_{p-1}})e^{i_1}\wedge \cdots \wedge e^{i_{p-1}}\\&\quad =\frac{-1}{(p-1)!}\sum _{i,i_1,\ldots ,i_{p-1}} (R(X,e_i)\omega )(e_i,e_{i_1},\ldots ,e_{i_n})e^{i_1}\wedge \cdots \wedge e^{i_{p-1}}\\&\qquad -\frac{1}{(p-1)!}\sum _{i,i_1,\ldots ,i_{p-1}} \sum _{l=1}^{p-1} \omega (e_i,e_{i_1},\cdots ,R(X,e_i)e_{i_l},\ldots ,e_{i_{p-1}})e^{i_1}\wedge \cdots \wedge e^{i_{p-1}}\\&\quad =-\sum _{i=1}^n\iota (e_i)(R(X,e_i)\omega )\\&\qquad -\frac{1}{(p-1)!}\sum _{i,i_1,\ldots ,i_{p-1}} \sum _{l=1}^{p-1} \omega (e_i,e_{i_1},\cdots ,R(X,e_i)e_{i_l},\ldots ,e_{i_{p-1}})e^{i_1}\wedge \cdots \wedge e^{i_{p-1}} \end{aligned}\nonumber \\ \end{aligned}$$

(5)

We calculate the second term.

$$\begin{aligned} \begin{aligned}&-\frac{1}{(p-1)!}\sum _{i,i_1,\ldots ,i_{p-1}} \sum _{l=1}^{p-1} \omega (e_i,e_{i_1},\cdots ,R(X,e_i)e_{i_l},\ldots ,e_{i_{p-1}})e^{i_1}\wedge \cdots \wedge e^{i_{p-1}}\\&\quad =\frac{1}{(p-1)!}\sum _{l=1}^{p-1} \sum _{i,j,i_1,\ldots ,i_{p-1}}\langle R(e_j,e_{l_l})X,e_i\rangle \omega (e_i,e_j,e_{i_1},\cdots ,\widehat{e_{i_l}},\ldots ,e_{i_{p-1}})\\&\qquad \qquad e^{i_l}\wedge e^{i_1}\wedge \cdots \wedge \widehat{e^{i_l}}\wedge \cdots \wedge e^{i_{p-1}}\\&\quad =\sum _{j,k} e^k\wedge \iota (e_j)\iota (R(e_j,e_{k})X)\omega \\&\quad =\sum _{j,k} e^k\wedge \iota (e_j)R(e_j,e_{k})(\iota (X)\omega )-\sum _{j,k} e^k\wedge \iota (e_j)\iota (X)R(e_j,e_{k})\omega \\&\quad ={\mathcal {R}}_{p-1}(\iota (X)\omega )-\iota (X){\mathcal {R}}_{p}\omega -\sum _{j=1}^n \iota (e_j)(R(X,e_j)\omega ) \end{aligned} \end{aligned}$$

Combining this and (5), we get (i).

Let us show (ii). We have

$$\begin{aligned} \nabla ^*\nabla \iota (X)\omega =\iota (\nabla ^*\nabla X)\omega -2\sum _{i} \iota (\nabla _{e_i}X)\nabla _{e_i}\omega +\iota (X)\nabla ^*\nabla \omega . \end{aligned}$$

Thus, by (i), we get

$$\begin{aligned} \begin{aligned} \Delta ( \iota (X)\omega )&=\nabla ^*\nabla \iota (X)\omega +{\mathcal {R}}_{p-1}\iota (X)\omega \\&=\iota (\Delta X)\omega +\iota (X)\Delta \omega +2\sum _{i=1}^n\iota (e_i) (R(X,e_i)\omega )-2\sum _{i=1}^n\iota (\nabla _{e_i}X) (\nabla _{e_i}\omega ). \end{aligned} \end{aligned}$$

This gives (ii).

Finally, we show (iii). We have

$$\begin{aligned} \begin{aligned} \sum _{i=1}^n \iota (e_i)(R(X,e_i)\omega )&=\sum _{i=1}^n \iota (e_i)\left( \nabla _{X}\nabla _{e_i}\omega -\nabla _{e_i}\nabla _X\omega -\nabla _{\nabla _{X} e_i}\omega +\nabla _{\nabla _{e_i}X}\omega \right) \\&=-\nabla _{X}d^*\omega +d^*\nabla _X\omega +\sum _{i,j=1}^n \langle \nabla _{e_j} X, e_i\rangle \iota (e_j)\nabla _{e_i}\omega . \end{aligned} \end{aligned}$$

This gives (iii). $\square$

When $\omega$ is parallel, we have the following corollary.

Corollary 2.9

Let (M, g) be an n-dimensional Riemannian manifold. Take a vector field $X\in \Gamma (TM)$ and a parallel p-form $\omega \in \Gamma (\bigwedge ^p T^*M)$ $(p\ge 1)$.

• (i) We have

  $$\begin{aligned} {\mathcal {R}}_{p-1}(\iota (X)\omega )=\iota ({\mathrm {Ric}}(X))\omega . \end{aligned}$$

• (ii) We have

  $$\begin{aligned} \Delta (\iota (X)\omega )=\iota (\Delta X)\omega . \end{aligned}$$

Finally, we give some easy equations for later use. Let (M, g) be an n-dimensional Riemannian manifold. Take a local orthonormal basis $\{e_1,\ldots ,e_n\}$ of TM. Let $\{e^1,\ldots ,e^n\}$ be its dual. For any $\omega ,\eta \in \Gamma (\bigwedge ^k T^*M)$, we have

$$\begin{aligned} \sum _{i=1}^n \langle e^i\wedge \omega , e^i\wedge \eta \rangle =(n-k)\langle \omega ,\eta \rangle , \quad \sum _{i=1}^n \langle \iota (e_i)\omega , \iota (e_i) \eta \rangle =k\langle \omega ,\eta \rangle . \end{aligned}$$

For any $\alpha _1,\ldots ,\alpha _k\in \Gamma (T^*M)$, we have

$$\begin{aligned} Q_1(\alpha _1\wedge \cdots \wedge \alpha _k)=\left( \frac{1}{k}\right) ^{1/2}\sum _{i=1}^k(-1)^{i-1}\alpha _i\otimes \alpha _1\wedge \cdots \wedge \widehat{\alpha _i}\wedge \cdots \wedge \alpha _k. \end{aligned}$$

Since $Q_1$ preserves the norms, we have

$$\begin{aligned} \begin{aligned}&k\left| \alpha _1\wedge \cdots \wedge \alpha _k\right| ^2\\&\quad =\left| \sum _{i=1}^k(-1)^{i-1}\alpha _i\otimes \alpha _1\wedge \cdots \wedge \widehat{\alpha _i}\wedge \cdots \wedge \alpha _k\right| ^2 \end{aligned} \end{aligned}$$

(6)

for any $\alpha _1,\ldots ,\alpha _k\in \Gamma (T^*M)$.

Suppose that M is oriented. For any k, the Hodge star operator $*:\bigwedge ^k T^*M\rightarrow \bigwedge ^{n-k} T^*M$ is defined so that

$$\begin{aligned} \langle *\omega ,\eta \rangle V_g=\omega \wedge \eta \end{aligned}$$

for all $\omega \in \Gamma (\bigwedge ^k T^*M)$ and $\eta \in \Gamma (\bigwedge ^{n-k} T^*M)$, where $V_g$ denotes the volume form on (M, g). For any $\alpha \in \Gamma (T^*M)$, $\omega \in \Gamma (\bigwedge ^k T^*M)$ and $\eta \in \Gamma (\bigwedge ^{k-1} T^*M)$, we have

$$\begin{aligned} \langle *(\omega \wedge \alpha ),\eta \rangle V_g&=\omega \wedge \alpha \wedge \eta ,\\ \langle \iota (\alpha )*\omega ,\eta \rangle V_g=\langle *\omega ,\alpha \wedge \eta \rangle V_g&=\omega \wedge \alpha \wedge \eta . \end{aligned}$$

Thus, we get

$$\begin{aligned} *(\omega \wedge \alpha )=\iota (\alpha )*\omega . \end{aligned}$$

(7)

Therefore, for any $\alpha ,\beta \in \Gamma (T^*M)$ and $\omega ,\eta \in \Gamma (\bigwedge ^k T^*M)$, we have

$$\begin{aligned} \begin{aligned}&\langle \iota (\alpha )\omega ,\iota (\beta )\eta \rangle =\langle \omega ,\alpha \wedge \iota (\beta )\eta \rangle \\&=-\langle \beta \wedge \omega ,\alpha \wedge \eta \rangle +\langle \alpha ,\beta \rangle \langle \omega ,\eta \rangle =-\langle \iota (\beta )*\omega ,\iota (\alpha )*\eta \rangle +\langle \alpha ,\beta \rangle \langle \omega ,\eta \rangle , \end{aligned} \end{aligned}$$

and so

$$\begin{aligned} \langle \iota (\alpha )\omega ,\iota (\beta )\eta \rangle +\langle \iota (\beta )*\omega ,\iota (\alpha )*\eta \rangle =\langle \alpha ,\beta \rangle \langle \omega ,\eta \rangle . \end{aligned}$$

(8)

3 Almost Parallel p-form

In this section, we show Main Theorems 1 and 3.

3.1 Parallel p-form

In this subsection, we show some easy results when the Riemannian manifold has a non-trivial parallel p-form. We first give an easy proof of what Grosjean called a new Bochner-Reilly formula [12, Proposition 3.1] for closed Riemannian manifolds with a non-trivial parallel p-form $\omega$. Similarly, we also get the formula [12, Proposition 3.1] for Riemannian manifold with boundary. In the next subsection, we estimate the error terms when $\omega$ is not parallel.

Proposition 3.1

(Bochner-Reilly-Grosjean formula [12]) Let (M, g) be an n-dimensional closed Riemannian manifold. For any $f\in C^\infty (M)$ and any parallel p-form $\omega$ $(1\le p \le n-1)$ on M, we have

$$\begin{aligned} \begin{aligned}&\int _M |T (\iota (\nabla f)\omega )|^2\,d\mu _g\\&\quad =\frac{p-1}{p}\int _M\langle \iota (\nabla f)\omega , \iota (\nabla \Delta f)\omega \rangle \,d\mu _g-\int _M \langle \iota ({\mathrm {Ric}}(\nabla f))\omega ,\iota (\nabla f)\omega \rangle \,d\mu _g. \end{aligned} \end{aligned}$$

See Sect. 2.2 for the definition of $T:\Gamma (\bigwedge ^{p-1}T^*M)\rightarrow \Gamma (T^{p-1,1}M)$.

Proof

Since $d^*\iota (\nabla f) \omega =-d^*d^*(f\omega )=0$, we have

$$\begin{aligned} \begin{aligned}&\int _M \langle \iota ({\mathrm {Ric}}(\nabla f))\omega ,\iota (\nabla f)\omega \rangle \,d\mu _g\\&=\quad \int _M \langle {\mathcal {R}}_{p-1}(\iota (\nabla f)\omega ),\iota (\nabla f)\omega \rangle \,d\mu _g\\&\quad =\int _M \langle d(\iota (\nabla f)\omega ),d(\iota (\nabla f)\omega )\rangle \,d\mu _g -\int _M \langle \nabla (\iota (\nabla f)\omega ),\nabla (\iota (\nabla f)\omega )\rangle \,d\mu _g \end{aligned} \end{aligned}$$

(9)

by Corollary 2.9 (i), Bochner-Weitzenböck formula and the divergence theorem. By (4) and Corollary 2.9 (ii), we have

$$\begin{aligned} \begin{aligned}&\int _M \langle d(\iota (\nabla f)\omega ),d(\iota (\nabla f)\omega )\rangle \,d\mu _g -\int _M \langle \nabla (\iota (\nabla f)\omega ),\nabla (\iota (\nabla f)\omega )\rangle \,d\mu _g\\&\quad =\frac{p-1}{p}\int _M \langle \iota (\nabla \Delta f)\omega ),\iota (\nabla f)\omega \rangle \,d\mu _g-\int _M |T(\iota (\nabla f)\omega )|^2\,d\mu _g \end{aligned} \end{aligned}$$

(10)

By (9) and (10), we get the proposition. $\square$

Based on Proposition 3.1, Grosjean showed Theorem 1.3. Assuming more strong condition on eigenvalues, we remove the assumption that the manifold is simply connected from Theorem 1.3.

Corollary 3.2

Let (M, g) be an n-dimensional closed Riemannian manifold. Assume that ${\mathrm {Ric}}\ge (n-p-1)g$ and there exists a non-trivial parallel p-form on M $(2\le p< n/2)$. If $\lambda _{n-p+1}(g)= n-p,$ then (M, g) is isometric to a product $S^{n-p}\times (X,g')$, where $(X,g')$ is some p-dimensional closed Riemannian manifold.

Proof

Let $f_k$ be the k-th eigenfunction of the Laplacian on $S^{n-p}$. Note that the functions $f_1,\ldots ,f_{n-p+1}$ are height functions.

By Theorem 1.3, the universal cover $({\widetilde{M}},{\tilde{g}})$ of (M, g) is isometric to a product $S^{n-p}\times (X,g')$, where $(X,g')$ is some p-dimensional closed Riemannian manifold. We regard the function $f_i$ as a function on ${\widetilde{M}}$. Since $\lambda _{n-p+1}(g)= n-p$, each $f_i\in C^\infty ({\widetilde{M}})$ ($i=1,\ldots ,n-p+1$) is a pull back of some function on M. Thus, the covering transformation preserves $f_1,\ldots ,f_{n-p+1}$. Therefore, the covering transformation does not act on $S^{n-p}$, and so we get the corollary. $\square$

The almost version of this corollary is Main Theorem 2.

Finally, we show that the assumption of Corollary 3.2 is optimal in some sense by giving an example.

Take a positive odd integer p with $p\ge 3$ and a positive integer n with $n> 2p$. Put $a:=\sqrt{(p-1)/(n-p-1)}$. We define an equivalence relation $\sim$ on $S^{n-p}\times S^p(a)$ as follows:

$$\begin{aligned} \begin{aligned}&((x_0,\ldots ,x_{n-p}),(y_0,\ldots ,y_p))\sim ((x'_0,\ldots ,x'_{n-p}),(y'_0,\ldots ,y'_p))\\ \Leftrightarrow&\text { there exists }k\in {\mathbb {Z}}\text { such that}\\&((x'_0,\ldots ,x'_{n-p}),(y'_0,\ldots ,y'_p))=((-1)^k x_0, x_1,\ldots ,x_{n-p}),(-1)^k(y_0,\ldots ,y_p)) \end{aligned} \end{aligned}$$

for any $((x_0,\ldots ,x_{n-p}),(y_0,\ldots ,y_p)), ((x'_0,\ldots ,x'_{n-p}),(y'_0,\ldots ,y'_p))\in S^{n-p}\times S^p(a)$. Then, we have the following:

Proposition 3.3

We have the following properties:

• $(M,g)=(S^{n-p}\times S^p(a))/\sim$ is an n-dimensional closed Riemannian manifold with a non-trivial parallel p-form.

• ${\mathrm {Ric}}= (n-p-1)g$.

• $\lambda _{n-p}(g)=n-p$.

• (M, g) is not isometric to any product Riemannian manifolds.

Proof

Let $\omega$ be the volume form on $S^p(a)$. Since the action on $S^{n-p}\times S^p(a)$ preserves $\omega$, there exists a non-trivial parallel p-form on (M, g). We also denote it by $\omega$. Since the action on $S^{n-p}\times S^p(a)$ preserves the function

$$\begin{aligned} x_i :S^{n-p}\times S^p(a)\rightarrow {\mathbb {R}},\,((x_0,\ldots ,x_{n-p}),(y_0,\ldots ,y_p))\mapsto x_i \end{aligned}$$

for each $i=1,\ldots ,n-p$, we have $\lambda _{n-p}(g)=n-p$.

Suppose that (M, g) is isometric to a product $(M^{n-k}_1,g_1)\times (M^{k}_2,g_2)$ ($k\le n-k$) for some $(n-k)$ and k-dimensional closed Riemannian manifolds $(M_1,g_1)$ and $(M_2,g_2)$. Since we have the irreducible decomposition $T_{[(x,y)]} M\cong T_x S^{n-p}\oplus T_y S^p(a)$ of the restricted holonomy action, we get $k=p$. Since $\lambda _1(g)=n-p$, we have that $(M_1,g_1)$ is isometric to $S^{n-p}$. Thus, we get $\lambda _{n-p+1}(g)=n-p$. However the action on $S^{n-p}\times S^p(a)$ does not preserve the function

$$\begin{aligned} x_0:S^{n-p}\times S^p(a)\rightarrow {\mathbb {R}},\,((x_0,\ldots ,x_{n-p}),(y_0,\ldots ,y_p))\mapsto x_0, \end{aligned}$$

and so $\lambda _{n-p+1}(g)\ne n-p$. This is a contradiction. $\square$

3.2 Error Estimates

In this subsection, we give error estimates about Proposition 3.1. Lemma 3.8 (vii) corresponds to Proposition 3.1.

We list the assumptions of this subsection.

Assumption 3.4

In this subsection, we assume the following:

• (M, g) is an n-dimensional closed Riemannian manifold with ${\mathrm {Ric}}_g\ge -Kg$ and ${\mathrm {diam}}(M)\le D$ for some positive real numbers $K>0$ and $D>0$.

• $1\le k \le n-1$.

• A k-form $\omega \in \Gamma (\bigwedge ^k T^*M)$ satisfies $\Vert \omega \Vert _2=1$, $\Vert \omega \Vert _\infty \le L_1$ and $\Vert \nabla \omega \Vert _2^2\le \lambda$ for some $L_1>0$ and $0\le \lambda \le 1$.

• A function $f\in C^\infty (M)$ satisfies $\Vert f\Vert _{\infty }\le L_2\Vert f\Vert _2$, $\Vert \nabla f\Vert _{\infty }\le L_2\Vert f\Vert _2$ and $\Vert \Delta f\Vert _2\le L_2\Vert f\Vert _2$ for some $L_2>0$.

Note that we have

$$\begin{aligned} \Vert \nabla ^2 f\Vert _2^2=\Vert \Delta f\Vert _2^2-\frac{1}{{\mathrm {Vol}}(M)}\int _M {\mathrm {Ric}}(\nabla f,\nabla f)\,d\mu _g\le (1+K)L^2_2\Vert f\Vert _2^2 \end{aligned}$$

(11)

by the Bochner formula.

We first show the following:

Lemma 3.5

There exists a positive constant $C(n,K,D)>0$ such that $\Vert |\omega |-1\Vert _2\le C \lambda ^{1/2}$ holds.

Proof

Put ${\overline{\omega }}:=\int _M |\omega | \,d\mu _g/{\mathrm {Vol}}(M).$ Since we have $|\omega |\in W^{1,2}(M)$, we get

$$\begin{aligned} \Vert |\omega |-{\overline{\omega }}\Vert _2^2\le \frac{1}{\lambda _1(g)}\Vert \nabla |\omega |\Vert _2^2\le \frac{1}{\lambda _1(g)}\Vert \nabla \omega \Vert _2^2\le \frac{\lambda }{\lambda _1(g)}\le C\lambda \end{aligned}$$

by the Kato inequality and the Li-Yau estimate [22, p.116]. Therefore, we get

$$\begin{aligned} |1-{\overline{\omega }}|=\left| \Vert \omega \Vert _2-\Vert {\overline{\omega }}\Vert _2\right| \le \Vert |\omega |-{\overline{\omega }}\Vert _2\le C\lambda ^{1/2}, \end{aligned}$$

and so $\Vert |\omega |-1\Vert _2\le C\lambda ^{1/2}.$ $\square$

Let us give error estimates about Proposition 3.1.

Lemma 3.6

There exists a positive constant $C=C(n,k,K,D,L_1,L_2)>0$ such that the following properties hold:

• (i) We have

  $$\begin{aligned} \frac{1}{{\mathrm {Vol}}(M)} \int _M |d^{*}(\iota (\nabla f)\omega )|^2\,d\mu _g \le C\Vert f\Vert _2^2\lambda . \end{aligned}$$

• (ii) We have

  $$\begin{aligned}&\left| \frac{1}{{\mathrm {Vol}}(M)}\int _M \Big (\langle \iota ({\mathrm {Ric}}(\nabla f))\omega ,\iota (\nabla f)\omega \rangle -\langle {\mathcal {R}}_{k-1}(\iota (\nabla f)\omega ),\iota (\nabla f)\omega \rangle \Big )\,d\mu _g\right| \\&\quad \le C\Vert f\Vert _2^2\lambda ^{1/2}. \end{aligned}$$

• (iii) We have

  $$\begin{aligned}&\left| \frac{1}{{\mathrm {Vol}}(M)} \int _M \Big (\langle \Delta (\iota (\nabla f)\omega ),\iota (\nabla f)\omega \rangle -\langle \iota (\nabla \Delta f)\omega ,\iota (\nabla f)\omega \rangle \Big ) \,d\mu _g\right| \\&\quad \le C\Vert f\Vert _2^2\lambda ^{1/2}. \end{aligned}$$

• (iv) We have

  $$\begin{aligned} \frac{1}{{\mathrm {Vol}}(M)}\int _M \left| \nabla (\iota (\nabla f)\omega )-\sum _{i=1}^n e^i\otimes \iota (\nabla _{e_i}\nabla f)\omega \right| ^2 \,d\mu _g\le C\Vert f\Vert _2^2\lambda . \end{aligned}$$

• (v) We have

  $$\begin{aligned} \frac{1}{{\mathrm {Vol}}(M)}\int _M \left| d (\iota (\nabla f)\omega )-\sum _{i=1}^n e^i\wedge \iota (\nabla _{e_i}\nabla f)\omega \right| ^2 \,d\mu _g\le C\Vert f\Vert _2^2\lambda . \end{aligned}$$

• (vi) We have

  $$\begin{aligned} \frac{1}{{\mathrm {Vol}}(M)}\int _M |\nabla (\iota (\nabla f)\omega ) |^2\,d\mu _g\le C\Vert f\Vert _2^2. \end{aligned}$$

• (vii) We have

  $$\begin{aligned}&\Bigg |\frac{1}{{\mathrm {Vol}}(M)}\int _M \langle \iota ({\mathrm {Ric}}(\nabla f))\omega ,\iota (\nabla f)\omega \rangle \,d\mu _g\\&\qquad - \frac{k-1}{k} \frac{1}{{\mathrm {Vol}}(M)}\int _M \langle \iota (\nabla \Delta f)\omega ,\iota (\nabla f)\omega \rangle \,d\mu _g+\Vert T(\iota (\nabla f)\omega )\Vert _2^2\Bigg | \\&\quad \le C\Vert f\Vert _2^2\lambda ^{1/2}. \end{aligned}$$

• (viii) If M is oriented and $1\le k\le n/2$, then we have

  $$\begin{aligned}&\frac{1}{{\mathrm {Vol}}(M)}\int _M {\mathrm {Ric}}(\nabla f,\nabla f)|\omega |^2\,d\mu _g\\&\quad \le \frac{n-k-1}{n-k} \frac{1}{{\mathrm {Vol}}(M)}\int _M \langle \nabla \Delta f,\nabla f\rangle |\omega |^2 \,d\mu _g -\Vert T(\iota (\nabla f)\omega )\Vert _2^2\\&\qquad -\Vert T(\iota (\nabla f)*\omega )\Vert _2^2\\&\qquad -\left( \frac{n-k-1}{n-k} -\frac{k-1}{k} \right) \Vert d(\iota (\nabla f)\omega )\Vert ^2_2 +C\Vert f\Vert _2^2\lambda ^{1/2}. \end{aligned}$$

Although an orthonormal basis $\{e_1,\ldots ,e_n\}$ of TM is defined only locally, $\sum _{i=1}^n e^i\otimes \iota (\nabla _{e_i}\nabla f)\omega$ and $\sum _{i=1}^n e^i\wedge \iota (\nabla _{e_i}\nabla f)\omega$ are well-defined as tensors.

Proof

We first prove (i). Since $d^*(f\omega )=-\iota (\nabla f)\omega +f d^*\omega$ and $d^*\circ d^*=0$, we have $d^*(\iota (\nabla f)\omega )=-\iota (\nabla f)d^*\omega .$ Thus, we get

$$\begin{aligned} \frac{1}{{\mathrm {Vol}}(M)} \int _M |d^{*}(\iota (\nabla f)\omega )|^2\,d\mu _g \le C\Vert \nabla f\Vert _{\infty }^2 \Vert \nabla \omega \Vert _2^2 \le C\Vert f\Vert _2^2 \lambda . \end{aligned}$$

To prove (ii) and (iii), we estimate following terms:

$$\begin{aligned}&\frac{1}{{\mathrm {Vol}}(M)} \int _M \langle \iota (\nabla f)\Delta \omega ,\iota (\nabla f) \omega \rangle \,d\mu _g,\\&\frac{1}{{\mathrm {Vol}}(M)} \int _M \langle \iota (\nabla f)\nabla ^*\nabla \omega ,\iota (\nabla f) \omega \rangle \,d\mu _g,\\&\frac{1}{{\mathrm {Vol}}(M)} \int _M \langle \iota (\nabla f) {\mathcal {R}}_k \omega , \iota (\nabla f)\omega \rangle \,d\mu _g,\\&\frac{1}{{\mathrm {Vol}}(M)} \int _M \langle \sum _{i=1}^n\iota (\nabla _{e_i}\nabla f) (\nabla _{e_i}\omega ),\iota (\nabla f)\omega \rangle \,d\mu _g,\\&\frac{1}{{\mathrm {Vol}}(M)} \int _M \langle \sum _{i=1}^n\iota (e_i)(R(\nabla f,e_i)\omega ),\iota (\nabla f) \omega \rangle \,d\mu _g. \end{aligned}$$

We have

$$\begin{aligned}&\int _M \langle \iota (\nabla f)\Delta \omega ,\iota (\nabla f) \omega \rangle \,d\mu _g\\&\quad =\int _M \langle d \omega ,d (d f \wedge \iota (\nabla f) \omega )\rangle \,d\mu _g+\int _M \langle d^*\omega ,d^*(d f \wedge \iota (\nabla f) \omega )\rangle \,d\mu _g \end{aligned}$$

and

$$\begin{aligned}&|\langle d \omega ,d (d f \wedge \iota (\nabla f) \omega )\rangle |\\&=\quad |\langle d \omega , \sum _{i=1}^n d f\wedge e^i \wedge \left( \iota (\nabla _{e_i}\nabla f) \omega +\iota (\nabla f) \nabla _{e_i}\omega \right) \rangle |\\&\quad \le C|\nabla \omega ||\nabla f|(|\nabla ^2 f||\omega |+|\nabla f||\nabla \omega |),\\&|\langle d^*\omega ,d^*(d f \wedge \iota (\nabla f) \omega )\rangle |\\&\quad =|\langle d^*\omega , \sum _{i=1}^n \iota (e_i)\left( \nabla _{e_i} d f\wedge \iota (\nabla f) \omega + d f\wedge \iota (\nabla _{e_i} \nabla f) \omega + d f\wedge \iota (\nabla f) \nabla _{e_i}\omega \right) \rangle |\\&\quad \quad \le C|\nabla \omega ||\nabla f|(|\nabla ^2 f||\omega |+|\nabla f||\nabla \omega |). \end{aligned}$$

Thus, we get

$$\begin{aligned} \left| \frac{1}{{\mathrm {Vol}}(M)} \int _M \langle \iota (\nabla f)\Delta \omega ,\iota (\nabla f) \omega \rangle \,d\mu _g\right| \le C\Vert f\Vert _2^2\lambda ^{1/2}. \end{aligned}$$

(12)

We have

$$\begin{aligned} \int _M \langle \iota (\nabla f)\nabla ^*\nabla \omega ,\iota (\nabla f) \omega \rangle \,d\mu _g =\int _M \langle \nabla \omega ,\nabla (d f \wedge \iota (\nabla f) \omega )\rangle \,d\mu _g \end{aligned}$$

and $|\langle \nabla \omega ,\nabla (d f \wedge \iota (\nabla f) \omega )\rangle | \le C|\nabla \omega ||\nabla f|(|\nabla ^2 f||\omega |+|\nabla f||\nabla \omega |).$ Thus, we get

$$\begin{aligned} \left| \frac{1}{{\mathrm {Vol}}(M)} \int _M \langle \iota (\nabla f)\nabla ^*\nabla \omega ,\iota (\nabla f) \omega \rangle \,d\mu _g\right| \le C\Vert f\Vert _2^2\lambda ^{1/2}. \end{aligned}$$

(13)

By Theorem 2.6, (12) and (13), we have

$$\begin{aligned} \begin{aligned}&\left| \frac{1}{{\mathrm {Vol}}(M)} \int _M \langle \iota (\nabla f) {\mathcal {R}}_k \omega , \iota (\nabla f)\omega \rangle \,d\mu _g\right| \\&\quad \le \frac{1}{{\mathrm {Vol}}(M)}\left( \left| \int _M \langle \iota (\nabla f)\Delta \omega , \iota (\nabla f) \omega \rangle \,d\mu _g\right| \right. \\&\qquad \left. + \left| \int _M \langle \iota (\nabla f)\nabla ^*\nabla \omega , \iota (\nabla f)\omega \rangle \,d\mu _g\right| \right) \\&\quad \le C\Vert f\Vert _2^2\lambda ^{1/2}. \end{aligned} \end{aligned}$$

(14)

Since $|\langle \sum _{i=1}^n\iota (\nabla _{e_i}\nabla f) (\nabla _{e_i}\omega ),\iota (\nabla f)\omega \rangle | \le C|\omega ||\nabla f| |\nabla \omega ||\nabla ^2 f|,$ we have

$$\begin{aligned} \left| \frac{1}{{\mathrm {Vol}}(M)} \int _M \langle \sum _{i=1}^n\iota (\nabla _{e_i}\nabla f) (\nabla _{e_i}\omega ),\iota (\nabla f)\omega \rangle \,d\mu _g\right| \le C \Vert f\Vert _2^2\lambda ^{1/2}. \end{aligned}$$

(15)

Let us estimate

$$\begin{aligned} \frac{1}{{\mathrm {Vol}}(M)} \int _M \langle \sum _{i=1}^n\iota (e_i)(R(\nabla f,e_i)\omega ),\iota (\nabla f) \omega \rangle \,d\mu _g. \end{aligned}$$

We have

$$\begin{aligned} \begin{aligned}&\left| \frac{1}{{\mathrm {Vol}}(M)} \int _M \langle \nabla _{\nabla f} d^*\omega ,\iota (\nabla f) \omega \rangle \,d\mu _g\right| \\&\quad =\left| \frac{1}{{\mathrm {Vol}}(M)} \int _M \langle d^*\omega ,\nabla ^*(d f\otimes \iota (\nabla f) \omega )\rangle \,d\mu _g\right| \\&\quad \le C\Vert f\Vert _2^2\lambda ^{1/2},\\&\left| \frac{1}{{\mathrm {Vol}}(M)} \int _M \langle d^*\nabla _{\nabla f}\omega ,\iota (\nabla f) \omega \rangle \,d\mu _g\right| \\&\quad =\left| \frac{1}{{\mathrm {Vol}}(M)} \int _M \langle \nabla \omega , d f \otimes d (\iota (\nabla f) \omega )\rangle \,d\mu _g\right| \\&\quad \le C\Vert f\Vert _2^2\lambda ^{1/2} \end{aligned} \end{aligned}$$

and

$$\begin{aligned}&\left| \frac{1}{{\mathrm {Vol}}(M)} \int _M \langle \sum _{i,j=1}^n \langle \nabla _{e_j} \nabla f, e_i\rangle \iota (e_j)\nabla _{e_i}\omega ,\iota (\nabla f)\omega \rangle \,d\mu _g\right| \\&\quad \le C\Vert f\Vert _2^2\lambda ^{1/2}. \end{aligned}$$

Thus, by Lemma 2.8 (iii), we get

$$\begin{aligned} \left| \frac{1}{{\mathrm {Vol}}(M)} \int _M \langle \sum _{i=1}^n\iota (e_i)(R(\nabla f,e_i)\omega ),\iota (\nabla f) \omega \rangle \,d\mu _g\right| \le C\Vert f\Vert _2^2\lambda ^{1/2}. \end{aligned}$$

(16)

By (12), (14), (15), (16) and Lemma 2.8, we get (ii) and (iii).

Since $\nabla (\iota (\nabla f)\omega )-\sum _{i=1}^n e^i\otimes \iota (\nabla _{e_i}\nabla f)\omega =\sum _{i=1}^n e^i\otimes \iota (\nabla f)\nabla _{e_i}\omega ,$ we get (iv) and (vi).

Since $d (\iota (\nabla f)\omega )-\sum _{i=1}^n e^i\wedge \iota (\nabla _{e_i}\nabla f)\omega =\sum _{i=1}^n e^i\wedge \iota (\nabla f)\nabla _{e_i}\omega ,$ we get (v).

By Theorem 2.6 and (4), we have

$$\begin{aligned} \begin{aligned}&\frac{1}{{\mathrm {Vol}}(M)}\int _M \langle {\mathcal {R}}_{k-1}(\iota (\nabla f)\omega ),\iota (\nabla f)\omega \rangle \,d\mu _g\\&\quad =\frac{1}{{\mathrm {Vol}}(M)}\int _M \langle (\Delta -\nabla ^*\nabla )(\iota (\nabla f)\omega ),\iota (\nabla f)\omega \rangle \,d\mu _g\\&\quad = \frac{k-1}{k} \frac{1}{{\mathrm {Vol}}(M)}\int _M \langle \Delta (\iota (\nabla f)\omega ), \iota (\nabla f)\omega \rangle \,d\mu _g\\&\qquad +\left( \frac{n-k+1}{n-k+2}-\frac{k-1}{k} \right) \Vert d^*(\iota (\nabla f)\omega )\Vert _2^2-\Vert T(\iota (\nabla f)\omega )\Vert _2^2. \end{aligned} \end{aligned}$$

Thus, by (i), (ii) and (iii), we get (vii)

Finally, we prove (viii). Suppose that M is oriented and $1\le k\le n/2$. Since $\nabla (*\omega )=*\nabla \omega$, we have

$$\begin{aligned}&\frac{1}{{\mathrm {Vol}}(M)}\int _M \langle \iota ({\mathrm {Ric}}(\nabla f))*\omega ,\iota (\nabla f)*\omega \rangle \,d\mu _g\\&\le \frac{n-k-1}{n-k} \frac{1}{{\mathrm {Vol}}(M)}\int _M \langle \iota (\nabla \Delta f)*\omega ,\iota (\nabla f)*\omega \rangle \,d\mu _g-\Vert T(\iota (\nabla f)*\omega )\Vert _2^2\\&\qquad +C\Vert f\Vert _2^2\lambda ^{1/2} \end{aligned}$$

by (vii). Thus, by (8), (i), (iii) and (vii), we get

$$\begin{aligned}&\frac{1}{{\mathrm {Vol}}(M)}\int _M {\mathrm {Ric}}(\nabla f,\nabla f)|\omega |^2\,d\mu _g\\&\quad \le \frac{n-k-1}{n-k} \frac{1}{{\mathrm {Vol}}(M)}\int _M \langle \nabla \Delta f,\nabla f\rangle |\omega |^2 \,d\mu _g -\Vert T(\iota (\nabla f)\omega )\Vert _2^2\\&\qquad -\Vert T(\iota (\nabla f)*\omega )\Vert _2^2\\&\qquad -\left( \frac{n-k-1}{n-k} -\frac{k-1}{k} \right) \frac{1}{{\mathrm {Vol}}(M)}\int _M \langle \iota (\nabla \Delta f)\omega ,\iota (\nabla f)\omega \rangle \,d\mu _g +C\Vert f\Vert _2^2\lambda ^{1/2}\\&\le \frac{n-k-1}{n-k} \frac{1}{{\mathrm {Vol}}(M)}\int _M \langle \nabla \Delta f,\nabla f\rangle |\omega |^2 \,d\mu _g -\Vert T(\iota (\nabla f)\omega )\Vert _2^2\\&\qquad -\Vert T(\iota (\nabla f)*\omega )\Vert _2^2\\&\qquad -\left( \frac{n-k-1}{n-k} -\frac{k-1}{k} \right) \Vert d(\iota (\nabla f)\omega )\Vert ^2_2 +C\Vert f\Vert _2^2\lambda ^{1/2}. \end{aligned}$$

This gives (viii). $\square$

3.3 Eigenvalue Estimate

In this subsection, we complete the proofs of Main Theorems 1 and 3. Recall that $\lambda _1(\Delta _{C,p})$ denotes the first eigenvalue of the connection Laplacian $\Delta _{C,p}$ acting on p-forms:

$$\begin{aligned} \Delta _{C,p}:=\nabla ^*\nabla :\Gamma (\bigwedge ^p T^*M)\rightarrow \Gamma (\bigwedge ^p T^*M). \end{aligned}$$

It is enough to show Main Theorem 1 when $\lambda _1(\Delta _{C,p})\le 1$. Note that we always have $\lambda _1(\Delta _{C,1})\ge 1$ if ${\mathrm {Ric}}_g\ge (n-1)g$.

We need the following $L^\infty$ estimates.

Lemma 3.7

Take an integer $n\ge 2$ and positive real numbers $K>0$, $D>0$, $\Lambda >0$. Let (M, g) be an n-dimensional closed Riemannian manifold with ${\mathrm {Ric}}\ge -Kg$ and ${\mathrm {diam}}(M)\le D$. Then, we have the following:

• (i) For any function $f\in C^\infty (M)$ and any $\lambda \ge 0$ with $\Delta f=\lambda f$ and $\lambda \le \Lambda$, then we have $\Vert \nabla f\Vert _\infty \le C(n,K,D,\Lambda )\Vert f\Vert _2$ and $\Vert f\Vert _\infty \le C(n,K,D,\Lambda )\Vert f\Vert _2$.

• (ii) For any p-form $\omega \in \Gamma \left( \bigwedge ^p T^*M\right)$ and any $\lambda \ge 0$ with $\Delta _{C,p} \omega =\lambda \omega$ and $\lambda \le \Lambda$, then we have $\Vert \omega \Vert _\infty \le C(n,K,D,\Lambda )\Vert \omega \Vert _2$.

Proof

By the gradient estimate for eigenfunctions [19, Theorem 7.3], we get (i).

Let us show (ii). Since we have

$$\begin{aligned} \Delta |\omega |^2=2\langle \Delta _{C,p} \omega , \omega \rangle -2|\nabla \omega |^2\le 2 \Lambda |\omega |^2, \end{aligned}$$

we get $\Vert \omega \Vert _\infty \le C$ by [20, Proposition 9.2.7] (see also Propositions 7.1.13 and 7.1.17 in [20]). Note that our sign convention of the Laplacian is different from [20]. $\square$

We use the following proposition not only for the proofs of Main Theorems 1 and 3 but also for other main theorems.

Proposition 3.8

For given integers $n\ge 4$ and $2\le p \le n/2$, there exists a constant $C(n,p)>0$ such that the following property holds. Let (M, g) be an n-dimensional closed oriented Riemannian manifold with ${\mathrm {Ric}}_g\ge (n-p-1)g$. Suppose that an integer $i\in {\mathbb {Z}}_{>0}$ satisfies $\lambda _i(g)\le n-p+1$, and there exists an eigenform $\omega$ of the connection Laplacian $\Delta _{C,p}$ acting on p-forms with $\Vert \omega \Vert _2=1$ corresponding to the eigenvalue $\lambda$ with $0\le \lambda \le 1$. Then, we have

$$\begin{aligned}&\frac{n-p-1}{n-p}\lambda _i(g)\left( \lambda _i(g)-(n-p)\right) \Vert f_i\Vert ^2\\&\quad \ge \Vert T(\iota (\nabla f_i)\omega )\Vert _2^2 +\Vert T(\iota (\nabla f_i)*\omega )\Vert _2^2\\&\qquad +\left( \frac{n-p-1}{n-p} -\frac{p-1}{p} \right) \Vert d(\iota (\nabla f_i)\omega )\Vert ^2_2 -C\lambda ^{1/2}\Vert f_i\Vert _2^2, \end{aligned}$$

where $f_i$ denotes the i-th eigenfunction of the Laplacian acting on functions.

Proof

By Lemma 3.6 (viii), we have

$$\begin{aligned}&\frac{n-p-1}{{\mathrm {Vol}}(M)}\int _M \langle \nabla f_i,\nabla f_i\rangle |\omega |^2\,d\mu _g\\&\quad \le \frac{1}{{\mathrm {Vol}}(M)}\int _M {\mathrm {Ric}}(\nabla f_i,\nabla f_i)|\omega |^2\,d\mu _g\\&\quad \le \frac{n-p-1}{n-p}\frac{\lambda _i(g)}{{\mathrm {Vol}}(M)}\int _M \langle \nabla f_i,\nabla f_i\rangle |\omega |^2\,d\mu _g -\Vert T(\iota (\nabla f_i)\omega )\Vert _2^2\\&\qquad -\Vert T(\iota (\nabla f_i)*\omega )\Vert _2^2\\&\qquad -\left( \frac{n-p-1}{n-p} -\frac{p-1}{p} \right) \Vert d(\iota (\nabla f_i)\omega )\Vert ^2_2 +C\lambda ^{1/2}\Vert f_i\Vert _2^2. \end{aligned}$$

Thus, we get the proposition by Lemma 3.5. $\square$

Proof of Main Theorem 1

If M is orientable, we get the theorem immediately by Proposition 3.8. If M is not orientable, we get the theorem by considering the two-sheeted orientable Riemannian covering $\pi :({\widetilde{M}},{\tilde{g}})\rightarrow (M,g)$ because we have $\lambda _1(g)\ge \lambda _1({\tilde{g}})$ and $\lambda _1(\Delta _{C,p},g)\ge \lambda _1(\Delta _{C,p},{\tilde{g}}).$ $\square$

Similarly, we get Main Theorem 3 because $\lambda _1(\Delta _{C,p},g)=\lambda _1(\Delta _{C,n-p},g)$ holds if the manifold is orientable.

4 Pinching

In this section, we show the remaining main theorems. Main Theorem 2 is proved in Sect. 4.5 except for the orientability, and the orientability is proved in Sect. 4.7. Main Theorem 4 is proved in Sect. 4.8.

We list assumptions of this section.

Assumption 4.1

Throughout in this section, we assume the following:

• $n\ge 5$, $2\le p < n/2$ and $1\le k\le n-p+1$.

• (M, g) is an n-dimensional closed Riemannian manifold with ${\mathrm {Ric}}_g\ge (n-p-1)g$.

• $C=C(n,p)>0$ denotes a positive constant depending only on n and p.

• $\delta >0$ satisfies $\delta \le \delta _0$ for sufficiently small $\delta _0=\delta _0(n,p)>0$.

• $f_i\in C^\infty (M)$ ($i\in \{1,\ldots ,k\}$) is an eigenfunction of the Laplacian acting on functions with $\Vert f_i\Vert _2^2=1/(n-p+1)$ corresponding to the eigenvalue $\lambda _i$ with $0<\lambda _i\le n-p+\delta$ such that

  $$\begin{aligned} \int _M f_i f_j\,d\mu _g=0 \end{aligned}$$

  holds for any $i\ne j$.

Note that, for given real numbers a, b with $0<b<a$ and a positive constant $C>0$, we can assume that $C \delta ^a\le \delta ^b.$ At the beginning of each subsections, we add either one of the following assumptions if necessary.

Assumption 4.2

There exists an eigenform $\omega \in \Gamma (\bigwedge ^p T^*M)$ of the connection Laplacian $\Delta _{C,p}$ with $\Vert \omega \Vert _2=1$ corresponding to the eigenvalue $\lambda$ with $0\le \lambda \le \delta$.

Assumption 4.3

There exists an eigenform $\xi \in \Gamma (\bigwedge ^{n-p} T^*M)$ of the connection Laplacian $\Delta _{C,n-p}$ with $\Vert \xi \Vert _2=1$ corresponding to the eigenvalue $\lambda$ with $0\le \lambda \le \delta$.

Under our assumptions, we have $\Vert \omega \Vert _\infty \le C$, $\Vert \xi \Vert _{\infty } \le C$, $\Vert f_i\Vert _\infty \le C$ and $\Vert \nabla f_i\Vert _\infty \le C$ for all i by Lemma 3.7. By Main Theorems 1 and 3, we have $\lambda _i\ge n-p-C(n,p)\delta ^{1/2}$ for all i. Note that we do not assume that $\lambda _i=\lambda _i(g)$.

4.1 Useful Techniques

In this subsection, we list some useful techniques for our pinching problems. Although we suppose that Assumption 4.1 holds, most assertions hold under weaker assumptions.

The following lemma is a variation of the Cheng-Yau estimate. See [1, Lemma 2.10] for the proof (see also [6, Theorem 7.1]).

Lemma 4.4

Take a positive real number $0<\epsilon _1 \le 1$. For any function $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots ,f_k\}$ and any point $x\in M$, we have

$$\begin{aligned} |\nabla f|^2(x)\le \frac{C}{\epsilon _1}\left( f(p)-f(x)+\epsilon _1\Vert f\Vert _2\right) ^2, \end{aligned}$$

where $p\in M$ denotes a maximum point of f.

The following theorem is an easy consequence of the Bishop-Gromov inequality.

Theorem 4.5

For any $p\in M$ and $0<r\le {\mathrm {diam}}(M)+1$, we have $r^n {\mathrm {Vol}}(M)\le C{\mathrm {Vol}}(B_r(p))$.

The following theorem is due to Cheeger-Colding [7] (see also [20, Theorem 7.1.10]). By this theorem, we get integral pinching conditions along the geodesics under the integral pinching condition for a function on M.

Theorem 4.6

(segment inequality) For any non-negative measurable function $h:M\rightarrow {\mathbb {R}}_{\ge 0}$, we have

$$\begin{aligned} \frac{1}{{\mathrm {Vol}}(M)^2}\int _{M\times M} \frac{1}{d(y_1,y_2)}\int _0^{d(y_1,y_2)} h\circ \gamma _{y_1,y_2}(s) \,dsdy_1dy_2\le \frac{C}{{\mathrm {Vol}}(M)}\int _M h\,d\mu _g. \end{aligned}$$

Remark 4.1

The book [20] deals with the segment $c_{y_1,y_2}:[0,1]\rightarrow M$ for each $y_1,y_2\in M$, defined to be $c_{y_1,y_2}(0)=y_1$, $c_{x,y}(1)=y_2$ and $\nabla _{\partial /\partial t} {\dot{c}}=0$. We have $c_{x,y}(t)=\gamma _{x,y}(t d(x,y))$ for all $t\in [0,1]$ and

$$\begin{aligned} d(y_1,y_2)\int _0^1 h\circ c_{y_1,y_2}(t) \,d t=\int _0^{d(y_1,y_2)} h\circ \gamma _{y_1,y_2}(s) \,d s. \end{aligned}$$

After getting integral pinching conditions along the geodesics, we use the following lemma to get $L^\infty$ error estimate along them. The proof is standard (c.f. [7, Lemma 2.41]).

Lemma 4.7

Take positive real numbers $l,\epsilon >0$ and a non-negative real number $r\ge 0$. Suppose that a smooth function $u:[0,l]\rightarrow {\mathbb {R}}$ satisfies

$$\begin{aligned} \int _0^l |u''(t)+r^2 u(t)| \,dt\le \epsilon . \end{aligned}$$

Then, we have

$$\begin{aligned} \begin{aligned} \left| u(t)-u(0) \cos r t- \frac{u'(0)}{r} \sin r t\right|&\le \epsilon \frac{\sinh rt}{r},\\ \left| u'(t)+ r u(0)\sin r t- u'(0)\cos r t\right|&\le \epsilon +\int _0^t\left| u(s)-u(0)\cos r s-\frac{u'(0)}{r}\sin r s\right| \,ds, \end{aligned} \end{aligned}$$

for all $t\in [0,l]$, where we defined $\frac{1}{r}\sin r t:=t,$ $\frac{1}{r}\sinh r t:=t$ if $r=0$.

The following lemma is standard.

Lemma 4.8

For all $t\in {\mathbb {R}}$, we have

$$\begin{aligned} 1-\frac{1}{2}t^2\le \cos t\le 1-\frac{1}{2}t^2+\frac{1}{24}t^4. \end{aligned}$$

For any $t\in [-\pi ,\pi ]$, we have $\cos t\le 1-\frac{1}{9}t^2$, and so $|t|\le 3(1-\cos t)^{1/2}$. For any $t_1,t_2 \in [0,\pi ]$, we have $|t_1-t_2|\le 3|\cos t_1-\cos t_2|^{1/2}$.

Finally, we recall some facts about the geodesic flow. Let UM denotes the sphere bundle defined by

$$\begin{aligned} U M:=\{u\in TM:|u|=1\}. \end{aligned}$$

There exists a natural Riemannian metric G on UM, which is the restriction of the Sasaki metric on TM (see [21, p.55]). The Riemannian volume measure $\mu _G$ satisfies

$$\begin{aligned} \int _{UM} F\,d\mu _G=\int _M \int _{U_p M} F(u)\, d\mu _0(u) \,d\mu _g(p) \end{aligned}$$

for any $F\in C^\infty (U M)$, where $\mu _0$ denotes the standard measure on $U_p M\cong S^{n-1}$. The geodesic flow $\phi _t:U M\rightarrow U M$ ($t\in {\mathbb {R}}$) is defined by

$$\begin{aligned} \phi _t(u):=\left. \frac{\partial }{\partial s}\right| _{s=t}\gamma _u (s)\in U_{\gamma _u(t)} M \end{aligned}$$

for any $u\in U M$. Though $\phi _t$ does not preserve the metric G in general, it preserves the measure $\mu _G$. This is an easy consequence of [21, Lemma 4.4], which asserts that the geodesic flow on TM preserve the natural symplectic structure on TM. We can easily show the following lemma.

Lemma 4.9

For any $f\in C^\infty (M)$ and $l>0$, we have

$$\begin{aligned} \frac{1}{{\mathrm {Vol}}(M)}\int _M f \,d\mu _g=\frac{1}{l{\mathrm {Vol}}(UM)}\int _{UM}\int _0^l f\circ \gamma _u(t)\,d t\,d\mu _G(u). \end{aligned}$$

This kind of lemma was used by Colding [10] to prove that the almost equality of the Bishop comparison theorem implies the Gromov-Hausdorff closeness to the standard sphere.

4.2 Estimates for the Segments

In this subsection, we suppose that Assumption 4.2 holds. The goal is to give error estimates along the geodesics. We first list some basic consequences of our pinching condition.

Lemma 4.10

For any $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots ,f_{k}\}$, we have

• (i) $\Vert \iota (\nabla f)\omega \Vert _2^2\le C\delta ^{1/2}\Vert f\Vert _2^2$,

• (ii) $\Vert \nabla (\iota (\nabla f)\omega )\Vert _2^2\le C\delta ^{1/2}\Vert f\Vert _2^2$,

• (iii) $\Vert (|\nabla ^2 f|^2-\frac{1}{n-p}|\Delta f|^2)|\omega |^2\Vert _1\le C\delta ^{1/4}\Vert f\Vert _2^2$.

Proof

It is enough to consider the case when M is orientable.

We first assume that $f=f_i$ for some $i=1,\ldots ,k$. Then, we have

$$\begin{aligned} \begin{aligned}&\Vert d(\iota (\nabla f)\omega )\Vert ^2_2\le C\delta ^{1/2}\Vert f\Vert _2^2,\\&\Vert d^*(\iota (\nabla f)\omega )\Vert ^2_2 \le C\delta ^{1/2}\Vert f\Vert _2^2,\quad \Vert T(\iota (\nabla f)\omega )\Vert _2^2\le C\delta ^{1/2}\Vert f\Vert _2^2,\\&\Vert d^*(\iota (\nabla f)*\omega )\Vert ^2_2 \le C\delta ^{1/2}\Vert f\Vert _2^2,\quad \Vert T(\iota (\nabla f)*\omega )\Vert _2^2 \le C\delta ^{1/2}\Vert f\Vert _2^2 \end{aligned} \end{aligned}$$

(17)

by Lemma 3.6 (i) and Proposition 3.8. Thus, by (4), we get

$$\begin{aligned} \Vert \nabla (\iota (\nabla f)\omega )\Vert ^2_2\le C\delta ^{1/2}\Vert f\Vert _2^2 \end{aligned}$$

(18)

and

$$\begin{aligned} \Vert \nabla (\iota (\nabla f)*\omega )\Vert ^2_2\le \frac{1}{n-p} \Vert d (\iota (\nabla f)*\omega )\Vert ^2_2+C\delta ^{1/2}\Vert f\Vert _2^2. \end{aligned}$$

(19)

Moreover, by Lemma 3.6 (iii), we have

$$\begin{aligned} \begin{aligned} \Vert \iota (\nabla f)\omega \Vert _2^2&=\frac{1}{\lambda _i}\frac{1}{{\mathrm {Vol}}(M)}\int _M \langle \iota (\nabla \Delta f)\omega , \iota (\nabla f)\omega \rangle \,d\mu _g\\&\le C\Vert d(\iota (\nabla f)\omega )\Vert ^2_2+C\Vert d^*(\iota (\nabla f)\omega )\Vert ^2_2+C\delta ^{1/2}\Vert f\Vert _2^2\\&\le C\delta ^{1/2}\Vert f\Vert _2^2. \end{aligned} \end{aligned}$$

(20)

For any $f=a_1 f_1+\cdots + a_k f_k\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots ,f_{k}\}$, we have (17), (18), (19), (20). For example, we have

$$\begin{aligned} \Vert \nabla (\iota (\nabla f)\omega )\Vert _2\le \sum _{i=1}^k |a_k|\Vert \nabla (\iota (\nabla f_i)\omega )\Vert _2\le C\delta ^{1/4}\sum _{i=1}^k |a_k|\Vert f_i\Vert _2\le C\delta ^{1/4}\Vert f\Vert _2. \end{aligned}$$

Thus, we get (i) and (ii) by (18) and (20).

Finally, we prove (iii). Take arbitrary $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots ,f_{k}\}$. We have

$$\begin{aligned} \begin{aligned}&\left| \sum _{i=1}^n e^i\otimes \iota (\nabla _{e_i}\nabla f)*\omega \right| ^2\\&\quad =\sum _{i=1}^n \langle \iota (\nabla _{e_i} \nabla f)*\omega ,\iota (\nabla _{e_i} \nabla f)*\omega \rangle =|\nabla ^2 f|^2|\omega |^2-\left| \sum _{i=1}^n e^i\otimes \iota (\nabla _{e_i}\nabla f)\omega \right| ^2. \end{aligned}\nonumber \\ \end{aligned}$$

(21)

Thus, we have

$$\begin{aligned} \begin{aligned}&\frac{1}{{\mathrm {Vol}}(M)}\int _M \left| |\nabla (\iota (\nabla f)*\omega )|^2-|\nabla ^2 f|^2|\omega |^2\right| \,d\mu _g\\&\quad \le \frac{1}{{\mathrm {Vol}}(M)}\int _M \left| |\nabla (\iota (\nabla f)*\omega )|^2-\left| \sum _{i=1}^n e^i\otimes \iota (\nabla _{e_i}\nabla f)*\omega \right| ^2\right| \,d\mu _g\\&\qquad +\frac{1}{{\mathrm {Vol}}(M)}\int _M \left| \sum _{i=1}^n e^i\otimes \iota (\nabla _{e_i}\nabla f)\omega \right| ^2\,d\mu _g, \end{aligned} \end{aligned}$$

and so we get

$$\begin{aligned} \frac{1}{{\mathrm {Vol}}(M)}\int _M \left| |\nabla (\iota (\nabla f)*\omega )|^2-|\nabla ^2 f|^2|\omega |^2\right| \,d\mu _g \le C\delta ^{1/2}\Vert f\Vert _2^2 \end{aligned}$$

(22)

by (ii) and Lemma 3.6 (iv) and (vi). We have

$$\begin{aligned} \begin{aligned}&\left| \sum _{i=1}^n e^i\wedge \iota (\nabla _{e_i}\nabla f)*\omega \right| ^2\\&\quad =\sum _{i=1}^n |\iota (\nabla _{e_i} \nabla f)*\omega |^2-\sum _{i,j=1}^n \langle \iota (e_i)\iota (\nabla _{e_j}\nabla f)*\omega , \iota (e_j)\iota (\nabla _{e_i}\nabla f)*\omega \rangle \\&\quad =|\nabla ^2 f|^2|\omega |^2-\left| \sum _{i=1}^n e^i\otimes \iota (\nabla _{e_i}\nabla f)\omega \right| ^2\\&\qquad -\sum _{i,j,k,l=1}^n \nabla ^2 f(e_i,e_k)\nabla ^2 f(e_j,e_l)\langle e^i\wedge e^l \wedge \omega , e^j\wedge e^k \wedge \omega \rangle \end{aligned} \end{aligned}$$

(23)

by (21) and (7). Since

$$\begin{aligned} \langle e^i\wedge e^l \wedge \omega , e^j\wedge e^k \wedge \omega \rangle&=(\delta _{i j}\delta _{k l}-\delta _{i k}\delta _{j l})|\omega |^2 -\delta _{i j}\langle \iota (e_k)\omega ,\iota (e_l)\omega \rangle \\&\quad +\delta _{i k}\langle \iota (e_j)\omega ,\iota (e_l)\omega \rangle +\langle e^l\wedge \omega ,e^j\wedge e^k\wedge \iota (e_i)\omega \rangle , \end{aligned}$$

we have

$$\begin{aligned} \begin{aligned}&\sum _{i,j,k,l=1}^n \nabla ^2 f(e_i,e_k)\nabla ^2 f(e_j,e_l)\langle e^i\wedge e^l \wedge \omega , e^j\wedge e^k \wedge \omega \rangle \\&\quad =|\nabla ^2 f|^2|\omega |^2-(\Delta f)^2|\omega |^2 -\sum _{i=1}^n | \iota (\nabla _{e_i}\nabla f)\omega |^2\\&\qquad -\sum _{i=1}^n\Delta f \langle \iota (\nabla _{e_i}\nabla f)\omega ,\iota (e_i)\omega \rangle \\&\qquad +\sum _{j,k,l=1}^n \nabla ^2 f(e_j,e_l)\langle e^l\wedge \omega ,e^j\wedge e^k\wedge \iota (\nabla _{e_k} \nabla f)\omega \rangle . \end{aligned} \end{aligned}$$

(24)

By (23) and (24), we get

$$\begin{aligned} \begin{aligned} \left| \sum _{i=1}^n e^i\wedge \iota (\nabla _{e_i}\nabla f)*\omega \right| ^2&=(\Delta f)^2|\omega |^2+\sum _{i=1}^n\Delta f \langle \iota (\nabla _{e_i}\nabla f)\omega ,\iota (e_i)\omega \rangle \\&\quad -\sum _{j,k,l=1}^n \nabla ^2 f(e_j,e_l)\langle e^l\wedge \omega ,e^j\wedge e^k\wedge \iota (\nabla _{e_k} \nabla f)\omega \rangle , \end{aligned} \end{aligned}$$

and so

$$\begin{aligned} \left| \left| \sum _{i=1}^n e^i\wedge \iota (\nabla _{e_i}\nabla f)*\omega \right| ^2-(\Delta f)^2|\omega |^2\right| \le C|\nabla ^2 f| |\omega |\left| \sum _{i=1}^n e^i\otimes \iota (\nabla _{e_i}\nabla f)\omega \right| \nonumber \\ \end{aligned}$$

(25)

By (25), (ii) and Lemma 3.6, we get

$$\begin{aligned} \frac{1}{{\mathrm {Vol}}(M)}\int _M \left| |d (\iota (\nabla f)*\omega )|^2-(\Delta f)^2|\omega |^2\right| \,d\mu _g \le C\delta ^{1/4}\Vert f\Vert _2^2. \end{aligned}$$

(26)

Since we have $|\nabla (\iota (\nabla f)*\omega )|^2\ge |d (\iota (\nabla f)*\omega )|^2/(n-p)$ at each point by (4), we get (iii) by (19), (22) and (26). $\square$

We use the following notation.

Notation 4.11

Take $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots ,f_{k}\}$ with $\Vert f\Vert _2^2=1/(n-p+1)$ and put

$$\begin{aligned} h_0&:=|\nabla ^2 f|^2, \quad h_1:=||\omega |^2-1|, \quad h_2:=|\nabla \omega |^2,\\ h_3&:=|\iota (\nabla f)\omega |^2, \quad h_4:=|\nabla (\iota (\nabla f)\omega )|^2,\quad h_5:=\left| \sum _{i=1}^n e^i\otimes \iota (\nabla _{e_i}\nabla f)\omega \right| ^2\\ h_6&:=\left| |\nabla ^2 f|^2-\frac{1}{n-p}(\Delta f)^2\right| |\omega |^2. \end{aligned}$$

For each $y_1\in M$, we define

$$\begin{aligned} D_f(y_1)&:=\Big \{y_2\in I_{y_1}\setminus \{y_1\}:\frac{1}{d(y_1,y_2)}\int _{0}^{d(y_1,y_2)} h_0\circ \gamma _{y_1,y_2}(s)\,d s\le \delta ^{-1/50} \text { and}\\&\qquad \qquad \frac{1}{d(y_1,y_2)}\int _{0}^{d(y_1,y_2)} h_i\circ \gamma _{y_1,y_2}(s)\,d s\le \delta ^{1/5} \text { for all }i=1,\ldots ,6 \Big \},\\ Q_f&:=\{y_1\in M: {\mathrm {Vol}}(M\setminus D_f(y_1))\le \delta ^{1/100}{\mathrm {Vol}}(M)\},\\ E_f(y_1)&:=\Big \{u\in U_{y_1} M: \frac{1}{\pi }\int _{0}^{\pi } h_0\circ \gamma _u(s)\,d s\le \delta ^{-1/50} \text { and }\\&\qquad \qquad \frac{1}{\pi }\int _{0}^{\pi } h_i\circ \gamma _u (s)\,d s\le \delta ^{1/5} \\&\quad \qquad \qquad \text { for all }i=1,\ldots ,6 \Big \},\\ R_f&:=\{y_1\in M: {\mathrm {Vol}}(U_{y_1} M\setminus E_f(y_1))\le \delta ^{1/100}{\mathrm {Vol}}(U_{y_1}M)\}. \end{aligned}$$

Now, we use the segment inequality and Lemma 4.9. We show that we have the integral pinching condition along most geodesics.

Lemma 4.12

Take $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots ,f_{k}\}$ with $\Vert f\Vert _2^2=1/(n-p+1)$. Then, we have the following properties:

• (i) ${\mathrm {Vol}}(M\setminus Q_f)\le C\delta ^{1/100}{\mathrm {Vol}}(M).$

• (ii) ${\mathrm {Vol}}(M\setminus R_f)\le C\delta ^{1/100}{\mathrm {Vol}}(M).$

Proof

We have $\Vert h_i\Vert _1\le C\delta ^{1/4}$ for all $i=1,\ldots ,6$ by the assumption, Lemmas 3.5, 3.6 (iv) and 4.10, and we have $\Vert h_0\Vert _1\le C$ by (11).

For any $y_1\in M\setminus Q_f$, we have ${\mathrm {Vol}}(M\setminus D_f(y_1))>\delta ^{1/100}{\mathrm {Vol}}(M)$, and so we have either

$$\begin{aligned} \frac{1}{{\mathrm {Vol}}(M)}\int _M\frac{1}{d(y_1,y_2)}\int _0^{d(y_1,y_2)}h_0\circ \gamma _{y_1,y_2}(s)\,d s \,d y_2\ge \frac{1}{7}\delta ^{-1/100} \end{aligned}$$

or

$$\begin{aligned} \frac{1}{{\mathrm {Vol}}(M)}\int _M\frac{1}{d(y_1,y_2)}\int _0^{d(y_1,y_2)}h_i\circ \gamma _{y_1,y_2}(s)\,d s \,d y_2\ge \frac{1}{7}\delta ^{21/100} \end{aligned}$$

for some $i=1,\ldots ,6$. Thus, we get either

$$\begin{aligned}&\frac{1}{{\mathrm {Vol}}(M)}\int _M \int _M\frac{1}{d(y_1,y_2)}\int _0^{d(y_1,y_2)}h_0\circ \gamma _{y_1,y_2}(s)\,d s \,d y_1\,d y_2\\&\quad \ge \frac{1}{49}\delta ^{-1/100}{\mathrm {Vol}}(M\setminus Q_f) \end{aligned}$$

or

$$\begin{aligned}&\frac{1}{{\mathrm {Vol}}(M)}\int _M \int _M\frac{1}{d(y_1,y_2)}\int _0^{d(y_1,y_2)}h_i\circ \gamma _{y_1,y_2}(s)\,d s \,d y_1\,d y_2\\&\quad \ge \frac{1}{49}\delta ^{21/100}{\mathrm {Vol}}(M\setminus Q_f) \end{aligned}$$

for some $i=1,\ldots ,6$. Therefore, we get (i) by the segment inequality (Theorem 4.6).

Similarly, we get (ii) by Lemma 4.9. $\square$

Under the pinching condition along the geodesic, we get the following:

Lemma 4.13

Take $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots ,f_{k}\}$ with $\Vert f\Vert _2^2=1/(n-p+1)$. Suppose that a geodesic $\gamma :[0,l]\rightarrow M$ satisfies one of the following:

• There exist $x\in M$ and $y\in D_f(x)$ such that $l=d(x,y)$ and $\gamma =\gamma _{x,y}$,

• There exist $x\in M$ and $u\in E_f(x)$ such that $l=\pi$ and $\gamma =\gamma _u$.

Then, we have

$$\begin{aligned} ||\omega |^2(s)-1|\le C\delta ^{1/10},\quad |\iota (\nabla f)\omega |(s)\le C\delta ^{1/10} \end{aligned}$$

for all $s\in [0,l]$, and at least one of the following:

• (i) $\frac{1}{l}\int _0^l|\nabla ^2 f|\circ \gamma (s)\,d s\le C\delta ^{1/250}$,

• (ii) There exists a parallel orthonormal basis $\{E^1(s),\ldots ,E^n(s)\}$ of $T_{\gamma (s)}^*M$ along $\gamma$ such that

  $$\begin{aligned} |\omega -E^{n-p+1}\wedge \cdots \wedge E^n|(s)\le C\delta ^{1/25} \end{aligned}$$

  for all $s\in [0,l]$, and

  $$\begin{aligned} \frac{1}{l}\int _0^l|\nabla ^2 f+f\sum _{i=1}^{n-p}E^i\otimes E^i|(s)\, d s\le C\delta ^{1/200}, \end{aligned}$$

  where we write $|\cdot |(s)$ instead of $|\cdot |\circ \gamma (s)$.

In particular, for both cases, there exists a parallel orthonormal basis $\{E^1(s),\ldots ,E^n(s)\}$ of $T_{\gamma (s)}^*M$ along $\gamma$ such that

$$\begin{aligned} \frac{1}{l}\int _0^l|\nabla ^2 f+f\sum _{i=1}^{n-p}E^i\otimes E^i|(s)\, d s\le C\delta ^{1/250}. \end{aligned}$$

Moreover, if we put ${\dot{\gamma }}^E:=\sum _{i=1}^{n-p} \langle {\dot{\gamma }},E_i\rangle E_i,$ where $\{E_1,\ldots ,E_n\}$ denotes the dual basis of $\{E^1,\ldots ,E^n\}$, then $|{\dot{\gamma }}^E|$ is constant along $\gamma$, and

$$\begin{aligned} \begin{aligned}&\left| f\circ \gamma (s)-f(\gamma (s_0))\cos (|{\dot{\gamma }}^E|(s-s_0))-\frac{1}{|{\dot{\gamma }}^E|}\langle \nabla f,{\dot{\gamma }}(s_0)\rangle \sin (|{\dot{\gamma }}^{E}|(s-s_0))\right| \\&\quad \le C\delta ^{1/250},\\&\left| \langle \nabla f, {\dot{\gamma }}(s)\rangle +f(\gamma (s_0))|{\dot{\gamma }}^{E}|\sin (|{\dot{\gamma }}^{E}|(s-s_0))-\langle \nabla f,{\dot{\gamma }}(s_0)\rangle \cos (|{\dot{\gamma }}^{E}|(s-s_0))\right| \\&\quad \le C\delta ^{1/250} \end{aligned} \end{aligned}$$

for all $s,s_0\in [0,l]$.

Proof

Let us show the first assertion. Since $\frac{d}{d s}|\omega |^2(s)=2\langle \nabla _{{\dot{\gamma }}}\omega ,\omega \rangle$, we have

$$\begin{aligned} \left| |\omega |^2(s)-|\omega |^2(0)\right|&=\left| \int _0^s \frac{d}{d s}|\omega |^2(t)\,d t\right| \\&\quad \le 2 \left( \int _0^s |\nabla \omega |^2 (t)\,d t\right) ^{1/2} \left( \int _0^s |\omega |^2 (t)\, d t\right) ^{1/2} \le C\delta ^{1/10} \end{aligned}$$

for all $s\in [0,l]$. Since we have $\int _0^l||\omega |^2-1|\, d t \le \delta ^{1/5}$, we get $||\omega |^2(s)-1|\le C\delta ^{1/10}$. In particular, $|\omega |(s)\ge 1/2$, and so

$$\begin{aligned} \frac{1}{l}\int _0^l\left| |\nabla ^2 f|^2-\frac{1}{n-p}(\Delta f)^2\right| (s)\,d s\le 2\delta ^{1/5}. \end{aligned}$$

(27)

Similarly, we have $|\iota (\nabla f)\omega |(s)\le C\delta ^{1/10}$ for all $s\in [0,l]$.

We show the remaining assertions. Put

$$\begin{aligned} A_1&:=\left\{ s\in [0,l]:\left| \sum _{i=1}^n e^i\otimes \iota (\nabla _{e_i}\nabla f)\omega \right| ^2(s)>\delta ^{1/10}\right\} ,\\ A_2&:=\left\{ s\in [0,l]:\left| |\nabla ^2 f|^2-\frac{1}{n-p}(\Delta f)^2\right| (s)>\delta ^{1/10}\right\} ,\\ A_3&:=\left\{ s\in [0,l]:|\nabla ^2 f|(s)<\delta ^{1/250}\right\} . \end{aligned}$$

Then, we have $H^1(A_1)\le \delta ^{1/10}l$ and $H^1(A_2)\le 2\delta ^{1/10} l$, where $H^1$ denotes the one dimensional Hausdorff measure. We consider the following two cases:

• (a) $[0,l]=A_1\cup A_2\cup A_3$,

• (b) $[0,l]\ne A_1\cup A_2\cup A_3$.

We first consider the case (a). Since $H^1([0,l]\setminus A_3)\le 3 \delta ^{1/10} l,$ we have

$$\begin{aligned} \int _{[0,l]\setminus A_3}|\nabla ^2 f|(s)\,d s&\le \left( \int _{[0,l]\setminus A_3}|\nabla ^2 f|^2(s)\,d s\right) ^{1/2}H^1 ([0,l]\setminus A_3)^{1/2}\\&\le C \delta ^{-1/100}\delta ^{1/20} l=C\delta ^{1/25}l. \end{aligned}$$

On the other hand, we have $\int _{A_3} |\nabla ^2 f|(s)\,d s\le \delta ^{1/250} l.$ Therefore, we get (i). Moreover, since $|\Delta f|\le \sqrt{n}|\nabla ^2 f|$ and $\left\| \Delta f-(n-p)f\right\| _{\infty }\le C\delta ^{1/2}$, we get

$$\begin{aligned} \frac{1}{l}\int _0^l|\nabla ^2 f+f\sum _{i=1}^{n-p}E^i\otimes E^i|(s)\, d s\le C\delta ^{1/250}, \end{aligned}$$

where $\{E^1(s),\ldots ,E^n(s)\}$ is any parallel orthonormal basis of $T_{\gamma (s)}^*M$ along $\gamma$.

We next consider the case (b). There exists $t\in [0,l]$ such that

$$\begin{aligned} \left| \sum _{i=1}^n e^i\otimes \iota (\nabla _{e_i}\nabla f)\omega \right| ^2(t)&\le \delta ^{1/10},\\ \left| |\nabla ^2 f|^2-\frac{1}{n-p}(\Delta f)^2\right| (t)&\le \delta ^{1/10},\quad |\nabla ^2 f|(t)\ge \delta ^{1/250}. \end{aligned}$$

Take an orthonormal basis $\{e_1,\ldots ,e_n\}$ of $T_{\gamma (t)}M$ such that $\nabla ^2 f(e_i,e_j)=\mu _i\delta _{i j}\, (\mu _i\in {\mathbb {R}})$ for all $i,j=1,\ldots ,n$. Let $\{e^1,\ldots ,e^n\}$ be the dual basis of $T_{\gamma (t)}^*M$. Then, we have

$$\begin{aligned} \delta ^{1/10}\ge \left| \sum _{i=1}^n e^i\otimes \iota (\nabla _{e_i}\nabla f)\omega \right| ^2(t) =\sum _{i=1}^n\mu _i^2 |\iota (e_i)\omega |^2(t). \end{aligned}$$

Thus, for each $i=1,\ldots ,n$, we have at least one of the following:

• (1) $|\mu _i|\le \delta ^{1/100}$,

• (2) $|\iota (e_i)\omega |(t)\le \delta ^{1/25}$.

Since $|\omega |(t)\ge 1/2$, we have ${\mathrm {Card}}\{i: |\iota (e_i)\omega |(t)\le \delta ^{1/25}\}\le n-p,$ and so ${\mathrm {Card}}\{i: |\mu _i|\le \delta ^{1/100}\}\ge p.$ Therefore, we can assume $|\mu _i|\le \delta ^{1/100}$ for all $i=n-p+1,\ldots , n$. Then, we get

$$\begin{aligned}&\left| \nabla ^2 f+\frac{\Delta f}{n-p}\sum _{i=1}^{n-p} e^i\otimes e^i \right| ^2(t) \\&\quad =|\nabla ^2 f|^2(t)+\frac{2}{n-p}(\Delta f)(t)\sum _{i=1}^{n-p}\mu _i +\frac{(\Delta f)^2(t)}{n-p}\\&\quad =|\nabla ^2 f|^2(t)-\frac{(\Delta f)^2(t)}{n-p}-\frac{2}{n-p}(\Delta f)(t)\sum _{i=n-p+1}^{n}\mu _i\\&\quad \le C\delta ^{1/100}. \end{aligned}$$

Putting $e_i\otimes e_i$ into the inside of the left hand side, we get $\left| \mu _i+\Delta f(t)/(n-p)\right| ^2\le C\delta ^{1/100}$ for all $i=1,\ldots , n-p$, and so

$$\begin{aligned} |\mu _i|\ge \frac{|\Delta f(t)|}{n-p}-C\delta ^{1/200}&\ge \left( \frac{|\nabla ^2 f|^2(t)-\delta ^{1/10}}{n-p}\right) ^{1/2}-C\delta ^{1/200}\\&\ge \left( \frac{\delta ^{1/125}-\delta ^{1/10}}{n-p}\right) ^{1/2}-C\delta ^{1/200} >\delta ^{1/100}. \end{aligned}$$

Thus, we have $|\iota (e_i)\omega |(t)\le \delta ^{1/25}$ for all $i=1,\ldots ,n-p$. Therefore, we get either $|\omega (t)-e^{n-p+1}\wedge \cdots \wedge e^n|\le C\delta ^{1/25}$ or $|\omega (t)+e^{n-p+1}\wedge \cdots \wedge e^n|\le C\delta ^{1/25}$ by $||\omega |^2(t)-1|\le C\delta ^{1/10}$. We can assume that $|\omega (t)-e^{n-p+1}\wedge \cdots \wedge e^n|\le C\delta ^{1/25}$.

Let $\{E_1,\ldots ,E_n\}$ be the parallel orthonormal basis of TM along $\gamma$ such that $E_i(t)=e_i$, and let $\{E^1,\ldots ,E^n\}$ be its dual. Because

$$\begin{aligned} \int _0^l \left| \frac{d}{d s}|\omega -E^{n-p+1}\wedge \cdots \wedge E^n|^2(s)\right| \,d s \le C\delta ^{1/10}, \end{aligned}$$

we get $|\omega -E^{n-p+1}\wedge \cdots \wedge E^n|(s)\le C\delta ^{1/25}$ for all $s\in [0,l]$. Thus, we get $|\langle \iota (E_i)\omega ,\iota (E_j)\omega \rangle |\le C\delta ^{1/25}$ for all $i=1,\cdots ,n$ and $j=1,\ldots ,n-p$, and $|\langle \iota (E_i)\omega ,\iota (E_j)\omega \rangle -\delta _{i j}|\le C\delta ^{1/25}$ for all $i,j=n-p+1,\cdots ,n$. Therefore, we get

$$\begin{aligned}&\left| \left| \sum _{i=1}^n E^i\otimes \iota (\nabla _{E_i}\nabla f)\omega \right| ^2-\sum _{i=1}^n\sum _{j=n-p+1}^n(\nabla ^2 f(E_i,E_j))^2 \right| \\&\quad =\left| \sum _{i,j,k=1}^n \nabla ^2 f(E_i,E_j)\nabla ^2 f(E_i,E_k)\langle \iota (E_j)\omega ,\iota (E_k)\omega \rangle \right. \\&\qquad \left. -\sum _{i=1}^n\sum _{j=n-p+1}^n(\nabla ^2 f (E_i,E_j))^2 \right| \\&\quad \le C |\nabla ^2 f|^2 \delta ^{1/25}. \end{aligned}$$

Thus, for all $i=1,\cdots ,n$ and $j=1,\ldots ,n-p$, we get

$$\begin{aligned} |\nabla ^2 f(E_i,E_j)|^2\le \left| \sum _{k=1}^n E^k\otimes \iota (\nabla _{E_k}\nabla f)\omega \right| ^2+C |\nabla ^2 f|^2 \delta ^{1/25}, \end{aligned}$$

and so

$$\begin{aligned} \frac{1}{l}\int _0^l|\nabla ^2 f (E_i,E_j)|^2(s)\,d s \le \delta ^{1/5}+C\delta ^{-1/50}\delta ^{1/25} \le C\delta ^{1/50}. \end{aligned}$$

This gives

$$\begin{aligned} \frac{1}{l}\int _0^l|\nabla ^2 f (E_i,E_j)|(s)\,d s \le C\delta ^{1/100} \end{aligned}$$

for all $i=1,\cdots ,n$ and $j=1,\ldots ,n-p$. Because

$$\begin{aligned} \left| \nabla ^2 f+\frac{\Delta f}{n-p}\sum _{i=1}^{n-p}E^i\otimes E^i \right| ^2&=|\nabla ^2 f|^2-\frac{(\Delta f)^2}{n-p}\\&\quad -2\frac{\Delta f}{n-p}\sum _{i=n-p+1}^{n}\nabla ^2 f (E_i,E_i), \end{aligned}$$

we have

$$\begin{aligned} \frac{1}{l}\int _0^l\left| \nabla ^2 f+\frac{\Delta f}{n-p}\sum _{i=1}^{n-p}E^i\otimes E^i \right| ^2\,d s \le 2\delta ^{1/5}+C\delta ^{1/100}\le C\delta ^{1/100} \end{aligned}$$

by (27). Since $\left\| f-\Delta f/(n-p)\right\| _{\infty }\le C\delta ^{1/2}$, we get (ii).

Let us show the final assertion. It is trivial that $|{\dot{\gamma }}^E|$ is constant along $\gamma$. Since we have

$$\begin{aligned} \left( \nabla ^2 f+f \sum _{i=1}^{n-p}E^i\otimes E^i\right) ({\dot{\gamma }},{\dot{\gamma }}) =\frac{d^2}{d s^2} f\circ \gamma + |{\dot{\gamma }}^E|^2 f\circ \gamma , \end{aligned}$$

we get

$$\begin{aligned} \int _0^l\left| \frac{d^2}{d s^2} f\circ \gamma (s) + |{\dot{\gamma }}^E|^2 f\circ \gamma (s)\right| \,d s\le C\delta ^{1/250}. \end{aligned}$$

Thus, we get the lemma by Lemma 4.7. $\square$

4.3 Almost Parallel $(n-p)$-form I

In this subsection, we suppose that Assumption 4.3 holds instead of 4.2. If M is orientable, then Assumption 4.3 implies 4.2, and so we assume that M is not orientable. We use the following notation.

Notation 4.14

Take $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots , f_k\}$ with $\Vert f\Vert _2^2=1/(n-p+1)$. Let $\pi :({\widetilde{M}},{\tilde{g}})\rightarrow (M,g)$ be the two-sheeted oriented Riemannian covering. Put ${\tilde{f}}:=f\circ \pi \in C^\infty ({\widetilde{M}})$, ${\widetilde{\xi }}:=\pi ^*\xi \in \Gamma (\bigwedge ^{n-p}T^*{\widetilde{M}})$ and $\omega :=*{\widetilde{\xi }}\in \Gamma (\bigwedge ^{p}T^*{\widetilde{M}})$. Define $h_0,\ldots ,h_6$, $Q_{{\tilde{f}}}$, $D_{{\tilde{f}}}({\tilde{y}}_1)$, $R_{{\tilde{f}}}$ and $E_{{\tilde{f}}}(\tilde{y_1})$ as Notation 4.11 for ${\tilde{f}}$, $\omega$ and ${\tilde{y}}_1\in {\widetilde{M}}$. Put

$$\begin{aligned} Q_f&:=M\setminus \pi \left( {\widetilde{M}}\setminus Q_{{\tilde{f}}}\right) ,\quad D_f(y_1):=M\setminus \pi \left( {\widetilde{M}}\setminus \bigcap _{{\tilde{y}}\in \pi ^{-1}(y_1)} D_{{\tilde{f}}}({\tilde{y}})\right) ,\\ R_f&:=M\setminus \pi \left( {\widetilde{M}}\setminus R_{{\tilde{f}}}\right) ,\quad E_f(y_1):=U_{y_1} M\setminus \bigcup _{{\tilde{y}}\in \pi ^{-1}(y_1)}\pi _*\left( U_{{\tilde{y}}}{\widetilde{M}}\setminus E_{{\tilde{f}}}({\tilde{y}})\right) \end{aligned}$$

for each $y_1\in M$.

We immediately have the following lemmas by Lemmas 4.12 and 4.13.

Lemma 4.15

We have the following:

• (i) ${\mathrm {Vol}}(M\setminus Q_f)\le C\delta ^{1/100}{\mathrm {Vol}}(M)$, and ${\mathrm {Vol}}(M\setminus D_f(y_1))\le 2\delta ^{1/100}{\mathrm {Vol}}({\widetilde{M}})=4\delta ^{1/100}{\mathrm {Vol}}(M)$ for each $y_1\in Q_f$.

• (ii) ${\mathrm {Vol}}(M\setminus R_f)\le C\delta ^{1/100}{\mathrm {Vol}}(M)$, and ${\mathrm {Vol}}(U_{y_1} M\setminus E_f(y_1))\le 2\delta ^{1/100}{\mathrm {Vol}}(U_{y_1}M)$ for each $y_1\in R_f$.

• (iii) Take $y_1\in M$ and $y_2\in D_f(y_1)$ and one of the lift of $\gamma _{y_1,y_2}$:

  $$\begin{aligned} {\tilde{\gamma }}_{y_1,y_2}:[0,d(y_1,y_2)]\rightarrow {\widetilde{M}}. \end{aligned}$$

  Put ${\tilde{y}}_1:={\tilde{\gamma }}_{y_1,y_2}(0)\in {\widetilde{M}}$ and ${\tilde{y}}_2:={\tilde{\gamma }}_{y_1,y_2}(d(y_1,y_2))\in {\widetilde{M}}$. Then, we have ${\tilde{y}}_2\in D_{{\tilde{f}}}({\tilde{y}}_1)$.

• (iv) Take $y_1\in M$ and $u\in E_f(y_1)$ and one of the lift of $\gamma _u$:

  $$\begin{aligned} {\tilde{\gamma }}_{u}:[0,\pi ]\rightarrow {\widetilde{M}}. \end{aligned}$$

  Put ${\tilde{y}}_1:={\tilde{\gamma }}_{u}(0)\in {\widetilde{M}}$ and ${\tilde{u}}:=\dot{{\tilde{\gamma }}}_{u}(0)\in U_{{\tilde{y}}_1}{\widetilde{M}}$. Then, we have ${\tilde{u}}\in E_{{\tilde{f}}}({\tilde{y}}_1)$.

Lemma 4.16

Suppose that a geodesic $\gamma :[0,l]\rightarrow M$ satisfies one of the following:

• There exist $x\in M$ and $y\in D_f(x)$ such that $l=d(x,y)$ and $\gamma =\gamma _{x,y}$,

• There exist $x\in M$ and $u\in E_f(x)$ such that $l=\pi$ and $\gamma =\gamma _u$.

Let ${\tilde{\gamma }}:[0,l]\rightarrow {\widetilde{M}}$ be one of the lift of $\gamma$. Then, we have

$$\begin{aligned} ||\omega |^2({\tilde{\gamma }}(s))-1|\le C\delta ^{1/10},\quad |\iota (\nabla {\tilde{f}})(\omega )|\circ {\tilde{\gamma }}(s)\le C\delta ^{1/10} \end{aligned}$$

for all $s\in [0,l]$, and at least one of the following:

• (i) $\frac{1}{l}\int _0^l|\nabla ^2 f|\circ \gamma (s)\,d s\le C\delta ^{1/250}$,

• (ii) There exists a parallel orthonormal basis $\{E^1(s),\ldots ,E^n(s)\}$ of $T_{\gamma (s)}^*M$ along $\gamma$ such that

  $$\begin{aligned} |\xi -E^{1}\wedge \cdots \wedge E^{n-p}|(s)\le C\delta ^{1/25} \end{aligned}$$

  for all $s\in [0,s]$, and

  $$\begin{aligned} \frac{1}{l}\int _0^l|\nabla ^2 f+f\sum _{i=1}^{n-p}E^i\otimes E^i|(s)\, d s\le C\delta ^{1/200}. \end{aligned}$$

4.4 Eigenfunction and Distance

In this subsection, we suppose that either Assumption 4.2 or 4.3 holds. In the following, Lemma 4.12 (resp. 4.13) shall be replaced by Lemma 4.15 (resp. 4.16) under Assumption 4.3. The following proposition, which asserts that our function is an almost cosine function in some sense, is the goal of this subsection. See Notation 4.11 (under Assumption 4.2) and Notation 4.14 (under Assumption 4.3) for the definitions of $D_f$, $Q_f$, $E_f$ and $R_f$.

Proposition 4.17

Take $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots ,f_{k}\}$ with $\Vert f\Vert _2^2=1/(n-p+1)$. There exists a point $p_f\in Q_f$ such that the following properties hold:

• (i) $\sup _M f\le f(p_f)+C\delta ^{1/100n}$ and $|f(p_f)-1|\le C\delta ^{1/800n}$,

• (ii) For any $x\in D_f(p_f)$ with $|\nabla f|(x)\le \delta ^{1/800n}$, we have $||f(x)|-1|\le C\delta ^{1/800n}.$

• (iii) For any $x\in D_f(p_f)\cap Q_f\cap R_f$, we have $|f(x)^2+|\nabla f|^2(x)-1|\le C \delta ^{1/800n}.$

• (iv) Put $A_f:=\{x\in M: |f(x)-1|\le \delta ^{1/900n}\}.$ Then, we have

  $$\begin{aligned} |f(x)-\cos d(x,A_f)|\le C\delta ^{1/2000n} \end{aligned}$$

  for all $x\in M$, and $\sup _{x\in M}d(x,A_f)\le \pi + C\delta ^{1/100n}.$

Proof

Take a maximum point ${\tilde{p}}\in M$ of f. Then, by the Bishop-Gromov theorem and Lemma 4.12, there exists a point $p_f\in Q_f$ with $d({\tilde{p}},p_f)\le C \delta ^{1/100n}$. By Lemmas 4.4 and 3.7, we have

$$\begin{aligned} |\nabla f|(p_f)\le C\delta ^{1/200n}. \end{aligned}$$

(28)

Claim 4.18

For any $x\in D_f(p_f)$ with $|\nabla f|(x)\le C\delta ^{1/800n}$, we have

$$\begin{aligned} ||f(x)|-|f(p_f)||\le C\delta ^{1/800n}. \end{aligned}$$

Proof of Claim 4.18

Since $|\nabla f|(p_f)\le C\delta ^{1/200n}$ and $|\nabla f|(x)\le C\delta ^{1/800n},$ we get

$$\begin{aligned} |f\circ \gamma _{p_f,x}(s)-f(p_f)\cos ( |{\dot{\gamma }}_{p_f,x}^E| s)|&\le C\delta ^{1/200n},\\ |f\circ \gamma _{p_f,x}(d(p_f,x)-s)-f(x)\cos ( |{\dot{\gamma }}_{p_f,x}^E|s)|&\le C\delta ^{1/800n} \end{aligned}$$

for all $s\in [0,d(p_f,x)]$ by Lemma 4.13. Thus, we have

$$\begin{aligned} |f(x)-f(p_f)\cos ( |{\dot{\gamma }}_{p_f,x}^E| d(p_f,x))|&\le C\delta ^{1/200n},\\ |f(p_f)-f(x)\cos ( |{\dot{\gamma }}_{p_f,x}^E|d(p_f,x))|&\le C\delta ^{1/800n}, \end{aligned}$$

and so we get $||f(x)|-|f(p_f)||\le C\delta ^{1/800n}$. $\square$

Similarly to $p_f$, we take a point $q_f\in Q_{f}(x)$ with $d({\tilde{q}},q_f)\le C\delta ^{1/100n}$, where ${\tilde{q}}\in M$ is minimum point of f. By $\Vert f\Vert _{\infty }\ge \Vert f\Vert _2=1/\sqrt{n-p+1}$, we have $\max \{|f(p_f)|,|f(q_f)|\}\ge 1/\sqrt{n-p+1}-C\delta ^{1/100n}$. Since $|\nabla f|(q_f)\le C\delta ^{1/200n}$, we have $|f(p_f)|\ge |f(q_f)|-C\delta ^{1/800n}$ by Claim 4.18. Therefore, we get

$$\begin{aligned} f(p_f)\ge \frac{1}{\sqrt{n-p+1}}-C\delta ^{1/800n}\ge \frac{1}{2\sqrt{n-p+1}}. \end{aligned}$$

(29)

Claim 4.19

Take $x\in M$ and $y\in D_f(x)$. Let $\{E^1,\ldots ,E^n\}$ be a parallel orthonormal basis along $\gamma _{x,y}$ in Lemma 4.13. If (i) holds in the lemma, we can assume that $E_1={\dot{\gamma }}_{x,y}$. Then, we have

$$\begin{aligned} |\langle \nabla f,{\dot{\gamma }}_{x,y}(s)\rangle -\langle \nabla f,{\dot{\gamma }}_{x,y}^{E}(s)\rangle |&\le C\delta ^{1/25}, \end{aligned}$$

(30)

$$\begin{aligned} |\langle \nabla f,{\dot{\gamma }}_{x,y}(s)\rangle |&\le |\nabla f(\gamma _{x,y}(s))||{\dot{\gamma }}_{x,y}^{E}|+C\delta ^{1/25} \end{aligned}$$

(31)

and

$$\begin{aligned} \begin{aligned} \left| f\circ \gamma _{x,y}(s)-f(x)\cos (|{\dot{\gamma }}_{x,y}^{E}|s)-\frac{1}{|{\dot{\gamma }}_{x,y}^{E}|}\langle \nabla f(x),{\dot{\gamma }}_{x,y}(0)\rangle \sin (|{\dot{\gamma }}_{x,y}^{E}|s)\right|&\le C\delta ^{1/250},\\ \left| \langle \nabla f, {\dot{\gamma }}_{x,y}(s)\rangle +f(x)|{\dot{\gamma }}_{x,y}^{E}|\sin (|{\dot{\gamma }}_{x,y}^{E}|s)-\langle \nabla f(x),{\dot{\gamma }}_{x,y}(0)\rangle \cos (|{\dot{\gamma }}_{x,y}^{E}|s)\right|&\le C\delta ^{1/250} \end{aligned} \end{aligned}$$

for all $s\in [0,d(x,y)]$.

Proof of Claim 4.19

If (i) holds in the lemma, ${\dot{\gamma }}_{x,y}={\dot{\gamma }}_{x,y}^E$, and so (30) and (31) are trivial. If (ii) in the lemma holds, we have $|\iota (\nabla f)(E^{n-p+1}\wedge \cdots \wedge E^n)|\le C\delta ^{1/25}$, and so $|\langle \nabla f(x),E_i\rangle |\le C\delta ^{1/25}$ for all $i=n-p+1,\ldots ,n$. This gives (30) and (31). We get the remaining part of the claim by Lemma 4.13 putting $s_0=0$. $\square$

Claim 4.20

For any $x\in Q_f\cap R_f$ with $|\nabla f|(x)\ge \delta ^{1/800n}$, we have

$$\begin{aligned} |f(x)^2+|\nabla f|^2(x)-f(p_f)^2|\le C\delta ^{1/800n}. \end{aligned}$$

Moreover, there exists a point $y\in D_f(p_f)\cap D_f(x)$ such that the following properties hold.

• (a) $d(x,y)< \pi$,

• (b) $|f(p_f)-f(y)|\le C \delta ^{1/800n}$,

• (c) $|f(x)-f(p_f)\cos d(x,y)|\le C \delta ^{1/800n},$

• (d) For any $z\in M$ with $d(x,z)\le d(x,y)-\delta ^{1/2000n}$, we have $f(p_f)-f(z)\ge \frac{1}{C}\delta ^{1/1000n}$.

Proof of Claim 4.20

Take $x\in Q_f\cap R_f$ with $|\nabla f|(x)\ge \delta ^{1/800n}$. By the definition of $R_f$, there exists a vector $u\in E_f(x)$ with

$$\begin{aligned} \left| \frac{\nabla f}{|\nabla f|}(x)-u \right| \le C \delta ^{1/100n}. \end{aligned}$$

Thus, we have

$$\begin{aligned} \Big |\langle \nabla f(x),{\dot{\gamma }}_u(0)\rangle -|\nabla f|(x)\Big |=|\nabla f|(x)-\langle \nabla f(x), u\rangle \le C\delta ^{1/100n}. \end{aligned}$$

(32)

Let $\{E^1,\ldots ,E^n\}$ be a parallel orthonormal basis along $\gamma _{u}$ in Lemma 4.13. We first suppose that (ii) holds in the lemma. Then, for all $i=n-p+1,\ldots , n$, we have $|\langle \nabla f, E_i\rangle |\le C\delta ^{1/25}$, and so

$$\begin{aligned} |\langle u,E_i\rangle |\le & {} \left| u-\frac{\nabla f}{|\nabla f|}(x)\right| +\left| \langle \frac{\nabla f}{|\nabla f|}(x), E_i\rangle \right| \\\le & {} C\delta ^{1/100n}+C\delta ^{1/25}\delta ^{-1/800n}\le C\delta ^{1/100n}. \end{aligned}$$

Thus, we get $|{\dot{\gamma }}_u^E|^2=|u^E|^2=1-\sum _{i=n-p+1}^n\langle u, E_i\rangle ^2\ge 1-C\delta ^{1/100n}.$ If (i) holds in the lemma, we can assume $u=E_1$, and so $|{\dot{\gamma }}_u^E|=|u^E|=1$. For both cases, we get

$$\begin{aligned} \begin{aligned} |f\circ \gamma _u(s)-f(x)\cos s-|\nabla f|(x)\sin s|&\le C\delta ^{1/100n}\\ |\langle \nabla f,{\dot{\gamma }}_u(s)\rangle +f(x)\sin s-|\nabla f|(x)\cos s|&\le C\delta ^{1/100n} \end{aligned} \end{aligned}$$

(33)

for all $s\in [0,\pi ]$ by (32). Take $s_0\in [0,\pi ]$ such that

$$\begin{aligned} \frac{f(x)}{(f(x)^2+|\nabla f|^2(x))^{1/2}}&=\cos s_0,\\ \frac{|\nabla f|(x)}{(f(x)^2+|\nabla f|^2(x))^{1/2}}&=\sin s_0. \end{aligned}$$

Since $\sin s_0\ge \frac{1}{C}\delta ^{1/800n}$ by the assumption, we have

$$\begin{aligned} \frac{1}{C}\delta ^{1/800n}\le s_0\le \pi -\frac{1}{C}\delta ^{1/800n}. \end{aligned}$$

(34)

By the definition of $s_0$ and the formulas for $\cos (s-s_0)$ and $\sin (s-s_0)$, we have

$$\begin{aligned} \begin{aligned} (f(x)^2+|\nabla f|^2(x))^{1/2}\cos (s-s_0)&=f(x)\cos s+|\nabla f|(x)\sin s,\\ (f(x)^2+|\nabla f|^2(x))^{1/2}\sin (s-s_0)&=f(x)\sin s-|\nabla f|(x)\cos s, \end{aligned} \end{aligned}$$

and so we get

$$\begin{aligned} \begin{aligned} |f\circ \gamma _u(s_0)-(f(x)^2+|\nabla f|^2(x))^{1/2}|&\le C\delta ^{1/100n},\\ |\langle \nabla f,{\dot{\gamma }}_u(s_0)\rangle |&\le C\delta ^{1/100n} \end{aligned} \end{aligned}$$

(35)

by (33). Take $y\in D_f(p_f)\cap D_f(x)$ with $d(\gamma _u(s_0),y)\le C\delta ^{1/100n}$. We have

$$\begin{aligned} d(x,y)\le d(x,\gamma _u(s_0))+d(\gamma _u(s_0),y)\le s_0+C\delta ^{1/100n}. \end{aligned}$$

(36)

By (35), we get

$$\begin{aligned} |f(y)-(f(x)^2+|\nabla f|^2(x))^{1/2}|\le C\delta ^{1/100n} \end{aligned}$$

(37)

Take a parallel orthonormal basis $\{\widetilde{E^1},\ldots ,\widetilde{E^n}\}$ of $T^*M$ along $\gamma _{x,y}$ in Lemma 4.13. By (34) and (36), we get (a) and

$$\begin{aligned} \frac{1}{C}\delta ^{1/800n}\le |{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|d(x,y)+s_0 \le 2\pi -\frac{1}{C}\delta ^{1/800n}, \end{aligned}$$

and so

$$\begin{aligned} \cos (|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|d(x,y)+s_0)\le 1-\frac{1}{C}\delta ^{1/400n}. \end{aligned}$$

(38)

If $|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|\le \delta ^{1/100}$, we have $|f(y)-f(x)|\le C\delta ^{1/250}$ by Claim 4.19, and so $(f(x)^2+|\nabla f|^2(x))^{1/2}-f(x)\le C\delta ^{1/100n}$ by (37). This contradicts to $|\nabla f|(x)\ge \delta ^{1/800n}.$ Thus, we get $|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|\ge \delta ^{1/100}$. Then, we have

$$\begin{aligned} \frac{1}{|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|}|\langle \nabla f(x),{\dot{\gamma }}_{x,y}(0)\rangle |\le |\nabla f|(x)+C\delta ^{3/100} \end{aligned}$$

(39)

and

$$\begin{aligned}&(f(x)^2+|\nabla f|^2(x))^{1/2}\\&\quad \le f(y)+C\delta ^{1/100n}\\&\quad \le f(x)\cos (|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|d(x,y))\\&\qquad +\frac{1}{|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|}\langle \nabla f(x),{\dot{\gamma }}_{x,y}(0)\rangle \sin (|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|d(x,y))+C\delta ^{1/100n}\\&\quad \le \left( f(x)^2+\frac{1}{|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|^2}\langle \nabla f(x),{\dot{\gamma }}_{x,y}(0)\rangle ^2\right) ^{1/2}+C\delta ^{1/100n} \end{aligned}$$

by Claim 4.19 and (37). Thus,

$$\begin{aligned} |\nabla f|^2(x) \le \frac{1}{|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|^2}\langle \nabla f(x),{\dot{\gamma }}_{x,y}(0)\rangle ^2 +C\delta ^{1/100n}. \end{aligned}$$

(40)

By (39) and (40), we get

$$\begin{aligned} \left| \frac{1}{|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|^2}\langle \nabla f(x),{\dot{\gamma }}_{x,y}(0)\rangle ^2-|\nabla f|^2(x)\right| \le C\delta ^{1/100n}. \end{aligned}$$

(41)

This gives

$$\begin{aligned} \begin{aligned}&\left| \frac{1}{|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|}|\langle \nabla f(x),{\dot{\gamma }}_{x,y}(0)\rangle |-|\nabla f|(x)\right| \\&\quad \le \left| \frac{1}{|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|^2}\langle \nabla f(x),{\dot{\gamma }}_{x,y}(0)\rangle ^2-|\nabla f|^2(x)\right| \delta ^{-1/800n}\le C\delta ^{7/800n}. \end{aligned} \end{aligned}$$

(42)

We show that $\langle \nabla f(x),{\dot{\gamma }}_{x,y}(0)\rangle > 0$. If $\langle \nabla f(x),{\dot{\gamma }}_{x,y}(0)\rangle \le 0$, we get

$$\begin{aligned} \left| f(y)-f(x)\cos (|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|d(x,y))+ |\nabla f|\sin (|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|d(x,y))\right| \le C\delta ^{7/800n} \end{aligned}$$

by (42) and Claim 4.19, and so

$$\begin{aligned} \left| f(y)-(f(x)^2+|\nabla f|^2(x))^{1/2}\cos (|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|d(x,y)+s_0)\right| \le C\delta ^{7/800n}. \end{aligned}$$

Thus, we get

$$\begin{aligned} \begin{aligned}&(f(x)^2+|\nabla f|^2(x))^{1/2}\\&\quad \le f(y)+C\delta ^{1/100n}\\&\quad \le (f(x)^2+|\nabla f|^2(x))^{1/2} \cos (|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|d(x,y)+s_0)+C\delta ^{7/800n}\\&\quad \le (f(x)^2+|\nabla f|^2(x))^{1/2} -\frac{1}{C}\delta ^{3/800n} \end{aligned} \end{aligned}$$

by (37), (38) and $|\nabla f|(x)\ge \delta ^{1/800n}$. This is a contradiction. Therefore, we get $\langle \nabla f(x),{\dot{\gamma }}_{x,y}(0)\rangle >0$. Thus,

$$\begin{aligned} \begin{aligned} \left| f(y)-(f(x)^2+|\nabla f|^2(x))^{1/2}\cos (|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|d(x,y)-s_0)\right|&\le C\delta ^{7/800n},\\ \left| \langle \nabla f(y), {\dot{\gamma }}_{x,y}\rangle +|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|(f(x)^2+|\nabla f|^2(x))^{1/2}\sin (|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|d(x,y)-s_0)\right|&\le C\delta ^{7/800n} \end{aligned}\nonumber \\ \end{aligned}$$

(43)

by (42) and Claim 4.19. Then, we have

$$\begin{aligned} (f(x)^2+|\nabla f|^2(x))^{1/2} (1-\cos (|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|d(x,y)-s_0)) \le C\delta ^{7/800n} \end{aligned}$$

by (37), and so

$$\begin{aligned} 1-\cos (|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|d(x,y)-s_0)\le C\delta ^{3/400n}. \end{aligned}$$

by $|\nabla f|(x)\ge \delta ^{1/800n}$. Since $-\pi<|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|d(x,y)-s_0<\pi ,$ we get

$$\begin{aligned} \left| |{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|d(x,y)-s_0\right| \le C\delta ^{3/800n}. \end{aligned}$$

(44)

Thus, we have $s_0\le |{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|s_0+ C\delta ^{3/800n}$ by (36), and so

$$\begin{aligned} 1-|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}| \le C\delta ^{1/400n} \end{aligned}$$

(45)

by (34). Thus, we get

$$\begin{aligned} |d(x,y)-s_0|\le C\delta ^{1/400n}. \end{aligned}$$

(46)

By (43) and (44), we have

$$\begin{aligned} |\langle \nabla f(y), {\dot{\gamma }}_{x,y}(d(x,y))\rangle |\le C\delta ^{3/800n}. \end{aligned}$$

(47)

We have

$$\begin{aligned} \begin{aligned}&\frac{d}{d s}\left( |\nabla f|^2(s)-\langle \nabla f,{\dot{\gamma }}_{x,y}(s)\rangle ^2\right) \\&\quad =2\left( \langle \nabla _{{\dot{\gamma }}_{x,y}}\nabla f,\nabla f\rangle (s)-\langle \nabla _{{\dot{\gamma }}_{x,y}}\nabla f,{\dot{\gamma }}_{x,y}(s)\rangle \langle \nabla f,{\dot{\gamma }}_{x,y}(s)\rangle \right) \\&\quad =2\langle \nabla ^2 f+ f\sum _{i=1}^{n-p}{\widetilde{E}}^i\otimes {\widetilde{E}}^i,{\dot{\gamma }}_{x,y}\otimes \nabla f\rangle (s)-2f\langle \nabla f,{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}\rangle \\&\qquad -2\langle \nabla ^2 f+ f\sum _{i=1}^{n-p}{\widetilde{E}}^i\otimes {\widetilde{E}}^i,{\dot{\gamma }}_{x,y}\otimes {\dot{\gamma }}_{x,y} \rangle (s)\langle \nabla f,{\dot{\gamma }}_{x,y}(s)\rangle \\&\qquad +2f|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|^2\langle \nabla f,{\dot{\gamma }}_{x,y}(s)\rangle . \end{aligned} \end{aligned}$$

(48)

Thus, we get

$$\begin{aligned} \begin{aligned}&\left| \frac{d}{d s}\left( |\nabla f|^2(s)-\langle \nabla f,{\dot{\gamma }}_{x,y}(s)\rangle ^2\right) \right| \\&\quad \le C\left| \nabla ^2 f+ f\sum _{i=1}^{n-p}{\widetilde{E}}^i\otimes {\widetilde{E}}^i\right| +C\left| \langle \nabla f,{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}\rangle -|{\dot{\gamma }}_{x,y}^{{\widetilde{E}}}|^2\langle \nabla f,{\dot{\gamma }}_{x,y}(s)\rangle \right| .\\&\quad \le C\left| \nabla ^2 f+ f\sum _{i=1}^{n-p}{\widetilde{E}}^i\otimes {\widetilde{E}}^i\right| + C\delta ^{1/400n} \end{aligned} \end{aligned}$$

(49)

by (30) and (45). By integration, we get

$$\begin{aligned} \int _0^{d(x,y)} \left| \frac{d}{d s}\left( |\nabla f|^2(s)-\langle \nabla f,{\dot{\gamma }}_{x,y}(s)\rangle ^2\right) \right| \,d s \le C\delta ^{1/400n}, \end{aligned}$$

and so

$$\begin{aligned} \Big ||\nabla f|^2(y)-\langle \nabla f(y),{\dot{\gamma }}_{x,y}(d(x,y))\rangle ^2 -|\nabla f|^2(x)+\langle \nabla f(x),{\dot{\gamma }}_{x,y}(0)\rangle ^2 \Big |\le C\delta ^{1/400n}. \end{aligned}$$

Thus, we get

$$\begin{aligned} |\nabla f|(y)\le C\delta ^{1/800n}. \end{aligned}$$

by (41), (45) and (47). By Claim 4.18 and (29), we get

$$\begin{aligned} \left| |f(y)|-f(p_f)\right| \le C\delta ^{1/800n}. \end{aligned}$$

Since

$$\begin{aligned} f(y)\ge (f(x)^2+|\nabla f|^2(x))^{1/2}-C\delta ^{1/100n}\ge \delta ^{1/800n}-C\delta ^{1/100n}>0 \end{aligned}$$

by (37), we get (b). We get

$$\begin{aligned} |(f(x)^2+|\nabla f|^2(x))^{1/2}-f(p_f)| \le C\delta ^{1/800n} \end{aligned}$$

(50)

by (43), (44) and (b), and so we get (c) by the definition of $s_0$ and (46). (50) implies the first assertion.

Finally, we show (d). Suppose that a point $z\in M$ satisfies $d(x,z)\le d(x,y)-\delta ^{1/2000n}$. Then, $d(x,y)\ge \delta ^{1/2000n}$, and so

$$\begin{aligned} f(x)\le f(p_f)\cos d(x,y)+C\delta ^{1/800n}\le f(p_f)-\frac{1}{C}\delta ^{1/1000n} \end{aligned}$$

by (29). There exists $w\in D_f(x)$ with $d(z,w)\le C\delta ^{1/100n}$. Let $\{{\overline{E}}^1,\ldots ,{\overline{E}}^n\}$ be a parallel orthonormal basis along $\gamma _{x,w}$ in Lemma 4.13. If (i) holds in the lemma, we assume that ${\overline{E}}_1={\dot{\gamma }}_{x,w}$. If $|{\dot{\gamma }}_{x,w}^{{\overline{E}}}|\le \delta ^{1/100}$, we have

$$\begin{aligned} f(z)\le f(w)+C\delta ^{1/100n} \le f(x)+ C\delta ^{1/100n} \le f(p_f)-\frac{1}{C}\delta ^{1/1000n} \end{aligned}$$

by Claim 4.19. If $|{\dot{\gamma }}_{x,w}^{{\overline{E}}}|\ge \delta ^{1/100}$, we have

$$\begin{aligned} \begin{aligned} f(z)&\le f(w)+C\delta ^{1/100n}\\&\le f(x)\cos (|{\dot{\gamma }}_{x,w}^{{\overline{E}}}|d(x,z))+|\nabla f|(x)\sin (|{\dot{\gamma }}_{x,w}^{{\overline{E}}}|d(x,z))+C\delta ^{1/100n}\\&\le f(p_f)\cos (|{\dot{\gamma }}_{x,w}^{{\overline{E}}}|d(x,z)-d(x,y))+\delta ^{1/800n} \le f(p_f)-\frac{1}{C}\delta ^{1/1000n} \end{aligned} \end{aligned}$$

by Claim 4.19, (46), (50) and $-\pi \le |{\dot{\gamma }}_{x,w}^{{\overline{E}}}|d(x,z)-d(x,y)\le -\delta ^{1/2000n}$. For both cases, we get (d). $\square$

By Claims 4.18 and 4.20, we get

$$\begin{aligned} |f(x)^2+|\nabla f|^2(x)-f(p_f)^2|\le C\delta ^{1/800n} \end{aligned}$$

(51)

for all $x\in D_f(p_f)\cap Q_f\cap R_f$.

Claim 4.21

We have $|f(p_f)-1|\le C\delta ^{1/800n}.$

Proof of Claim 4.21

Since $\Vert f^2+|\nabla f|^2-f(p_f)^2\Vert _{\infty }\le C$ and ${\mathrm {Vol}}(M\setminus (D_f(p_f)\cap Q_f\cap R_f) )\le C\delta ^{1/100},$ we get

$$\begin{aligned} \frac{1}{{\mathrm {Vol}}(M)}\int _M|f(x)^2+|\nabla f|^2(x)-f(p_f)^2| \,d\mu _g\le C \delta ^{1/800n} \end{aligned}$$

by (51). By the assumption, we have

$$\begin{aligned} \frac{1}{{\mathrm {Vol}}(M)}\left| \int _M (f(x)^2+|\nabla f|^2(x)-1) \,d\mu _g\right| \le C \delta ^{1/2} \end{aligned}$$

Thus, we get $|f(p_f)^2-1|\le C\delta ^{1/800n}.$ Since $f(p_f)>0$, we get the claim. $\square$

By Claims 4.18, 4.21 and (51), we get (i), (ii) and (iii).

Finally, we prove (iv). Put $A_f:=\{x\in M: |f(x)-1|\le \delta ^{1/900n}\}.$ Since we have $|f(w)-\cos d(w,A_f)|\le \delta ^{1/900n}$ for all $w\in A_f$, we get (iv) on $A_f$.

Let us show (iv) on $M\setminus A_f$. Take $w\notin A_f$ and $x\in D_f(p_f)\cap Q_f\cap R_f$ with $d(w,x)\le C\delta ^{1/100n}$.

We first suppose that $|\nabla f|(x)\ge \delta ^{1/800n}$. Take $y\in D_f(p_f)\cap D_f(x)$ of Claim 4.20. Then, $|f(y)-1|\le C\delta ^{1/800n}$, and so $y\in A_f$. Thus,

$$\begin{aligned} d(x, A_f)\le d(x,y)<\pi . \end{aligned}$$

(52)

For all $z\in A_f$, we have $|f(p_f)-f(z)|\le C\delta ^{1/900n}$, and so $d(x,z)> d(x,y)-\delta ^{1/2000n}$ by Claim 4.20 (d). Thus,

$$\begin{aligned} d(x,A_f)\ge d(x,y)-\delta ^{1/2000n}. \end{aligned}$$

(53)

By (52) and (53), we get $|d(x,A_f)- d(x,y)|\le \delta ^{1/2000n}.$ Therefore, we have $|f(x)-\cos d(x,A_f)|\le C\delta ^{1/2000n}$ by Claim 4.20 (c), and so $|f(w)-\cos d(w,A_f)|\le C\delta ^{1/2000n}$. By (52), we have $d(w,A_f)\le \pi +C\delta ^{1/100n}$.

We next suppose that $|\nabla f|(x)\le \delta ^{1/800n}$. Then, $||f|(x)-1|\le C\delta ^{1/800n}$ by Claim 4.18. If $f(x)\ge 0$, then $w \in A_f$. This contradicts to $w\notin A_f$. Thus, we have $|f(x)+1|\le C\delta ^{1/800n}$. We see that (i) in Lemma 4.13 cannot occur for $\gamma _{p_f,x}$ because we have

$$\begin{aligned} |\nabla ^2 f|\ge \frac{1}{\sqrt{n}}|\Delta f|\ge \frac{n-p}{\sqrt{n}}|f|-C\delta ^{1/2}. \end{aligned}$$

Thus, there exists an orthonormal basis $\{e^1,\ldots ,e^n\}$ of $T_x^*M$ such that $|\omega (x)-e^{n-p+1}\wedge \cdots \wedge e^n|\le C\delta ^{1/25}$ if Assumption 4.2 holds, and $|\xi (x)-e^{1}\wedge \cdots \wedge e^{n-p}|\le C\delta ^{1/25}$ if Assumption 4.3 holds. Take $u\in E_f(x)$ with $|u-e_1|\le C\delta ^{1/100n}$. Then, we get $|f\circ \gamma _u(s)+\cos s|\le C\delta ^{1/800n}$ for all $s\in [0,\pi ]$ by Lemma 4.13. Thus, we get $\gamma _u(\pi )\in A_f$, and so

$$\begin{aligned} d(w,A_f)\le \pi +C\delta ^{1/100n}. \end{aligned}$$

(54)

For any $y\in A_f$, there exists $z\in D_f(x)$ with $d(y,z)\le C\delta ^{1/100n}$. Let $\{E^1,\ldots ,E^n\}$ be a parallel orthonormal basis of $T^*M$ along $\gamma _{x,z}$ of Claim 4.19. Then,

$$\begin{aligned} |1+\cos ( |{\dot{\gamma }}_{x,z}^{E}| d(x,z))|\le C\delta ^{1/900n} \end{aligned}$$

by Claim 4.19. Thus, we get $d(x,z)\ge \pi -C\delta ^{1/1800n}$, and so

$$\begin{aligned} d(w,A_f)\ge \pi - C\delta ^{1/1800n}. \end{aligned}$$

(55)

By (54) and (55), we get $|d(w,A_f)-\pi |\le C\delta ^{1/1800n}$, and so $|f(w)-\cos d(w,A_f)|\le C\delta ^{1/1800n}$.

For both cases, we get (iv). $\square$

4.5 Gromov-Hausdorff Approximation

In this subsection, we suppose that Assumption 4.1 for $k=n-p+1$ and either Assumption 4.2 or 4.3 hold. We construct a Gromov-Hausdorff approximation map, and show that the Riemannian manifold is close to the product metric space $S^{n-p}\times X$ in the Gromov-Hausdorff topology. The following proposition is based on [19, Lemma 5.2].

Lemma 4.22

Define ${\widetilde{\Psi }}:=(f_1,\dots ,f_{n-p+1}):M\rightarrow {\mathbb {R}}^{n-p+1}$. Then, we have

$$\begin{aligned} \Vert |{\widetilde{\Psi }}|^2-1\Vert _{\infty }\le C\delta ^{1/1000n^2}. \end{aligned}$$

Proof

We first prove the following claim:

Claim 4.23

For any $x\in M$, we have $|{\widetilde{\Psi }}|(x)\le 1+C\delta ^{1/800n}$

Proof of Claim 4.23

If $|{\widetilde{\Psi }}|(x)=0$, the claim is trivial. Thus, we assume that $|{\widetilde{\Psi }}|(x)\ne 0$. Put

$$\begin{aligned} f_x:=\frac{1}{|{\widetilde{\Psi }}|(x)}\sum _{i=1}^{n-p+1} f_i(x)f_i. \end{aligned}$$

Then, we have $\Vert f_x\Vert _2^2=1/(n-p+1).$ Thus, we get $|{\widetilde{\Psi }}|(x)=f_x(x)\le 1+ C\delta ^{1/800n}$ by Proposition 4.17 (i). $\square$

For $x\in M$ with $|{\widetilde{\Psi }}(x)|^2-1< 0$, we have $||{\widetilde{\Psi }}(x)|^2-1|=1-|{\widetilde{\Psi }}(x)|^2$. For $x\in M$ with $|{\widetilde{\Psi }}(x)|^2-1\ge 0$, we have $||{\widetilde{\Psi }}(x)|^2-1|=|{\widetilde{\Psi }}(x)|^2-1 \le 1-|{\widetilde{\Psi }}(x)|^2+C\delta ^{1/800n}$ by Claim 4.23. For both cases, we have $||{\widetilde{\Psi }}(x)|^2-1|\le 1-|{\widetilde{\Psi }}(x)|^2+C\delta ^{1/800n}$. Combining this and $\Vert {\widetilde{\Psi }}\Vert _2=1$, we get $\Vert |{\widetilde{\Psi }}|^2-1\Vert _1 \le C \delta ^{1/800n}.$ Therefore, we have

$$\begin{aligned} {\mathrm {Vol}}(\{x\in M:||{\widetilde{\Psi }}(x)|^2-1|\ge \delta ^{1/1000n^2}\})\le C\delta ^{1/800n}\delta ^{-1/1000n^2}\le C\delta ^{1/1000n} \end{aligned}$$

(note that we assumed $n\ge 5$). This and the Bishop-Gromov inequality imply that, for any $x\in M$, there exists $y\in \{x\in M:||{\widetilde{\Psi }}(x)|^2-1|< \delta ^{1/1000n^2}\}$ with $d(x,y)\le C\delta ^{1/1000n^2}$, and so $||{\widetilde{\Psi }}(x)|^2-1|\le C\delta ^{1/1000n^2}$ by $\Vert \nabla |{\widetilde{\Psi }}|^2\Vert _\infty \le C$. Thus, we get the lemma. $\square$

Notation 4.24

In the remaining part of this subsection, we use the following notation.

• Let $d_S$ denotes the intrinsic distance function on $S^{n-p}(1)$. Note that we have $\cos d_S(x,y)=x\cdot y$ and

  $$\begin{aligned} d_{{\mathbb {R}}^{n-p+1}}(x,y)\le d_{S}(x,y)\le 3 d_{{\mathbb {R}}^{n-p+1}}(x,y) \end{aligned}$$

  for all $x,y\in S^{n-p}\subset {\mathbb {R}}^{n-p+1}$.

• For each $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots , f_{n-p+1}\}$ with $\Vert f\Vert _2^2=1/(n-p+1)$, we use the notation $p_f$ and $A_f$ of Proposition 4.17. Recall that we defined $A_f:=\{x\in M: |f(x)-1|\le \delta ^{1/900n}\}.$

• Define ${\widetilde{\Psi }}:=(f_1,\dots ,f_{n-p+1}):M\rightarrow {\mathbb {R}}^{n-p+1}$ and

  $$\begin{aligned} \Psi :=\frac{{\widetilde{\Psi }}}{|{\widetilde{\Psi }}|}:M\rightarrow S^{n-p}. \end{aligned}$$

• For each $x\in M$, put

  $$\begin{aligned} f_x:=\frac{1}{|{\widetilde{\Psi }}|(x)}\sum _{i=1}^{n-p+1} f_i(x)f_i=\sum _{i=1}^{n-p+1} \Psi _i(x)f_i, \end{aligned}$$

  $p_x:=p_{f_x}$ and $A_x:=A_{f_x}$.

• For each $x\in M$ and $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots , f_{n-p+1}\}$ with $\Vert f\Vert _2^2=1/(n-p+1)$, choose $a_f(x)\in A_f$ such that $d(x,A_f)=d(x,a_f(x)).$

The goal of this subsection is to show that

$$\begin{aligned} \Phi _f :M\rightarrow S^{n-p}\times A_f,\,x\mapsto (\Psi (x),a_f(x)) \end{aligned}$$

is a Gromov-Hausdorff approximation map.

Lemma 4.25

For all $x,y\in M$, we have $|\Psi (x)-\Psi (y)|\le Cd(x,y).$

Proof

Since we have $\Vert \nabla f_i\Vert _{\infty }\le C$ for all $i\in \{1,\ldots ,n-p+1\}$, we get $|{\widetilde{\Psi }}(x)-{\widetilde{\Psi }}(y)|\le Cd(x,y)$ for all $x,y\in M$. Thus, we get the lemma by Lemma 4.22 ($|{\widetilde{\Psi }}|\ge 1/2$). $\square$

Lemma 4.26

Take $u\in S^{n-p}$ and put $f=\sum _{i=1}^{n-p+1}u_i f_i$. Then, we have

$$\begin{aligned} |d_S(\Psi (y),u)-d(y,A_{f})|\le C\delta ^{1/2000n^2} \end{aligned}$$

for all $y\in M$.

Proof

Since $f(y)=u\cdot {\widetilde{\Psi }}(y),$ we have $|u \cdot {\widetilde{\Psi }}(y)-\cos d(y,A_{f})|\le C\delta ^{1/2000n}$ by Proposition 4.17, and so

$$\begin{aligned} |u\cdot \Psi (y)-\cos d(y,A_{f})|\le C\delta ^{1/1000n^2} \end{aligned}$$

by Lemma 4.22. Since $\cos d_S(\Psi (y),u)=u\cdot \Psi (y)$, this and $d(y,A_{f})\le \pi +C\delta ^{1/100n}$ imply the lemma. $\square$

By the definition of $A_{y}$, we immediately get the following corollaries:

Corollary 4.27

Take $u\in S^{n-p}$ and put $f=\sum _{i=1}^{n-p+1}u_i f_i$. Then, we have $d_S(\Psi (p_f),u)\le C\delta ^{1/2000n^2}.$

Corollary 4.28

For each $y_1,y_2\in M$, we have

$$\begin{aligned} |d_S(\Psi (y_1),\Psi (y_2))-d(y_2,A_{y_1})|\le C\delta ^{1/2000n^2}. \end{aligned}$$

Corollary 4.29

For each $y\in M$, we have $d(y,A_{y})\le C\delta ^{1/2000n^2}.$

We need to show the almost Pythagorean theorem for our purpose. To do this, we regard $|{\dot{\gamma }}^E| s$ in Lemma 4.13 as a moving distance in $S^{n-p}$. We first approximate their cosine.

Lemma 4.30

Take $y_1\in M$, ${\tilde{y}}_1\in D_{f_{y_1}}(p_{y_1})\cap R_{f_{y_1}}\cap Q_{f_{y_1}}$ with $d(y_1,{\tilde{y}}_1)\le C\delta ^{1/100n}$ and $y_2\in D_{f_{y_1}}({\tilde{y}}_1)$ (note that we can take such ${\tilde{y}}_1$ for any $y_1$ by the Bishop-Gromov theorem). Let $\{E^1,\ldots ,E^n\}$ be a parallel orthonormal basis of $T^*M$ along $\gamma _{{\tilde{y}}_1,y_2}$ in Lemma 4.13 for $f_{y_1}$. Then, (ii) holds in the lemma, and

$$\begin{aligned} |\cos (|{\dot{\gamma }}_{{\tilde{y}}_1,y_2}^E|s)-\cos d_S(\Psi (y_1),\Psi (\gamma _{{\tilde{y}}_1,y_2}(s)))|\le C\delta ^{1/2000n^2} \end{aligned}$$

for all $s\in [0,d({\tilde{y}}_1,y_2)]$. In particular, we have

$$\begin{aligned} |\cos (|{\dot{\gamma }}_{{\tilde{y}}_1,y_2}^E|d({\tilde{y}}_1,y_2))-\cos d_S(\Psi (y_1),\Psi (y_2))|\le C\delta ^{1/2000n^2}. \end{aligned}$$

Proof

By Corollary 4.29, we have $d({\tilde{y}}_1,A_{y_1})\le C\delta ^{1/2000n^2},$ and so we get

$$\begin{aligned} f_{y_1}\circ \gamma _{{\tilde{y}}_1,y_2}(s)\ge & {} \cos d(\gamma _{{\tilde{y}}_1,y_2}(s),A_{y_1})- C\delta ^{1/2000n} \\\ge & {} \cos s- C\delta ^{1/2000n^2} \ge \frac{1}{\sqrt{2}}- C\delta ^{1/2000n^2} \end{aligned}$$

for all $s\le \min \{\pi /4,d({\tilde{y}}_1,y_2)\}$. Therefore, we have

$$\begin{aligned} |\nabla ^2 f_{y_1}|(\gamma _{{\tilde{y}}_1,y_2}(s)) \ge \frac{1}{\sqrt{n}}|\Delta f_{y_1}|(\gamma _{{\tilde{y}}_1,y_2}(s)) \ge \frac{n-p}{\sqrt{2n}}- C\delta ^{1/2000n^2}\nonumber \\ \end{aligned}$$

for all $s\le \min \{\pi /4,d({\tilde{y}}_1,y_2)\}$. Thus, (i) in Lemma 4.13 cannot occur, and so (ii) holds in the lemma.

Since we have $f_{y_1}(y_1)=|{\widetilde{\Psi }}(y_1)|$, we get

$$\begin{aligned} |f_{y_1}({\tilde{y}}_1)-1|\le C\delta ^{1/1000n^2} \end{aligned}$$

(56)

by Lemma 4.22 and $d(y_1,{\tilde{y}}_1)\le C\delta ^{1/100n}$. By (56) and Proposition 4.17 (iii), we have $|\nabla f_{y_1}|({\tilde{y}}_1)\le C\delta ^{1/2000n^2}.$ Thus, we get

$$\begin{aligned} |f_{y_1}(\gamma _{{\tilde{y}}_1,y_2}(s))-\cos (|{\dot{\gamma }}_{{\tilde{y}}_1,y_2}^E|s)|\le C\delta ^{1/2000n^2} \end{aligned}$$

for all $s\in [0,d({\tilde{y}}_1,y_2)]$ by Lemma 4.13. On the other hand, we have

$$\begin{aligned} |f_{y_1}(\gamma _{{\tilde{y}}_1,y_2}(s))-\cos d_S(\Psi (y_1),\Psi (\gamma _{{\tilde{y}}_1,y_2}(s)))|\le C\delta ^{1/2000n^2} \end{aligned}$$

for all $s\in [0,d({\tilde{y}}_1,y_2)]$ by Proposition 4.17 (iv) and Corollary 4.28. Thus, we get the lemma. $\square$

Notation 4.31

We use the following notation:

• For any $y_1,y_2\in M$ and $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots , f_{n-p+1}\}$ with $\Vert f\Vert _2^2=1/(n-p+1)$, define

  $$\begin{aligned}&G_f^{y_1}(y_2)\\&\quad :=\langle {\dot{\gamma }}_{y_2,y_1}(0),\nabla f(y_2)\rangle d(y_1,y_2)\sin d_S(\Psi (y_1),\Psi (y_2))\\&\qquad +\Big (\cos d(y_2, A_f)\cos d_S(\Psi (y_1),\Psi (y_2))-\cos d(y_1,A_f)\Big ) d_S(\Psi (y_1),\Psi (y_2)). \end{aligned}$$

• For any $y_1,y_2\in M$, define

  $$\begin{aligned} H^{y_1}(y_2):=\left\{ \begin{array}{ll} 1 &{}\qquad d(y_1,y_2)\le \pi ,\\ 0 &{}\qquad d(y_1,y_2)>\pi . \end{array}\right. \end{aligned}$$

• For any $y_1,y_2\in M$ and $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots , f_{n-p+1}\}$ with $\Vert f\Vert _2^2=1/(n-p+1)$, define

  $$\begin{aligned} C_f^{y_1}(y_2)&:=\Big \{y_3\in M : \gamma _{y_2,y_3}(s)\in I_{y_1}\setminus \{y_1\} \text { for almost all }s\in [0,d(y_2,y_3)], \text { and}\\&\qquad \qquad \int _{0}^{d(y_2,y_3)} |G_f^{y_1}H^{y_1}|(\gamma _{y_2,y_3}(s))\,d s\le \delta ^{1/12000n^2}\Big \},\\ P_f^{y_1}&:=\{y_2\in M: {\mathrm {Vol}}(M\setminus C_f^{y_1}(y_2))\le \delta ^{1/12000n^2}{\mathrm {Vol}}(M)\}. \end{aligned}$$

Pinching condition on $G_f^{y_1}$ plays a crucial role for our purpose. Let us estimate $G_f^{y_1}$.

Lemma 4.32

Take $\eta >0$ with $\eta \ge \delta ^{1/2000n}$, $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots , f_{n-p+1}\}$ with $\Vert f\Vert _2^2=1/(n-p+1)$, $y_1\in Q_f$ and $y_2\in D_f(y_1)$. Let $\{E^1,\ldots ,E^n\}$ be a parallel orthonormal basis of $T^*M$ along $\gamma _{y_1,y_2}$ in Lemma 4.13 for f. If

$$\begin{aligned} ||{\dot{\gamma }}_{y_1,y_2}^E|d(y_1,y_2)-d_S(\Psi (y_1),\Psi (y_2))|\le \eta , \end{aligned}$$

then $|G_f^{y_1}(y_2)|\le C\eta .$

Proof

We have

$$\begin{aligned}&\Big |f(y_1)-f(y_2)\cos (|{\dot{\gamma }}_{y_1,y_2}^E|d(y_1,y_2))\\&\quad -\frac{1}{|{\dot{\gamma }}_{y_1,y_2}^E|}\langle \nabla f(y_2),{\dot{\gamma }}_{y_2,y_1}(0)\rangle \sin (|{\dot{\gamma }}_{y_1,y_2}^E|d(y_1,y_2)) \Big | \le C\delta ^{1/250} \end{aligned}$$

by Lemma 4.13. Thus, by Proposition 4.17 (iv), we get

$$\begin{aligned}&\Big ||{\dot{\gamma }}_{y_1,y_2}^E|\cos d(y_1,A_f)-|{\dot{\gamma }}_{y_1,y_2}^E|\cos d(y_2, A_f)\cos (|{\dot{\gamma }}_{y_1,y_2}^E|d(y_1,y_2))\\&\quad -\langle \nabla f(y_2),{\dot{\gamma }}_{y_2,y_1}(0)\rangle \sin (|{\dot{\gamma }}_{y_1,y_2}^E|d(y_1,y_2)) \Big | \le C\delta ^{1/2000n}, \end{aligned}$$

and so we get the lemma. $\square$

The quantity $|{\dot{\gamma }}_{y_1,y_2}^E|$ in the above lemma is slightly different from that of Lemma 4.30. Comparing these two quantity, we get the following:

Corollary 4.33

Take $\eta >0$ with $\eta \ge \delta ^{1/2000n}$, $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots , f_{n-p+1}\}$ with $\Vert f\Vert _2^2=1/(n-p+1)$, $y_1\in M$, ${\tilde{y}}_1\in D_{f_{y_1}}(p_{y_1})\cap R_{f_{y_1}}\cap Q_{f_{y_1}}\cap Q_f$ with $d(y_1,{\tilde{y}}_1)\le C\delta ^{1/100n}$ and $y_2\in D_{f_{y_1}}({\tilde{y}}_1)\cap D_f({\tilde{y}}_1)$. Let $\{E^1,\ldots ,E^n\}$ be a parallel orthonormal basis of $T^*M$ along $\gamma _{{\tilde{y}}_1,y_2}$ in Lemma 4.13 for $f_{y_1}$. If

$$\begin{aligned} ||{\dot{\gamma }}_{{\tilde{y}}_1,y_2}^E|d({\tilde{y}}_1,y_2)-d_S(\Psi ({\tilde{y}}_1),\Psi (y_2))|\le \eta , \end{aligned}$$

then $|G_f^{{\tilde{y}}_1}(y_2)|\le C\eta .$

Proof

Let $\{{\widetilde{E}}^1,\ldots ,{\widetilde{E}}^n\}$ be a parallel orthonormal basis of $T^*M$ along $\gamma _{{\tilde{y}}_1,y_2}$ in Lemma 4.13 for f (if (i) holds, then we can assume that ${\widetilde{E}}^i=E^i$ for all i). We show that $\left| |{\dot{\gamma }}_{{\tilde{y}}_1,y_2}^E|-|{\dot{\gamma }}_{{\tilde{y}}_1,y_2}^{{\widetilde{E}}}|\right| \le C\delta ^{1/50}.$ Then, we immediately get the corollary by Lemma 4.32.

We first suppose that Assumption 4.2 holds. We have $|\omega (y_2)-E^{n-p+1}\wedge \cdots \wedge E^n|\le C\delta ^{1/25}$ by Lemmas 4.13 and 4.30. Since $|{\dot{\gamma }}_{{\tilde{y}}_1,y_2}^E|^2=1-|\iota ({\dot{\gamma }}_{{\tilde{y}}_1,y_2})(E^{n-p+1}\wedge \cdots \wedge E^n)|^2$, we get

$$\begin{aligned} \left| |{\dot{\gamma }}_{{\tilde{y}}_1,y_2}^E|^2-\left( 1-|\iota ({\dot{\gamma }}_{{\tilde{y}}_1,y_2})\omega |^2(y_2)\right) \right| \le C\delta ^{1/25}. \end{aligned}$$

(57)

Similarly, we get

$$\begin{aligned} \left| |{\dot{\gamma }}_{{\tilde{y}}_1,y_2}^{{\widetilde{E}}}|^2-\left( 1-|\iota ({\dot{\gamma }}_{{\tilde{y}}_1,y_2})\omega |^2(y_2)\right) \right| \le C\delta ^{1/25}. \end{aligned}$$

(58)

By (57) and (58), we get $\left| |{\dot{\gamma }}_{{\tilde{y}}_1,y_2}^E|-|{\dot{\gamma }}_{{\tilde{y}}_1,y_2}^{{\widetilde{E}}}|\right| \le C\delta ^{1/50}.$

We next suppose that Assumption 4.3 holds. Similarly, we have

$$\begin{aligned}&\left| |{\dot{\gamma }}_{{\tilde{y}}_1,y_2}^E|^2-|\iota ({\dot{\gamma }}_{{\tilde{y}}_1,y_2})\xi |^2(y_2)\right| \le C\delta ^{1/25},\\&\left| |{\dot{\gamma }}_{{\tilde{y}}_1,y_2}^{{\widetilde{E}}}|^2-|\iota ({\dot{\gamma }}_{{\tilde{y}}_1,y_2})\xi |^2(y_2)\right| \le C\delta ^{1/25}, \end{aligned}$$

and so $\left| |{\dot{\gamma }}_{{\tilde{y}}_1,y_2}^E|-|{\dot{\gamma }}_{{\tilde{y}}_1,y_2}^{{\widetilde{E}}}|\right| \le C\delta ^{1/50}.$

By the above two cases, we get the corollary. $\square$

Let us show the integral pinching condition.

Lemma 4.34

Take $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots , f_{n-p+1}\}$ with $\Vert f\Vert _2^2=1/(n-p+1)$, $y_1\in M$ and ${\tilde{y}}_1\in D_{f_{y_1}}(p_{y_1})\cap R_{f_{y_1}}\cap Q_{f_{y_1}}\cap Q_f$ with $d(y_1,{\tilde{y}}_1)\le C\delta ^{1/100n}$. Then, $\Vert G_f^{{\tilde{y}}_1} H_{{\tilde{y}}_1}\Vert _1\le C\delta ^{1/4000n^2}$ and ${\mathrm {Vol}}(M\setminus P_f^{{\tilde{y}}_1})\le C\delta ^{1/12000n^2}.$

Proof

Take arbitrary $y_2\in D_f({\tilde{y}}_1)\cap D_{f_{y_1}}({\tilde{y}}_1)$. Let $\{E^1,\ldots ,E^n\}$ be a parallel orthonormal basis of $T^*M$ along $\gamma _{{\tilde{y}}_1,y_2}$ in Lemma 4.13 for $f_{y_1}$. Then, we have $||{\dot{\gamma }}_{{\tilde{y}}_1,y_2}^E|d({\tilde{y}}_1,y_2)-d_S(\Psi ({\tilde{y}}_1),\Psi (y_2))|\le C\delta ^{1/4000n^2},$ if $d({\tilde{y}}_1,y_2)\le \pi$ by Lemmas 4.25 and 4.30. Thus, by Corollary 4.33, we have

$$\begin{aligned} \sup _{D_f({\tilde{y}}_1)\cap D_{f_{y_1}}({\tilde{y}}_1)}|G_f^{{\tilde{y}}_1} H^{{\tilde{y}}_1}|\le C\delta ^{1/4000n^2}. \end{aligned}$$

Since ${\mathrm {Vol}}(M\setminus (D_f({\tilde{y}}_1)\cap D_{f_{y_1}}({\tilde{y}}_1)))\le C\delta ^{1/100}{\mathrm {Vol}}(M)$ and $\Vert G_f^{{\tilde{y}}_1} H^{{\tilde{y}}_1}\Vert _\infty \le C$, we get $\Vert G_f^{{\tilde{y}}_1} H^{{\tilde{y}}_1}\Vert _1\le C\delta ^{1/4000n^2}.$ By the segment inequality (Theorem 4.6), we get the remaining part of the lemma. $\square$

Notation 4.35

We use the following notation.

$$\begin{aligned} \eta _0=\delta ^{1/12000n^3},\, \eta _1=\eta _0^{1/26}, \, \eta _2=\eta _1^{1/78}\text { and } L=\eta _2^{1/150}. \end{aligned}$$

We use Lemma 4.34 to give the almost Pythagorean theorem for the special case (see Lemma 4.43). For the general case, we need to estimate $\Vert G_f^{{\tilde{y}}_1}\Vert _1$. To do this, we show that $|{\dot{\gamma }}_{{\tilde{y}}_1,y_2}^E|d({\tilde{y}}_1,y_2)\le \pi +L$ under the assumption of Lemma 4.30 in Lemma 4.45. Then, we can estimate $\Vert G_f^{{\tilde{y}}_1}\Vert _1$ similarly to Lemma 4.34. After proving that, we use Lemma 4.38 again to give the almost Pythagorean theorem for the general case. The following lemma, which guarantees that an almost shortest pass from a point in M to $A_f$ almost corresponds to a geodesic in $S^{n-p}$ through $\Psi$ under some assumptions, is the first step to achieve these objectives.

Lemma 4.36

Take

• $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots , f_{n-p+1}\}$ with $\Vert f\Vert _2^2=1/(n-p+1)$,

• $u\in S^{n-p}$ with $f=\sum _{i=1}^{n-p+1}u_i f_i$,

• $x,y\in M$,

• $\eta >0$ with $\eta _0\le \eta \le L^{1/3n}$.

Suppose

• $d(y,A_f)\le C \eta$,

• $|d(x,A_f)-d(x,y)|\le C\eta$.

Then, we have the following for all $s,s'\in [0,d(x,y)]$:

• (i) $|d(\gamma _{y,x}(s),A_f)-s|\le C\eta$,

• (ii) $\left| |s-s'|-d_S\left( \Psi (\gamma _{y,x}(s)),\Psi (\gamma _{y,x}(s'))\right) \right| \le C\eta$,

• (iii) If in addition $d(x,A_f)\ge \frac{1}{C}\eta ^{1/26}$, there exists $v\in S^{n-p}$ such that $u\cdot v=0$ and

  $$\begin{aligned} d_S(\Psi (\gamma _{y,x}(s)),\gamma _v(s))\le C\eta ^{3/13} \end{aligned}$$

  for all $s\in [0,d(x,y)]$, where we define $\gamma _v(s):=(\cos s) u+(\sin s) v\in S^{n-p}$.

Proof

We first prove (i). We have $d(\gamma _{y,x}(s),A_f)\le s+ C\eta$ and

$$\begin{aligned} d(x,y)-C\eta \le d(x,A_f)\le d(\gamma _{y,x}(s),A_f)+d(x,y)-s. \end{aligned}$$

Thus, we get (i).

We next prove (ii). By Lemma 4.26, we have $d_S(\Psi (y),u)\le C\eta$ and $|d_S(\Psi (\gamma _{y,x}(s)),u)-d(\Psi (\gamma _{y,x}(s)),A_f)|\le C\delta ^{1/2000n^2}$, and so we get

$$\begin{aligned} |s-d_S(\Psi (\gamma _{y,x}(s)),\Psi (y))|\le C \eta \end{aligned}$$

(59)

for all $s\in [0,d(x,y)]$ by (i). Take arbitrary $s,s'\in [0,d(x,y)]$ with $s<s'$. Then,

$$\begin{aligned} \begin{aligned} s'-s=d(\gamma _{y,x}(s),\gamma _{y,x}(s'))&\ge d(\gamma _{y,x}(s),A_{\gamma _{y,x}(s')})-d(\gamma _{y,x}(s'),A_{\gamma _{y,x}(s')})\\&\ge d_S(\Psi (\gamma _{y,x}(s)),\Psi (\gamma _{y,x}(s')))-C\delta ^{1/2000n^2} \end{aligned} \end{aligned}$$

(60)

by Corollaries 4.28 and 4.29. On the other hand, we have

$$\begin{aligned} \begin{aligned} s'-C\eta&\le d_S(\Psi (\gamma _{y,x}(s')),\Psi (y))\\&\le d_S(\Psi (\gamma _{y,x}(s)),\Psi (\gamma _{y,x}(s')))+d_S(\Psi (\gamma _{y,x}(s)),\Psi (y))\\&\le d_S(\Psi (\gamma _{y,x}(s)),\Psi (\gamma _{y,x}(s'))) +s+C\eta \end{aligned} \end{aligned}$$

by (59), and so

$$\begin{aligned} s'-s\le d_S(\Psi (\gamma _{y,x}(s)),\Psi (\gamma _{y,x}(s'))) +C\eta . \end{aligned}$$

(61)

By (60) and (61), we get (ii).

Finally, we prove (iii). Since $d(x,A_f)\ge \frac{1}{C}\eta ^{1/26}$, there exists $s_0\in [0,d(x,y)]$ such that $\frac{1}{C}\eta ^{1/26}\le d(z,y)\le \pi - \frac{1}{C}\eta ^{1/26}$, where we put $z=\gamma _{y,x}(s_0)$. Then, there exists $v\in S^{n-p}$ with $u\cdot v=0$ and $t_1\in [0,\pi ]$ such that $\Psi (z)=(\cos t_1) u+(\sin t_1) v.$ We have

$$\begin{aligned} |\cos t_1-\cos d(z,y)|=&|\cos d_S(\Psi (z),u)-\cos s_0|\\ \le&|\cos d(z,A_f)-\cos s_0|+C\delta ^{1/2000n^2} \le C\eta \end{aligned}$$

by Lemma 4.26 and (i). This gives

$$\begin{aligned} |t_1-d(z,y)|\le C\eta ^{1/2}. \end{aligned}$$

(62)

Take arbitrary $s\in [0,d(x,y)]$. Then, there exist $w\in S^{n-p}$ and $x_1,x_2,x_3\in {\mathbb {R}}$ such that $w\perp {\mathrm {Span}}_{{\mathbb {R}}}\{u,v\}$, $x_1^2+x_2^2+x_3^2=1$ and $\Psi (\gamma _{y,x}(s))=x_1 u+x_2 v+ x_3 w.$ Since we have $|s-d_S(\Psi (\gamma _{y,x}(s)),u)|\le C\eta$ by (i) and Lemma 4.26, and $\cos d_S(\Psi (\gamma _{y,x}(s)),u)=x_1$, we get

$$\begin{aligned} |\cos s- x_1|\le C\eta . \end{aligned}$$

(63)

We have

$$\begin{aligned} \left| |d(z,y)-s|-d_S(\Psi (\gamma _{y,x}(s)),\Psi (z))\right| \le C\eta \end{aligned}$$

by (ii). Since $\cos d_S(\Psi (\gamma _{y,x}(s)),\Psi (z))=x_1 \cos t_1+x_2\sin t_1$, we get

$$\begin{aligned} |\cos (d(z,y)-s)- x_1 \cos d(z,y)-x_2\sin d(z,y)|\le C\eta ^{1/2} \end{aligned}$$

(64)

by (62). By (63) and (64), we have $\sin d(z,y)|\sin s- x_2|\le C\eta ^{1/2}.$ By the assumption, we have $\sin d(z,y)\ge \frac{1}{C}\eta ^{1/26},$ and so we get

$$\begin{aligned} |\sin s- x_2|\le C\eta ^{6/13}. \end{aligned}$$

(65)

By (63) and (65), we get

$$\begin{aligned} |\cos d_S(\Psi (\gamma _{y,x}(s)),\gamma _v(s))-1| =|x_1 \cos s+x_2\sin s-1|\le C\eta ^{6/13}. \end{aligned}$$

Thus, we get (iii). $\square$

The following lemma asserts that the differential of an almost shortest pass from a point in M to $A_f$ is in the direction of $\nabla f$ under some assumptions.

Lemma 4.37

Take

• $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots , f_{n-p+1}\}$ with $\Vert f\Vert _2^2=1/(n-p+1)$,

• $x\in D_f(p_f)\cap Q_f \cap R_f$,

• $y\in D_f(x)\cap D_f(p_f)\cap Q_f\cap R_f$,

• $\eta >0$ with $\eta _0\le \eta \le L^{1/3n}$.

Suppose

• $d(x,A_f)\ge \frac{1}{C}\eta ^{1/26}$,

• $d(y,A_f)\le C \eta$,

• $|d(x,A_f)-d(x,y)|\le C\eta$.

Let $\{E^1,\ldots ,E^n\}$ be a parallel orthonormal basis of $T^*M$ along $\gamma _{x,y}$ in Lemma 4.13 for f. Then, we have the following for all $s\in [0,d(x,y)]$:

• (i) $||{\dot{\gamma }}^E_{x,y}|-1|\le C \eta ^{6/13}$,

• (ii) $|\nabla f (\gamma _{y,x}(s))+(\sin s) {\dot{\gamma }}_{y,x}(s)|\le C\eta ^{3/26}$.

Proof

We first note that we have

$$\begin{aligned} d(x,y)\le \pi +C\eta \end{aligned}$$

(66)

by the assumption and Proposition 4.17 (iv).

Let us prove (i). By $d(y,A_f)\le C \eta$, we have $\cos d(y,A_f)\ge 1- C\eta ^2$. Thus, we have

$$\begin{aligned} |1-f(y)|\le C\eta ^2 \end{aligned}$$

(67)

by Proposition 4.17 (iv). By Proposition 4.17 (iii), we get $|\nabla f|(y)\le C\eta .$ Thus, we have

$$\begin{aligned} |f(x)-\cos (|{\dot{\gamma }}_{x,y}^E|d(x,y))|\le C\eta \end{aligned}$$

(68)

by Lemma 4.13, and so $||{\dot{\gamma }}_{x,y}^E|d(x,y)-d(x,A_f)|\le C\eta ^{1/2}$ by Proposition 4.17 (iv) and (66). By the assumptions, we get (i).

We next prove (ii). By Proposition 4.17, we have $||\nabla f|^2(x)-\sin ^2 d(x,A_f)|\le C\delta ^{1/2000n},$ and so $||\nabla f|(x)-|\sin d(x,A_f)||\le C\delta ^{1/4000n}.$ Since $\sin d(x, A_f)\ge -C\delta ^{1/100n}$ by Proposition 4.17 (iv), we have $||\nabla f|(x)-\sin d(x,A_f)|\le C\delta ^{1/4000n}.$ Thus, we get

$$\begin{aligned} ||\nabla f|(x)-\sin d(x,y)|\le C\eta \end{aligned}$$

(69)

by the assumption. On the other hand, by (i) and Lemma 4.13, we have $|f(y)-f(x)\cos d(x,y)-\langle \nabla f(x),{\dot{\gamma }}_{x,y}(0)\rangle \sin d(x,y)|\le C\eta ^{6/13},$ and so

$$\begin{aligned} |\sin ^2 d(x,y)-\langle \nabla f(x),{\dot{\gamma }}_{x,y}(0)\rangle \sin d(x,y)|\le C\eta ^{6/13} \end{aligned}$$

(70)

by (67) and (68).

We consider the following two cases:

• $d(x,y)\le \pi -\eta ^{3/13}$,

• $d(x,y)> \pi -\eta ^{3/13}$.

We first suppose that $d(x,y)\le \pi -\eta ^{3/13}$. We get $|\sin d(x,y)-\langle \nabla f(x),{\dot{\gamma }}_{x,y}(0)\rangle |\le C\eta ^{3/13}$ by the assumption and (70). By (69), we get

$$\begin{aligned} |\nabla f|(x)-\langle \nabla f(x),{\dot{\gamma }}_{x,y}(0)\rangle \le C\eta ^{3/13}. \end{aligned}$$

(71)

We next suppose that $d(x,y)> \pi -\eta ^{3/13}$. Then, we have $\cos d(x,A_f)\le -1+C\eta ^{6/13}$, and so $|\nabla f|(x)\le C\eta ^{3/13}$ by Proposition 4.17 (iii) and (iv). Thus, we also get (71) for this case.

By (i), (48) and Lemma 4.13, we have

$$\begin{aligned} \int _0^{d(x,y)} \left| \frac{d}{d s}\left( |\nabla f|^2(\gamma _{x,y}(s))-\langle \nabla f(\gamma _{x,y}(s)), {\dot{\gamma }}_{x,y}(s)\rangle ^2\right) \right| \,d s\le C\eta ^{6/13}. \end{aligned}$$

Thus, we get

$$\begin{aligned} |\nabla f|^2(\gamma _{x,y}(s))-\langle \nabla f(\gamma _{x,y}(s)), {\dot{\gamma }}_{x,y}(s)\rangle ^2\le C\eta ^{3/13} \end{aligned}$$

(72)

for all $s\in [0,d(x,y)]$ by (71). Since

$$\begin{aligned}&|\nabla f (\gamma _{x,y}(s))-\langle \nabla f(\gamma _{x,y}(s)), {\dot{\gamma }}_{x,y}(s)\rangle {\dot{\gamma }}_{x,y}(s)|^2\\&\quad =|\nabla f|^2(\gamma _{x,y}(s))-\langle \nabla f(\gamma _{x,y}(s)), {\dot{\gamma }}_{x,y}(s)\rangle ^2, \end{aligned}$$

we get $|\nabla f (\gamma _{x,y}(s))-\langle \nabla f(\gamma _{x,y}(s)), {\dot{\gamma }}_{x,y}(s)\rangle {\dot{\gamma }}_{x,y}(s)|\le C\eta ^{3/26}$ by (72). Since we have

$$\begin{aligned} |\langle \nabla f(\gamma _{x,y}(s)), {\dot{\gamma }}_{x,y}(s)\rangle +\cos d(x,y)\sin s-\sin d(x,y) \cos s|\le C\eta ^{3/13} \end{aligned}$$

by (68), (69), (71), (i) and Lemma 4.13, we get

$$\begin{aligned} |\nabla f (\gamma _{x,y}(s))-\sin (d(x,y)-s){\dot{\gamma }}_{x,y}(s)|\le C\eta ^{3/26} \end{aligned}$$

This gives (ii). $\square$

The following lemma is crucial to show the almost Pythagorean theorem.

Lemma 4.38

Take

• $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots , f_{n-p+1}\}$ with $\Vert f\Vert _2^2=1/(n-p+1)$,

• $x\in D_f(p_f)\cap Q_f \cap R_f$,

• $y\in D_f(x)\cap D_f(p_f)\cap Q_f\cap R_f$,

• $z\in M$,

• $\eta >0$ with $\eta _0\le \eta \le L^{1/3n}$ and $T\in [0, d(x,y)]$.

Suppose

• $d(y,A_f)\le C\eta$,

• $|d(x,A_f)-d(x,y)|\le C\eta$,

• $\gamma _{y,x}(s)\in I_z\setminus \{z\}$ for almost all $s\in [T,d(x,y)]$,

• $\int _T^{d(x,y)} |G_f^z(\gamma _{y,x}(s))|\,d s\le C\eta ^{3/26}$.

Then, we have

$$\begin{aligned} \left| d(z,x)^2-d_S(\Psi (z),\Psi (x))^2- d(z,\gamma _{y,x}(T))^2+d_S(\Psi (z),\Psi (\gamma _{y,x}(T)))^2 \right| \le C\eta ^{1/26}. \end{aligned}$$

Proof

If $d(x,A_f)\le \eta ^{1/26}$, then $d(x,y)\le C\eta ^{1/26}$, and so $d(x,\gamma _{y,x}(T))\le C\eta ^{1/26}$. Thus, we immediately get the lemma by Lemma 4.25 if $d(x,A_f)\le \eta ^{1/26}$. In the following, we assume that $d(x,A_f)\ge \eta ^{1/26}$. Take $u\in S^{n-p}$ with $f=\sum _{i=1}^{n-p+1}u_i f_i$, and $v\in S^{n-p}$ of Lemma 4.36 (iii). Define

$$\begin{aligned} r(s):=d_S(\Psi (z),\gamma _v(s)). \end{aligned}$$

Then, by the triangle inequality and Lemma 4.36 (iii), we have

$$\begin{aligned} |r(s)-d_S(\Psi (z),\Psi (\gamma _{y,x}(s)))|\le C\eta ^{3/13}. \end{aligned}$$

(73)

There exist $w\in S^{n-p}$ and $x_1,x_2,x_3\in {\mathbb {R}}$ such that $w\perp {\mathrm {Span}}_{{\mathbb {R}}}\{u,v\}$, $x_1^2+x_2^2+x_3^2=1$ and $\Psi (z)=x_1 u+x_2 v+ x_3 w.$ Then,

$$\begin{aligned} \cos r(s)=x_1\cos s+x_2\sin s \end{aligned}$$

(74)

by the definition of $\gamma _v$ in Lemma 4.36 (iii), and so

$$\begin{aligned} -x_1\sin s+x_2\cos s =\frac{d}{d s} \cos r(s) =-r'(s)\sin r(s). \end{aligned}$$

Thus, we get

$$\begin{aligned} \begin{aligned} -r'(s)\sin r(s) \sin s=-x_1\sin ^2 s+x_2\sin s\cos s=\cos r(s)\cos s-x_1 \end{aligned} \end{aligned}$$

(75)

by (74). Since $x_1=\Psi (z)\cdot u$ and $f(z)={\widetilde{\Psi }}(z)\cdot u$, we have

$$\begin{aligned} |x_1-\cos d(z,A_f)|\le C\delta ^{1/1000n^2} \end{aligned}$$

(76)

by Proposition 4.17 (iv) and Lemma 4.22. By Lemma 4.36, (73), (75) and (76), we get

$$\begin{aligned} \begin{aligned}&\Big |\Big (\cos d(\gamma _{y,x}(s),A_f)\cos d_S(\Psi (z),\Psi (\gamma _{y,x}(s)))\\&\qquad -\cos d(z,A_f)\Big )d_S(\Psi (z),\Psi (\gamma _{y,x}(s)))\\&\qquad +r'(s)r(s)\sin r(s) \sin s\Big |\le C\eta ^{3/13}. \end{aligned} \end{aligned}$$

(77)

Define

$$\begin{aligned} l(s):=d(z,\gamma _{y,x}(s)). \end{aligned}$$

Then, we have $l'(s)=\langle {\dot{\gamma }}_{z,\gamma _{y,x}(s)}(l(s)),{\dot{\gamma }}_{y,x}(s)\rangle$ for all $s\in [0,d(x,y)]$ with $\gamma _{y,x}(s)\in I_z\setminus \{z\}$, and so $|l'(s)\sin s+\langle {\dot{\gamma }}_{z,\gamma _{y,x}(s)}(l(s)),\nabla f(\gamma _{y,x}(s))\rangle |\le C\eta ^{3/26}$ by Lemma 4.37 (ii). Thus, for almost all $s\in [T,d(x,y)]$, we have

$$\begin{aligned} \begin{aligned}&\Big |\langle {\dot{\gamma }}_{\gamma _{y,x}(s),z}(0),\nabla f(\gamma _{y,x}(s))\rangle l(s)\sin d_S(\Psi (z),\Psi (\gamma _{y,x}(s)))\\&\quad -l'(s)l(s)\sin r(s)\sin s \Big |\le C\eta ^{3/26} \end{aligned} \end{aligned}$$

(78)

by (73). By the definition of $G_f^z$, (77) and (78), for almost all $s\in [T,d(x,y)]$, we have

$$\begin{aligned} \Big | G_f^z(\gamma _{y,x}(s))-l'(s)l(s)\sin r(s)\sin s+r'(s)r(s)\sin r(s) \sin s \Big |\le C\eta ^{3/26}. \end{aligned}$$

Thus, by the assumption, we get

$$\begin{aligned} \int _T^{d(x,y)}\left| \left( \frac{d}{d s}(l(s)^2-r(s)^2)\right) \sin r(s)\sin s\right| \,d s \le C\eta ^{3/26}. \end{aligned}$$

(79)

Define

$$\begin{aligned} I&:=\{s\in [T,d(x,y)]: \eta ^{1/26}\le s\le \pi -\eta ^{1/26}\text { and }\eta ^{1/26}\le r(s) \le \pi -\eta ^{1/26} \}\\ II&:=[T,d(x,y)]\setminus I. \end{aligned}$$

Then, we have

$$\begin{aligned} \int _I \left| \frac{d}{d s}(l(s)^2-r(s)^2)\right| \,d s \le C\eta ^{1/26} \end{aligned}$$

(80)

by (79). Let us estimate $H^1(II)$, where $H^1$ denotes the 1-dimensional Hausdorff measure. Suppose that

$$\begin{aligned} \{s\in [T,d(x,y)]: r(s)<\eta ^{1/26} \text { or } r(s)>\pi -\eta ^{1/26}\}\ne \emptyset , \end{aligned}$$

and take arbitrary $s\in [T,d(x,y)]$ such that $r(s)<\eta ^{1/26}$ or $r(s)>\pi -\eta ^{1/26}$. Then, we have

$$\begin{aligned} ||\cos r(s)|-1|\le C\eta ^{1/13}. \end{aligned}$$

(81)

Note that we have $r(s)\le \pi$ by ${\mathrm {diam}}(S^{n-p})=\pi$. By (74), we get

$$\begin{aligned} 1-C\eta ^{1/13}\le (x_1^2+x_2^2)^{1/2}\le 1. \end{aligned}$$

(82)

Take $s_1\in [0,2\pi ]$ such that

$$\begin{aligned} \cos s_1=&\frac{x_1}{(x_1^2+x_2^2)^{1/2}},\\ \sin s_1=&\frac{x_2}{(x_1^2+x_2^2)^{1/2}}. \end{aligned}$$

Then, we get $||\cos (s-s_1)|-1|\le C\eta ^{1/13}$ by (74), (81) and (82). Thus, there exists $n\in {\mathbb {Z}}$ such that $|s-s_1-n\pi |\le C\eta ^{1/26}.$ Then, we have $|n|\le 2$, and so

$$\begin{aligned} H^1\left( \{s\in [T,d(x,y)]: r(s)<\eta ^{1/26} \text { or } r(s)>\pi -\eta ^{1/26}\}\right) \le C\eta ^{1/26}. \end{aligned}$$

Note that we have $d(x,y)\le d(x,A_f)+C\eta \le \pi +C\eta$ by the assumption and Proposition 4.17 (iv). Since we have

$$\begin{aligned} H^1\left( \{s\in [T,d(x,y)]: s<\eta ^{1/26} \text { or } s>\pi -\eta ^{1/26}\}\right) \le C\eta ^{1/26}, \end{aligned}$$

we get $H^1(II)\le C\eta ^{1/26}$. Since $\left| \frac{d}{d s}(l(s)^2-r(s)^2)\right| \le C$ for almost all $s\in [T,d(x,y)]$, we get

$$\begin{aligned} \int _{II} \left| \frac{d}{d s}(l(s)^2-r(s)^2)\right| \,d s \le C\eta ^{1/26}. \end{aligned}$$

(83)

By (80) and (83), we get

$$\begin{aligned} \int _T^{d(x,y)}\left| \frac{d}{d s}(l(s)^2-r(s)^2)\right| \,d s \le C\eta ^{1/26}. \end{aligned}$$

Thus, we have $|l(d(x,y))^2-r(d(x,y))^2-l(T)^2+r(T)^2|\le C\eta ^{1/26}.$ By (73) and the definition of l, we get the lemma. $\square$

Definition 4.39

Take $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots , f_{n-p+1}\}$ with $\Vert f\Vert _2^2=1/(n-p+1)$. By Lemma 4.34 and the Bishop-Gromov inequality, for any triple $(x_1,x_2,x_3)\in M\times M\times M$, we can take points ${\tilde{x}}_1\in D_{f_{x_1}}(p_{x_1})\cap Q_{f_{x_1}} \cap R_{f_{x_1}}\cap Q_f$, ${\tilde{x}}_2\in D_f(p_f)\cap Q_f \cap R_f\cap P_f^{{\tilde{x}}_1}$ and ${\tilde{x}}_3\in D_f({\tilde{x}}_2)\cap D_f(p_f)\cap Q_f\cap R_f\cap C_f^{{\tilde{x}}_1}({\tilde{x}}_2)$ such that $d(x_1,{\tilde{x}}_1)\le C\delta ^{1/100n}$, $d(x_2,{\tilde{x}}_2)\le C\eta _0$, $d(x_3,{\tilde{x}}_3)\le C\eta _0$. We call the triple $({\tilde{x}}_1,{\tilde{x}}_2,{\tilde{x}}_3)$ a “$\Pi$-triple for $(x_1,x_2,x_3,f)$”.

Lemma 4.40

Take

• $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots , f_{n-p+1}\}$ with $\Vert f\Vert _2^2=1/(n-p+1)$,

• $x,y,z\in M$,

• $\eta >0$ with $\eta _0\le \eta \le L^{1/3n}$ and $T\in [0, d(x,y)]$.

Take a $\Pi$-triple $({\tilde{z}},{\tilde{x}},{\tilde{y}})$ for (z, x, y, f). Suppose

• $d(y,A_f)\le C\eta$,

• $|d(x,A_f)-d(x,y)|\le C\eta$,

• $d({\tilde{z}},\gamma _{{\tilde{y}},{\tilde{x}}}(s))\le \pi$ for all $s\in [T,d({\tilde{x}},{\tilde{y}})]$.

Then, we have

$$\begin{aligned} \left| d({\tilde{z}},{\tilde{x}})^2-d_S(\Psi ({\tilde{z}}),\Psi ({\tilde{x}}))^2- d({\tilde{z}},\gamma _{{\tilde{y}},{\tilde{x}}}(T))^2+d_S(\Psi ({\tilde{z}}),\Psi (\gamma _{{\tilde{y}},{\tilde{x}}}(T)))^2 \right| \le C\eta ^{1/26}. \end{aligned}$$

Proof

We have $(G^{{\tilde{z}}}_f H^{{\tilde{z}}})(\gamma _{{\tilde{y}},{\tilde{x}}}(s))=G^{{\tilde{z}}}_f(\gamma _{{\tilde{y}},{\tilde{x}}}(s))$ for all $s\in [T,d({\tilde{x}},{\tilde{y}})]$. Thus, we get the lemma immediately by the definition of $C_f^{{\tilde{z}}}({\tilde{x}})$ and Lemma 4.38. $\square$

The following lemma guarantees that if the images of two points in M under $\Phi _f$ are close to each other in $S^{n-p}\times A_f$, then their distance in M are close to each other under some assumptions.

Lemma 4.41

Take

• $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots , f_{n-p+1}\}$ with $\Vert f\Vert _2^2=1/(n-p+1)$,

• $x,y,z\in M$,

• $\eta >0$ with $\eta _0\le \eta \le L^{1/3n}$.

Suppose

• $d(x,A_f)\le \pi - \frac{1}{C}\eta ^{1/78}$ and $d(z,A_f)\le \pi - \frac{1}{C}\eta ^{1/78}$,

• $d(y,A_f)\le C\eta$,

• $|d(x,A_f)-d(x,y)|\le C\eta$ and $|d(z,A_f)-d(z,y)|\le C\eta$

• $d_S(\Psi (x),\Psi (z))\le C\eta$.

Then, we have $d(x,z)\le C\eta ^{1/52}.$

Proof

We first show the following claim.

Claim 4.42

If $x,y,z\in M$ satisfies:

• $d(x,A_f)\le \frac{1}{2}\pi - \frac{1}{C}\eta ^{1/2}$ and $d(z,A_f)\le \frac{1}{2}\pi - \frac{1}{C}\eta ^{1/2}$,

• $d(y,A_f)\le C\eta$,

• $|d(x,A_f)-d(x,y)|\le C\eta$ and $|d(z,A_f)-d(z,y)|\le C\eta$,

• $d_S(\Psi (x),\Psi (z))\le C\eta ^{1/52}$.

Then, we have $d(x,z)\le C\eta ^{1/52}.$

Proof of Claim 4.42

Take $u\in S^{n-p}$ with $f=\sum _{i=1}^{n-p+1} u_i f_i$. By the assumptions and Lemma 4.26, we have

$$\begin{aligned} d_S(u,\Psi (y))&\le C\eta ,\\ |d_S(\Psi (z),u)-d(z,A_f)|&\le C\delta ^{1/2000n^2}. \end{aligned}$$

Since we have $|d(z,A_f)-d(z,y)|\le C\eta$ by the assumptions, we get

$$\begin{aligned} |d_S(\Psi (z),\Psi (y))-d(z,y)|\le C\eta . \end{aligned}$$

(84)

Take a $\Pi$-triple $({\tilde{z}},{\tilde{x}},{\tilde{y}})$ for (z, x, y, f). Then, we have

$$\begin{aligned} d({\tilde{z}},\gamma _{{\tilde{y}},{\tilde{x}}}(s)) \le d(z,y)+d(y,x)+C\eta _0 \le \pi -\frac{1}{C}\eta ^{1/2}+C\eta \le \pi \end{aligned}$$

for all $s\in [0,d({\tilde{x}},{\tilde{y}})]$, and so

$$\begin{aligned} \left| d(z,x)^2-d_S(\Psi (z),\Psi (x))^2- d(z,y)^2+d_S(\Psi (z),\Psi (y))^2 \right| \le C\eta ^{1/26} \end{aligned}$$

by Lemmas 4.25 and 4.40. Thus, we get $d(x,z)\le C\eta ^{1/52}$ by (84). $\square$

Let us suppose that $x,y,z\in M$ satisfies the assumptions of the lemma. Take $u\in S^{n-p}$ with $f=\sum _{i=1}^{n-p+1}u_i f_i$. By the assumptions and Lemma 4.26, we have

$$\begin{aligned} |d(x,A_f)-d(z,A_f)| \le |d_S(\Psi (x),u)-d(\Psi (z),u)|+C\delta ^{1/2000n^2} \le C\eta \end{aligned}$$

(85)

Thus, if either $d(x,A_f)\le \eta ^{1/26}$ or $d(z,A_f)\le \eta ^{1/26}$ holds, then the lemma is trivial. In the following, we assume $d(x,A_f)\ge \eta ^{1/26}$ and $d(z,A_f)\ge \eta ^{1/26}$. Take a $\Pi$-triple $({\tilde{z}},{\tilde{x}},{\tilde{y}})$ for (z, x, y, f). By Lemma 4.36 (iii), we can take $v_1,v_2\in S^{n-p}$ such that $u\cdot v_i=0$ ($i=1,2$),

$$\begin{aligned} d_S(\Psi (\gamma _{{\tilde{y}},{\tilde{x}}}(s)),\gamma _{v_1}(s))\le C\eta ^{3/13} \end{aligned}$$

(86)

for all $s\in [0,d({\tilde{y}},{\tilde{x}})]$ and

$$\begin{aligned} d_S(\Psi (\gamma _{{\tilde{y}},{\tilde{z}}}(s)),\gamma _{v_2}(s))\le C\eta ^{3/13} \end{aligned}$$

(87)

for all $s\in [0,d({\tilde{y}},{\tilde{z}})]$, where $\gamma _{v_i}(s):=(\cos s) u+(\sin s) v_i \in S^{n-p}$ ($i=1,2$). By the assumptions and (85), we get

$$\begin{aligned} |d({\tilde{y}},{\tilde{x}})-d({\tilde{y}},{\tilde{z}})|\le C\eta , \end{aligned}$$

(88)

and so

$$\begin{aligned} \sin d({\tilde{y}},{\tilde{x}}) |v_1- v_2| \le&C d_S(\gamma _{v_1}(d({\tilde{y}},{\tilde{x}})),\gamma _{v_2}(d({\tilde{y}},{\tilde{x}})))\\ \le&Cd_S(\Psi ({\tilde{x}}),\Psi ({\tilde{z}}))+C\eta ^{3/13} \le C\eta ^{3/13} \end{aligned}$$

by (86) and (87). By $\eta ^{1/26}\le d(x,A_f)\le \pi -\frac{1}{C}\eta ^{1/78}$, we have $\sin d({\tilde{y}},{\tilde{x}})\ge \frac{1}{C}\eta ^{1/26}$. Thus, we get $|v_1-v_2|\le C\eta ^{1/26}.$ This gives

$$\begin{aligned} d_S(\gamma _{v_1}(s),\gamma _{v_2}(s))\le C\eta ^{1/26}. \end{aligned}$$

(89)

for all $s\in {\mathbb {R}}$.

Put $a:=\gamma _{{\tilde{y}},{\tilde{x}}}\left( d({\tilde{y}},{\tilde{x}})/2\right)$ and $b:=\gamma _{{\tilde{y}},{\tilde{z}}}\left( d({\tilde{y}},{\tilde{z}})/2\right) .$ By (86), (87), (88) and (89), we have $d_S(\Psi (a),\Psi (b))\le C\eta ^{1/26}.$ Moreover, other assumptions of Claim 4.42 hold for the pair (a, y, b) by Lemma 4.36 (i), and so $d(a,b)\le C\eta ^{1/52}.$ Therefore, we have

$$\begin{aligned} d({\tilde{z}}, \gamma _{{\tilde{y}},{\tilde{x}}}(s))\le & {} d({\tilde{z}},b)+d(a,b)+d(\gamma _{{\tilde{y}},{\tilde{x}}}(s),a)\\\le & {} \frac{1}{2}d({\tilde{x}},{\tilde{y}})+\frac{1}{2}d({\tilde{z}},{\tilde{y}})+C\eta ^{1/52}\le \pi \end{aligned}$$

for all $s\in [0,d({\tilde{y}},{\tilde{x}})]$, and so $d({\tilde{x}},{\tilde{z}})\le C\eta ^{1/52}$ similarly to Claim 4.42. Thus, we get the lemma. $\square$

Let us show the almost Pythagorean theorem for the special case. Recall that we defined $\eta _1:=\eta _0^{1/26}$.

Lemma 4.43

Take

• $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots , f_{n-p+1}\}$ with $\Vert f\Vert _2^2=1/(n-p+1)$,

• $x,y,z,w\in M$,

• $\eta >0$ with $\eta _1\le \eta \le L^{1/3n}$.

Suppose

• $d(x,z)\le C\eta$,

• $d(x,A_f)\le \pi - \frac{1}{C}\eta ^{1/2}$ and $d(z,A_f)\le \pi - \frac{1}{C}\eta ^{1/2}$,

• $d(y,A_f)\le C\eta _0$ and $d(w,A_f)\le C\eta _0$,

• $|d(x,A_f)-d(x,y)|\le C\eta _0$ and $|d(z,A_f)-d(z,w)|\le C\eta _0$.

Then, we have

$$\begin{aligned} |d(x,z)^2-d_S(\Psi (x),\Psi (z))^2-d(y,w)^2|\le C\eta _1. \end{aligned}$$

Proof

By Lemma 4.26, we have

$$\begin{aligned} d_S(\Psi (y),\Psi (w)) \le d(y,A_f)+d(w,A_f)+C\delta ^{1/2000n^2} \le C\eta _0. \end{aligned}$$

(90)

Put $a_0:=x$ and $b_0:=z$. In the following, we define $a_{i},b_{i}\in M$ ($i=1,2,3$) so that

• (i) $d(a_{i},b_{i})\le C\eta ^{1/2}$,

• (ii) $|d(a_{i},A_f)-d(a_{i},y)|\le C\eta _0$ and $|d(b_{i},A_f)-d(b_{i},w)|\le C\eta _0$,

• (iii) $d(a_{i},A_f)\le \frac{3-i}{3}\pi +C\eta _0$ and $d(b_{i},A_f)\le \frac{3-i}{3}\pi +C\eta _0$,

• (iv) $|d(a_{i+1},b_{i+1})^2-d_S(\Psi (a_{i+1}),\Psi (b_{i+1}))^2-d(a_{i},b_{i})^2+d_S(\Psi (a_{i}),\Psi (b_{i}))^2|\le C\eta _0^{1/26}$ ($i=0,1,2$),

• (v) $d(y,a_3)\le C\eta _0$ and $d(w,b_3)\le C\eta _0$.

If we succeed in defining such $a_i$ and $b_i$, we have

$$\begin{aligned} |d(x,z)^2-d_S(\Psi (x),\Psi (z))^2-d(y,w)^2+d_S(\Psi (y),\Psi (w))^2|\le C\eta _0^{1/26}=C\eta _1 \end{aligned}$$

by (iv) and (v), and so we get the lemma by (90).

Take arbitrary $i\in \{0,1,2\}$ and suppose that we have chosen $a_i,b_i\in M$ such that (i), (ii) and (iii) hold if $i\ge 1$. Let us define $a_{i+1},b_{i+1}\in M$ that satisfy our properties. Take a $\Pi$-triple $({\tilde{b}}_i,{\tilde{a}}_i, {\tilde{y}}_i)$ for $(b_i,a_i,y,f)$. Define

$$\begin{aligned} a_{i+1}:=\gamma _{{\tilde{y}}_i,{\tilde{a}}_i}\left( \frac{2-i}{3-i}d({\tilde{y}}_i,{\tilde{a}}_i)\right) . \end{aligned}$$

Since

$$\begin{aligned} d({\tilde{b}}_i, \gamma _{{\tilde{y}}_i,{\tilde{a}}_i}(s))\le d({\tilde{a}}_i,{\tilde{b}}_i) +d({\tilde{a}}_i,\gamma _{{\tilde{y}}_i,{\tilde{a}}_i}(s))\le \frac{\pi }{3}+C\eta ^{1/2} \end{aligned}$$

for all $s\in \left[ \frac{2-i}{3-i}d({\tilde{y}}_i,{\tilde{a}}_i),d({\tilde{y}}_i,{\tilde{a}}_i)\right]$ by the assumptions, we get

$$\begin{aligned} |d(a_{i+1},b_{i})^2-d_S(\Psi (a_{i+1}),\Psi (b_{i}))^2-d(a_{i},b_{i})^2+d_S(\Psi (a_{i}),\Psi (b_{i}))^2|\le C\eta _0^{1/26}\nonumber \\ \end{aligned}$$

(91)

by Lemmas 4.25 and 4.40. Take a $\Pi$-triple $({\overline{a}}_{i+1},{\overline{b}}_i,{\overline{w}}_i)$ for $(a_{i+1},b_i,w,f)$. Define

$$\begin{aligned} b_{i+1}:=\gamma _{{\overline{w}}_i,{\overline{b}}_i}\left( \frac{2-i}{3-i}d({\overline{w}}_i,{\overline{b}}_i)\right) . \end{aligned}$$

Since

$$\begin{aligned} d({\overline{a}}_{i+1},\gamma _{{\overline{w}}_i,{\overline{b}}_i}(s))\le d({\overline{a}}_{i+1},a_i)+d(a_i,{\overline{b}}_i)+d(\overline{b_i},\gamma _{{\overline{w}}_i,{\overline{b}}_i}(s))\le \frac{2}{3}\pi +C\eta ^{1/2} \end{aligned}$$

for all $s\in \left[ \frac{2-i}{3-i}d({\overline{w}}_i,{\overline{b}}_i),d({\overline{w}}_i,{\overline{b}}_i)\right]$ by the assumptions, we get

$$\begin{aligned}&|d(a_{i+1},b_{i+1})^2-d_S(\Psi (a_{i+1}),\Psi (b_{i+1}))^2\nonumber \\&\quad -d(a_{i+1},b_{i})^2+d_S(\Psi (a_{i+1}),\Psi (b_{i}))^2|\le C\eta _0^{1/26} \end{aligned}$$

(92)

by Lemmas 4.25 and 4.40. By (91) and (92), we get (iv).

By the assumptions and Lemma 4.36, we get (ii) for $a_{i+1}$ and $b_{i+1}$.

By the assumptions, we have

$$\begin{aligned} d(a_{i+1},A_f)&\le d(a_{i+1},{\tilde{y}}_i)+d(y,A_f)+C\eta _0\\&=\frac{2-i}{3-i}d({\tilde{a}}_{i},{\tilde{y}}_i)+C\eta _0 \le \frac{2-i}{3}\pi +C\eta _0. \end{aligned}$$

Similarly, we have $d(b_{i+1},A_f)\le \frac{2-i}{3}\pi +C\eta _0$. Thus, we get (iii) for $a_{i+1}$ and $b_{i+1}$.

By definition, we have $a_3={\tilde{y}}_3$ and $b_3={\overline{w}}_3.$ Thus, we get (v).

In the following, we prove (i) for $a_{i+1}$ and $b_{i+1}$. If $d(a_i,y)\le \eta _0^{1/26}$, then we have

$$\begin{aligned} d(b_i,w)\le d(b_i,A_f)+C\eta _0 \le d(a_i,A_f)+C\eta ^{1/2} \le C\eta ^{1/2}, \end{aligned}$$

and so $d(y,w)\le C\eta ^{1/2}$, $d(a_{i+1},y)\le C\eta ^{1/2}$ and $d(b_{i+1},w)\le C\eta ^{1/2}.$ Then, we have $d(a_{i+1},b_{i+1})\le C\eta ^{1/2}$. Similarly, if $d(b_i,w)\le \eta _0^{1/26}$, then $d(a_{i+1},b_{i+1})\le C\eta ^{1/2}$. Thus, in the following, we assume that $d(a_i,y)\ge \eta _0^{1/26}$ and $d(b_i,w)\ge \eta _0^{1/26}$. By Lemma 4.36, we can take $u,v_1,v_2\in S^{n-p}$ such that $f=\sum _{j=1}^{n-p+1}u_j f_j$, $u\cdot v_k=0$ ($k=1,2$),

$$\begin{aligned} d_S(\Psi (\gamma _{{\tilde{y}}_i,{\tilde{a}}_i}(s)),\gamma _{v_1}(s))\le C\eta _0^{3/13} \end{aligned}$$

(93)

for all $s\in [0,d({\tilde{a}}_i,{\tilde{y}}_i)]$ and

$$\begin{aligned} d_S(\Psi (\gamma _{{\overline{w}}_i,{\overline{b}}_i}(s)),\gamma _{v_2}(s))\le C\eta _0^{3/13} \end{aligned}$$

(94)

for all $s\in [0,d({\overline{b}}_i,{\overline{w}}_i)]$, where $\gamma _{v_k}(s):=(\cos s) u+(\sin s) v_k\in S^{n-p}$ ($k=1,2$). Since

$$\begin{aligned} |d({\tilde{a}}_i,{\tilde{y}}_i)-d({\overline{b}}_i,{\overline{w}}_i)|\le |d(a_i,A_f)-d(b_i,A_f)|+C\eta _0\le d(a_i,b_i)+C\eta _0, \end{aligned}$$

we have

$$\begin{aligned} \left| d_S(\Psi ({\tilde{a}}_i),\Psi ({\overline{b}}_i)) -d_S\left( \gamma _{v_1}(l_i),\gamma _{v_2} (l_i)\right) \right| \le d(a_i,b_i)+C\eta _0^{3/13} \end{aligned}$$

(95)

and

$$\begin{aligned}&\left| d_S(\Psi (a_{i+1}),\Psi (b_{i+1})) -d_S\left( \gamma _{v_1}\left( \frac{2-i}{3-i}l_i\right) ,\gamma _{v_2} \left( \frac{2-i}{3-i}l_i\right) \right) \right| \nonumber \\&\quad \le d(a_i,b_i)+C\eta _0^{3/13} \end{aligned}$$

(96)

by (93) and (94), where we put $l_i:=d({\tilde{a}}_i,{\tilde{y}}_i)$. By (95) and Lemma 4.25, we get

$$\begin{aligned} |v_1-v_2|\sin l_i \le C d_S\left( \gamma _{v_1}(l_i),\gamma _{v_2} (l_i)\right) \le Cd(a_i,b_i)+C\eta _0^{3/13}. \end{aligned}$$

(97)

We first suppose that $d(a_i,y)\le \pi /6$. Since $l_i\le \pi /2$, we have

$$\begin{aligned} \sin \left( \frac{2-i}{3-i}l_i\right) \le \sin l_i, \end{aligned}$$

and so

$$\begin{aligned} d_S(\Psi (a_{i+1}),\Psi (b_{i+1}))&\le d_S\left( \gamma _{v_1}\left( \frac{2-i}{3-i}l_i\right) ,\gamma _{v_2} \left( \frac{2-i}{3-i}l_i\right) \right) +C\eta ^{1/2}\\&\le C|v_1-v_2|\sin \left( \frac{2-i}{3-i}l_i\right) +C\eta ^{1/2}\\&\le C|v_1-v_2|\sin l_i+C\eta ^{1/2}\\&\le C d_S(\Psi ({\tilde{a}}_i),\Psi ({\overline{b}}_i))+C\eta ^{1/2} \le C\eta ^{1/2} \end{aligned}$$

by (95), (96) and $d(a_i,b_i)\le C\eta ^{1/2}$. Thus, we get $d(a_{i+1},b_{i+1})\le C\eta ^{1/2}$ by (iv).

We next suppose that $\pi /6\le d(a_i,y)\le 5\pi /6$. By (97) and $d(a_i,b_i)\le C\eta ^{1/2}$, we have $|v_1-v_2|\le C\eta ^{1/2}$. Thus, we get $d_S(\Psi (a_{i+1}),\Psi (b_{i+1})) \le C\eta ^{1/2}$ by (96). Thus, we get $d(a_{i+1},b_{i+1})\le C\eta ^{1/2}$ by (iv).

If $i\ge 1$, we have $d(a_i,y)\le 5\pi /6$, and so we get $d(a_{i+1},b_{i+1})\le C\eta ^{1/2}$ by the above two cases.

Finally, we suppose that $i=0$ and $d(x,y)\ge 5\pi /6$. By (97) and $d(a_0,b_0)\le C\eta$, we have $|v_1-v_2|\sin l_0\le C\eta$. By the definition of $l_0$, we have $|l_0-d(x,y)|\le C\eta _0.$ Thus, we have $\sin l_0\ge \frac{1}{C}(\pi - l_0)\ge \frac{1}{C}\eta ^{1/2}$, and so we get $|v_1-v_2|\le C\eta ^{1/2}$. This gives $d_S(\Psi (a_{i+1}),\Psi (b_{i+1})) \le C\eta ^{1/2}$ by (96). Thus, $d(a_{i+1},b_{i+1})\le C\eta ^{1/2}$ by (iv).

Therefore, we have (i) for all cases, and we get the lemma. $\square$

Let us show that the map $\Phi _f:M\rightarrow S^{n-p}\times A_f,\,x\mapsto (\Psi (x), a_f(x))$ is almost surjective.

Proposition 4.44

Take $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots , f_{n-p+1}\}$ with $\Vert f\Vert _2^2=1/(n-p+1)$. For any $(v,a)\in S^{n-p}\times A_f$, there exists $x\in M$ such that $d(\Phi _f(x),(v,a))\le C\eta _1^{1/2}$ holds.

Proof

Take arbitrary $(v,a)\in S^{n-p}\times A_f$. Take $u\in S^{n-p}$ with $f=\sum _{i=1}^{n-p+1} u_i f_i$. Since there exists ${\tilde{v}}\in S^{n-p}$ such that $d_S(u,{\tilde{v}})\le \pi -\eta _1^{1/2}$ and $d_S(v,{\tilde{v}})\le \eta _1^{1/2}$, it is enough to prove the proposition assuming $d_S(u,v)\le \pi -\eta _1^{1/2}$.

Put $F_v:=\sum _{i=1}^{n-p+1}v_i f_i$. Then, $|F_v(p_{F_v})-1|\le C\delta ^{1/800n}$ and $A_{F_v}=\{x\in M:|F_v(x)-1|\le \delta ^{1/900n}\}$ by Proposition 4.17. In the following, we show that $a_v:=a_{F_v}(a)\in A_{F_v}$ has the desired property. By Lemma 4.26, we get

$$\begin{aligned} \nonumber d_S(\Psi (a),u)\le&C\delta ^{1/2000n^2},\\ d_S(\Psi (a_v),v)\le&C\delta ^{1/2000n^2}. \end{aligned}$$

(98)

Thus, by Lemma 4.26, we get

$$\begin{aligned} |d(a,a_v)-d(a_f(a_v),a_v)|=&|d(a,A_{F_v})-d(a_v,A_f)|\\ \le&|d_S(\Psi (a),v)-d_S(\Psi (a_v),u)|+C\delta ^{1/2000n^2}\\ \le&C\delta ^{1/2000n^2}\le \eta _0 \end{aligned}$$

and

$$\begin{aligned} d(a_v,A_f)\le d_S(\Psi (a_v),u)+C\delta ^{1/2000n^2} \le d_S(u,v)+C\delta ^{1/2000n^2}\le \pi -\frac{1}{2}\eta _1^{1/2}. \end{aligned}$$

Since we have $d(a_v,A_f)=d(a_v,a_f(a_v))$, we get

$$\begin{aligned} |d(a_v,A_f)-d(a_v,a)|\le |d(a_v,A_f)-d(a_v,a_f(a_v))|+\eta _0=\eta _0, \end{aligned}$$

and so we get

$$\begin{aligned} d(a,a_f(a_v))\le C\eta _1^{1/2} \end{aligned}$$

(99)

by Lemma 4.43 putting $x=z=a_v$, $y=a$ and $w=a_f(a_v)$.

By (98) and (99), putting $x=a_v$, we get the proposition. $\square$

Now, we are in position to show $|{\dot{\gamma }}_{{\tilde{y}}_1,y_2}^E|d({\tilde{y}}_1,y_2)\le \pi +L$ under the assumption of Lemma 4.30. Note that we defined $\eta _2=\eta _1^{1/78}$ and $L=\eta _2^{1/150}$.

Lemma 4.45

Take $y_1\in M$, ${\tilde{y}}_1\in D_{f_{y_1}}(p_{y_1})\cap R_{f_{y_1}}\cap Q_{f_{y_1}}$ with $d(y_1,{\tilde{y}}_1)\le C\delta ^{1/100n}$ and $y_2\in D_{f_{y_1}}({\tilde{y}}_1)$. Let $\{E_1,\ldots ,E_n\}$ be a parallel orthonormal basis of TM along $\gamma _{{\tilde{y}}_1,y_2}$ in Lemma 4.13 for $f_{y_1}$. Then, $|{\dot{\gamma }}_{{\tilde{y}}_1,y_2}^E|d({\tilde{y}}_1,y_2)\le \pi + L$ and

$$\begin{aligned} ||{\dot{\gamma }}_{{\tilde{y}}_1,y_2}^E|d({\tilde{y}}_1,y_2)-d_S(\Psi (y_1),\Psi (y_2))|\le CL. \end{aligned}$$

Proof

We immediately get the second assertion by the first assertion and Lemma 4.30.

Let us show the first assertion by contradiction. Suppose that $|{\dot{\gamma }}_{{\tilde{y}}_1,y_2}^E|d({\tilde{y}}_1,y_2)>\pi + L.$ Put

$$\begin{aligned} f:=-f_{y_1},\,\gamma :=\gamma _{{\tilde{y}}_1,y_2},\,s_0:=\frac{1}{|{\dot{\gamma }}^E|}\eta _2^{1/104}\text { and }s_1:=\frac{1}{|{\dot{\gamma }}^E|}(\pi +L). \end{aligned}$$

Take $k\in {\mathbb {N}}$ to be $(s_1-s_0)/\eta _2^{-1}<k\le (s_1-s_0)/\eta _2^{-1}+1,$ and put $t_j:= s_0+ (s_1-s_0)j/k$ for each $j\in \{0,\ldots ,k\}$. Note that we have $t_0=s_0$, $t_k=s_1$ and

$$\begin{aligned} \frac{1}{C}\eta _2^{-1}\le k\le C\eta _2^{-1}. \end{aligned}$$

(100)

For all $s\in [s_0,s_1]$, we have

$$\begin{aligned} \cos d_S(\Psi (y_1),\Psi (\gamma (s))) \le \cos (|{\dot{\gamma }}^E| s)+C\delta ^{1/2000n^2} \le 1-\frac{1}{C}\eta _2^{1/52} \end{aligned}$$

for all $s\in [s_0,s_1]$ by Lemma 4.30. Since $f(\gamma (s))=-|{\widetilde{\Psi }}|(\gamma (s))\cos d_S(\Psi (y_1),\Psi (\gamma (s)))$ by the definitions of $f_{y_1}$ and f, we get $f(\gamma (s))\ge -1+\frac{1}{C}\eta _2^{1/52}$ for all $s\in [s_0,s_1]$ by Lemma 4.22. This gives

$$\begin{aligned} d(\gamma (s),A_f)\le \pi -\frac{1}{C}\eta _2^{1/104} \end{aligned}$$

(101)

$s\in [s_0,s_1]$ by Proposition 4.17. By the definition of $t_j$ and (101), we have

$$\begin{aligned} d(\gamma (t_j),\gamma (t_{j+1}))\le&\eta _2,\\ \nonumber d(\gamma (t_{j+\sigma }),A_f)\le&\pi -\frac{1}{C}\eta _2^{1/104}\le \pi -\eta _2^{1/2} \end{aligned}$$

(102)

for all $j\in \{0,\ldots ,k-1\}$ and $\sigma \in \{0,1\}$, and so we get

$$\begin{aligned}&|d(\gamma (t_j),\gamma (t_{j+1}))^2-d_S(\Psi (\gamma (t_j)),\Psi (\gamma (t_{j+1})))^2\nonumber \\&\quad -d(a_f(\gamma (t_j)),a_f(\gamma (t_{j+1})))^2|\le C\eta _1 \end{aligned}$$

(103)

by Lemma 4.43. In particular, we get

$$\begin{aligned} d(a_f(\gamma (t_j)),a_f(\gamma (t_{j+1})))\le C\eta _2 \end{aligned}$$

(104)

by (102).

Take $j_0\in \{1,\ldots , k-1\}$ to be $|{\dot{\gamma }}^E|t_{j_0}< \pi \le |{\dot{\gamma }}^E|t_{j_0+1}.$ Since

$$\begin{aligned} ||{\dot{\gamma }}^E|s-d_S(\Psi (y_1),\Psi (\gamma (s)))|\le C\delta ^{1/4000n^2} \end{aligned}$$

for all $s\in \left[ 0,\frac{1}{|{\dot{\gamma }}^E|}\pi \right]$ by Lemma 4.30, we get

$$\begin{aligned} \begin{aligned} d_S(\Psi (\gamma (t_j)),\Psi (\gamma (t_{j+1})))&\ge d_S(\Psi (y_1),\Psi (\gamma (t_{j+1})))- d_S(\Psi (y_1),\Psi (\gamma (t_{j})))\\&\ge |{\dot{\gamma }}^E|(t_{j+1}-t_j)-C\delta ^{1/4000n^2} \end{aligned} \end{aligned}$$

(105)

for all $j\in \{0,\ldots ,j_0-1\}$. Since

$$\begin{aligned} |2\pi -|{\dot{\gamma }}^E|s-d_S(\Psi (y_1),\Psi (\gamma (s)))|\le C\delta ^{1/4000n^2} \end{aligned}$$

for all $s\in \left[ \frac{1}{|{\dot{\gamma }}^E|}\pi ,s_1\right]$ by Lemma 4.30, we get

$$\begin{aligned} d_S(\Psi (\gamma (t_j)),\Psi (\gamma (t_{j+1}))) \ge |{\dot{\gamma }}^E|(t_{j+1}-t_j)-C\delta ^{1/4000n^2} \end{aligned}$$

(106)

for all $j\in \{j_0+1,\ldots ,k-1\}$. By (103), (105) and (106), we get

$$\begin{aligned} d(a_f(\gamma (t_j)),a_f(\gamma (t_{j+1})))^2 \le d(\gamma (t_j),\gamma (t_{j+1}))^2-|{\dot{\gamma }}^E|^2(t_{j+1}-t_j)^2+C\eta _1\nonumber \\ \end{aligned}$$

(107)

for all $j\in \{0,\ldots ,k-1\}\setminus \{j_0\}$.

Since we have

$$\begin{aligned} d_S(\Psi (\gamma (s_l)),\Psi (p_f)) \le d(\gamma (s_l),A_f)+C\delta ^{1/2000n^2} \le \pi -\frac{1}{C}\eta _2^{1/104} \end{aligned}$$

for each $l=0,1$ by Lemma 4.26, Corollary 4.27 and (101), we can take a curve $\beta :[0,K]\rightarrow S^{n-p}$ in $S^{n-p}$ with unit speed (K is some constant) such that

$$\begin{aligned} \beta (0)&=\Psi (\gamma (s_0)),\\ \beta (K)&=\Psi (\gamma (s_1)),\\ |d_S(\Psi (\gamma (s_0)),\Psi (\gamma (s_1)))-K|\le&C\eta _2^{1/104},\\ d_S(\beta (s),\Psi (p_f))\le&\pi -\frac{1}{C}\eta _2^{1/104} \end{aligned}$$

for all $s\in [0,K]$. Note that we can find such $\beta$ by taking an almost shortest pass in $\left\{ u\in S^{n-p}: d(u,\Psi (p_f))\le \pi -\frac{1}{C}\eta _2^{1/104}\right\} .$ By Proposition 4.44, there exists $x_j\in M$ such that

$$\begin{aligned} d\left( \Phi _f(x_j),\left( \beta \left( \frac{j}{k}K\right) ,a_f(\gamma (t_j))\right) \right) \le C\eta _1^{1/2} \end{aligned}$$

(108)

for each $j\in \{0,\ldots ,k\}$. Moreover, we can take $x_0:=\gamma (s_0)$ and $x_k:=\gamma (s_1)$. By (100), (104), (108), Lemma 4.26 and Corollary 4.27, we have

$$\begin{aligned} d(a_f(x_j),a_f(x_{j+1}))&\le C\eta _2,\nonumber \\ d_S(\Psi (x_j),\Psi (x_{j+1}))&\le \frac{1}{k}K+C\eta _1^{1/2}\le C\eta _2, \end{aligned}$$

(109)

$$\begin{aligned} d(x_j,A_f)&\le d_S(\Psi (x_j),\Psi (p_f))+C\delta ^{1/2000n^2} \nonumber \\&\le d_S\left( \beta \left( \frac{j}{k}K\right) ,\Psi (p_f)\right) +C\eta _1^{1/2} \le \pi -\frac{1}{C}\eta _2^{1/104} \end{aligned}$$

(110)

for all j, and so

$$\begin{aligned} d(x_j,x_{j+1})\le C\eta _2^{1/52} \end{aligned}$$

(111)

by Lemma 4.41 putting $x=x_j, y=a_f(x_j), z=x_{j+1}$ and $\eta =\eta _2$. By (110), (111) and Lemma 4.43 putting $x=x_j, y=a_f(x_j), z=x_{j+1}, w=a_f(x_{j+1})$ and $\eta =\eta _2^{1/52}$, we get

$$\begin{aligned} |d(x_j,x_{j+1})^2-d_S(\Psi (x_j),\Psi (x_{j+1}))^2-d(a_f(x_j),a_f(x_{j+1}))^2|\le C\eta _1 \end{aligned}$$

(112)

for all $j\in \{0,\ldots ,k-1\}$. By (107), (109) and (112), we have

$$\begin{aligned} \begin{aligned} d(x_j,x_{j+1})^2 \le&\frac{1}{k^2}K^2+d(a_f(x_j),a_f(x_{j+1}))^2+C\eta _1^{1/2}\\ \le&\frac{1}{k^2}K^2+d(\gamma (t_j),\gamma (t_{j+1}))^2-|{\dot{\gamma }}^E|^2(t_{j+1}-t_j)^2+C\eta _1^{1/2} \end{aligned} \end{aligned}$$

(113)

for all $j\in \{0,\ldots ,k-1\}\setminus \{j_0\}$. Since $K\le \pi +C\eta _2^{1/104}$, we have

$$\begin{aligned} \frac{1}{k^2}K^2\le \frac{\pi ^2}{k^2}+\frac{C}{k^2}\eta _2^{1/104}. \end{aligned}$$

(114)

Since

$$\begin{aligned} |{\dot{\gamma }}^E|(t_{j+1}-t_j)=\frac{|{\dot{\gamma }}^E|}{k}(s_1-s_0) =\frac{1}{k}(\pi +L-\eta _2^{1/104}) \ge \frac{1}{k}\left( \pi +\frac{1}{2}L\right) , \end{aligned}$$

we have

$$\begin{aligned} |{\dot{\gamma }}^E|^2(t_{j+1}-t_j)^2\ge \frac{\pi ^2}{k^2}+\frac{1}{k^2}L \end{aligned}$$

(115)

for all $j\in \{0,\ldots ,k-1\}$. By (114) and (115), we get

$$\begin{aligned} |{\dot{\gamma }}^E|^2(t_{j+1}-t_j)^2-\frac{1}{k^2}K^2\ge \frac{1}{k^2}L-\frac{C}{k^2}\eta _2^{1/104} \ge \frac{1}{2k^2}L \end{aligned}$$

for all $j\in \{0,\ldots ,k-1\}$. Thus, by (113), we have

$$\begin{aligned} d(x_j,x_{j+1})^2 \le d(\gamma (t_j),\gamma (t_{j+1}))^2-\frac{1}{2k^2}L+C\eta _1^{1/2} \le d(\gamma (t_j),\gamma (t_{j+1}))^2-\frac{1}{4k^2}L \end{aligned}$$

for all $j\in \{0,\ldots ,k-1\}\setminus \{j_0\}$. Since $d(\gamma (t_j),\gamma (t_{j+1}))+d(x_j,x_{j+1})\le 1$, we get

$$\begin{aligned} \frac{1}{4k^2}L\le d(\gamma (t_j),\gamma (t_{j+1}))^2-d(x_j,x_{j+1})^2 \le d(\gamma (t_j),\gamma (t_{j+1}))-d(x_j,x_{j+1})\nonumber \\ \end{aligned}$$

(116)

$j\in \{0,\ldots ,k-1\}\setminus \{j_0\}$. By (100), (111) and (116), we get

$$\begin{aligned} \begin{aligned} d(x_0,x_k)\le \sum _{i=0}^{k-1}d(x_j,x_{j+1}) \le&\sum _{i=0}^{k-1}d(\gamma (t_j),\gamma (t_{j+1}))-\frac{k-1}{4k^2}L+d(x_{j_0},x_{j_0+1})\\ \le&d(x_0,x_k)-\frac{1}{8k}L. \end{aligned}\nonumber \\ \end{aligned}$$

This is a contradiction. Thus, we get the lemma. $\square$

Notation 4.46

For all $y_1,y_2\in M$, define

$$\begin{aligned} {\overline{C}}_f^{y_1}(y_2)&=\Big \{y_3\in M : \gamma _{y_2,y_3}(s)\in I_{y_1}\setminus \{y_1\} \text { for almost all }s\in [0,d(y_2,y_3)],\text { and}\\&\qquad \int _{0}^{d(y_2,y_3)} |G_f^{y_1}|(\gamma _{y_2,y_3}(s))\,d s\le L^{1/3}\Big \},\\ {\overline{P}}_f^{y_1}&=\{y_2\in M: {\mathrm {Vol}}(M\setminus {\overline{C}}_f^{y_1}(y_2))\le L^{1/3}{\mathrm {Vol}}(M)\}. \end{aligned}$$

Let us complete the Gromov-Hausdorff approximation.

Theorem 4.47

Take $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots , f_{n-p+1}\}$ with $\Vert f\Vert _2^2=1/(n-p+1)$. Then, the map $\Phi _f:M\rightarrow S^{n-p}\times A_f$ is a $CL^{1/156n}$-Hausdorff approximation map. In particular, we have $d_{GH}(M, S^{n-p}\times A_f)\le CL^{1/156n}$.

Proof

Take arbitrary $y_1\in M$ and ${\tilde{y}}_1\in D_{f_{y_1}}(p_{y_1})\cap R_{f_{y_1}}\cap Q_{f_{y_1}}\cap Q_f$ with $d(y_1,{\tilde{y}}_1)\le C\delta ^{1/100n}$. By Lemmas 4.25, 4.45 and Corollary 4.33, we have $|G_f^{{\tilde{y}}_1}|(y_2)\le CL$ for all $y\in D_f(\tilde{y_1})\cap D_{f_{y_1}}({\tilde{y}}_1)$. Since ${\mathrm {Vol}}(M\setminus (D_f(\tilde{y_1})\cap D_{f_{y_1}}({\tilde{y}}_1)))\le C\delta ^{1/100}{\mathrm {Vol}}(M)$ and $\Vert G_f^{{\tilde{y}}_1}\Vert _\infty \le C$, we get $\Vert G_f^{{\tilde{y}}_1}\Vert _1\le CL.$ Thus, by the segment inequality, we get ${\mathrm {Vol}}(M\setminus {\overline{P}}^{{\tilde{y}}_1}_f)\le CL^{1/3}.$

Take arbitrary $x,z\in M$. By the Bishop-Gromov inequality, there exist ${\tilde{z}}\in D_{f_{z}}(p_{z})\cap Q_{f_{z}} \cap R_{f_{z}}\cap Q_f$, ${\tilde{x}}\in D_f(p_f)\cap Q_f \cap R_f\cap {\overline{P}}_f^{{\tilde{z}}}$ and ${\tilde{y}}\in D_f({\tilde{x}})\cap D_f(p_f)\cap Q_f\cap R_f\cap {\overline{C}}_f^{{\tilde{z}}}({\tilde{x}})$ such that $d(z,{\tilde{z}})\le C \delta ^{1/100n}$, $d(x,{\tilde{x}})\le CL^{1/3n}$ and $d(a_f(x),{\tilde{y}})\le CL^{1/3n}$. Here, we used the estimate ${\mathrm {Vol}}(M\setminus {\overline{P}}^{{\tilde{z}}}_f)\le CL^{1/3}$. Then, we get

$$\begin{aligned} \left| d({\tilde{z}},{\tilde{x}})^2-d_S(\Psi ({\tilde{z}}),\Psi ({\tilde{x}}))^2- d({\tilde{z}},{\tilde{y}})^2+d_S(\Psi ({\tilde{z}}),\Psi ({\tilde{y}}))^2 \right| \le CL^{1/78n} \end{aligned}$$

by Lemma 4.38. Thus, we get

$$\begin{aligned} \left| d(z,x)^2-d_S(\Psi (z),\Psi (x))^2- d(z,a_f(x))^2+d_S(\Psi (z),\Psi (a_f(x)))^2 \right| \le CL^{1/78n}\nonumber \\ \end{aligned}$$

(117)

by Lemma 4.25. Similarly, we have

$$\begin{aligned} \begin{aligned}&\left| d(a_f(x),z)^2-d_S(\Psi (a_f(x)),\Psi (z))^2- d(a_f(x),a_f(z))^2\right. \\&\qquad \left. + d_S(\Psi (a_f(x)),\Psi (a_f(z)))^2\right| \\&\quad \le CL^{1/78n}. \end{aligned} \end{aligned}$$

(118)

Since we have $d_S(\Psi (a_f(x)),\Psi (a_f(z)))\le C\delta ^{1/2000n^2}$ by Lemma 4.26, we get

$$\begin{aligned} \left| d(z,x)^2-d_S(\Psi (z),\Psi (x))^2- d(a_f(x),a_f(z))^2\right| \le CL^{1/78n}. \end{aligned}$$

by (117) and (118). This gives

$$\begin{aligned} \begin{aligned}&\left| d(z,x)-d(\Phi _f(z),\Phi _f(x))\right| \\&\quad =\left| d(z,x)-\left( d_S(\Psi (z),\Psi (x))^2+ d(a_f(x),a_f(z))^2\right) ^{1/2}\right| \le CL^{1/156n}. \end{aligned} \end{aligned}$$

Combining this and Proposition 4.43, we get the theorem. $\square$

By the above theorem, we get Main Theorem 2 except for the orientability, which is proved in Sect. 4.7.

4.6 Further Inequalities

In this subsection, we assume that Assumption 4.2 holds, and prepare two lemmas to prove the remaining part of main theorems.

Lemma 4.48

For any $f\in {\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots ,f_{k}\}$, we have

$$\begin{aligned} \left\| \sum _{i=1}^n e^i\otimes (\nabla _{e_i}d f+f e^i)\wedge \omega \right\| _2\le C\delta ^{1/8}\Vert f\Vert _2. \end{aligned}$$

Proof

We have

$$\begin{aligned} \begin{aligned}&\left| \sum _{i=1}^n e^i\otimes (\nabla _{e_i}d f+f e^i)\wedge \omega \right| ^2\\&\quad = |\nabla ^2 f|^2|\omega |^2-\frac{1}{n-p}(\Delta f)^2|\omega |^2 +2\Delta f \left( \frac{1}{n-p}\Delta f-f\right) |\omega |^2\\&\qquad -(n-p)\left( \left( \frac{\Delta f}{n-p}\right) ^2-f^2\right) |\omega |^2 -\left| \sum _{i=1}^n e^i \otimes \iota (\nabla _{e_i}\nabla f)\omega \right| ^2\\&\qquad -2\sum _{i=1}^n f \langle \omega , e^i\wedge \iota (\nabla _{e_i}\nabla f)\omega \rangle . \end{aligned} \end{aligned}$$

(119)

By the assumption, we have

$$\begin{aligned} \left\| \Delta f \left( \frac{1}{n-p}\Delta f-f\right) |\omega |^2\right\| _1&\le C\delta ^{1/2}\Vert f\Vert _2^2, \end{aligned}$$

(120)

$$\begin{aligned} \left\| \left( \left( \frac{\Delta f}{n-p}\right) ^2-f^2\right) |\omega |^2\right\| _1&\le C\delta ^{1/2}\Vert f\Vert _2^2. \end{aligned}$$

(121)

By Lemma 3.6 (iv) and Lemma 4.10 (ii), we have

$$\begin{aligned} \left\| \sum _{i=1}^n e^i \otimes \iota (\nabla _{e_i}\nabla f)\omega \right\| _2 \le \Vert \nabla (\iota (\nabla f)\omega )\Vert _2+ C\delta ^{1/2}\Vert f\Vert _2\le C\delta ^{1/4}\Vert f\Vert _2, \end{aligned}$$

(122)

and so

$$\begin{aligned} \left\| \sum _{i=1}^n f \langle \omega , e^i\wedge \iota (\nabla _{e_i}\nabla f)\omega \rangle \right\| _1\le C\Vert f\Vert _2\left\| \sum _{i=1}^n e^i \otimes \iota (\nabla _{e_i}\nabla f)\omega \right\| _2 \le C\delta ^{1/4}\Vert f\Vert _2^2.\nonumber \\ \end{aligned}$$

(123)

By Lemma 4.10, (119), (120), (121), (122) and (123), we get the lemma. $\square$

Lemma 4.49

Define $G=G(f_1,\ldots ,f_k)$ by

$$\begin{aligned} \begin{aligned}&G:=\Big \{x\in M: |f_i^2+|\nabla f_i|^2-1|(x)\le \delta ^{1/1600n}\text { for all }i=1,\ldots ,k,\text { and}\\&\quad \left| \frac{1}{2}(f_i\pm f_j)^2+\frac{1}{2}|\nabla f_i\pm \nabla f_j|^2-1\right| (x)\le \delta ^{1/1600n}\text { for all }i\ne j \Big \}. \end{aligned} \end{aligned}$$

Then, we have the following properties.

• (i) We have ${\mathrm {Vol}}(M\setminus G)\le C\delta ^{1/1600n}{\mathrm {Vol}}(M)$.

• (ii) For all $x\in G$ and i, j with $i\ne j$, we have $\left| f_i f_j+\langle \nabla f_i,\nabla f_j\rangle \right| (x)\le \delta ^{1/1600n}$.

Proof

By Proposition 4.17 (iii), we have

$$\begin{aligned} \begin{aligned} \Vert f_i^2+|\nabla f_i|^2-1\Vert _1&\le C\delta ^{1/800n},\\ \left\| \frac{1}{2}(f_i\pm f_j)^2+\frac{1}{2}|\nabla f_i\pm \nabla f_j|^2-1\right\| _1&leq C\delta ^{1/800n} \end{aligned} \end{aligned}$$

for all $i\ne j$. Therefore, we get

$$\begin{aligned}&{\mathrm {Vol}}\left( \left\{ x\in M: \left| f_i^2+|\nabla f_i|^2-1\right| (x)>\delta ^{1/1600n}\right\} \right) \\&\quad \le \delta ^{-1/1600n}\int _M \left| f_i^2+|\nabla f_i|^2-1\right| \,d\mu _g\le C\delta ^{1/1600n}{\mathrm {Vol}}(M) \end{aligned}$$

for all i. Similarly, we have

$$\begin{aligned}&{\mathrm {Vol}}\left( \left\{ x\in M: \left| \frac{1}{2}(f_i\pm f_j)^2+\frac{1}{2}|\nabla f_i\pm \nabla f_j|^2-1\right| (x)>\delta ^{1/1600n}\right\} \right) \\&\quad \le C\delta ^{1/1600n}{\mathrm {Vol}}(M) \end{aligned}$$

for all $i\ne j$. Thus, we get (i).

For all $x\in G$ and i, j with $i\ne j$, we have

$$\begin{aligned}&\left| f_i f_j+\langle \nabla f_i,\nabla f_j\rangle \right| (x)\\&\quad =\frac{1}{2}\left| \frac{1}{2}(f_i+f_j)^2+\frac{1}{2}|\nabla f_i+\nabla f_j|^2 -\frac{1}{2}(f_i-f_j)^2-\frac{1}{2}|\nabla f_i-\nabla f_j|^2\right| (x) \\&\quad \le \delta ^{1/1600n}. \end{aligned}$$

Thus, we get (ii). $\square$

4.7 Orientability

The goal of this subsection is to show the orientability of the manifold under the assumption of Main Theorem 2.

Theorem 4.50

If Assumption 4.1 for $k=n-p+1$ and Assumption 4.2 hold, then M is orientable.

Proof

To prove the theorem, we use the following claim:

Claim 4.51

Define

$$\begin{aligned} \lambda _1(\Delta _{C,n}):=\inf \left\{ \frac{\Vert \nabla \eta \Vert _2^2}{\Vert \eta \Vert _2^2}: \eta \in \Gamma (\bigwedge ^n T^*M)\text { with } \eta \ne 0\right\} . \end{aligned}$$

If $\lambda _1(\Delta _{C,n})< n(n-p-1)/(n-1)$ holds, then M is orientable.

Proof of Claim 4.51

Suppose that M is not orientable. Take the two-sheeted oriented Riemannian covering $\pi :({\widetilde{M}},{\tilde{g}})\rightarrow (M,g)$. Since we have ${\mathrm {Ric}}_{{\tilde{g}}}\ge (n-p-1){\tilde{g}}$, we get

$$\begin{aligned} \lambda _1(\Delta _{C,n},g)\ge \lambda _2(\Delta _{C,n},{\tilde{g}})=\lambda _1({\tilde{g}})\ge \frac{n}{n-1}(n-p-1) \end{aligned}$$

by the Lichnerowicz estimate (note that $\lambda _1(\Delta _{C,n},{\tilde{g}})=\lambda _0({\tilde{g}})=0$). This gives the claim. $\square$

Put

$$\begin{aligned} V:=\sum _{i=1}^{n-p+1} (-1)^{i-1} f_i d f_1\wedge \cdots \wedge \widehat{d f_i}\wedge \cdots \wedge d f_{n-p+1}\wedge \omega \in \Gamma (\bigwedge ^n T^*M). \end{aligned}$$

In the following, we show that $\Vert \nabla V\Vert _2^2/\Vert V\Vert _2^2< n(n-p+1)/(n-1)$.

Define a vector bundle $E:=T^*M\oplus {\mathbb {R}}e$, where ${\mathbb {R}}e$ denotes the trivial bundle of rank 1 with a nowhere vanishing section e. We consider an inner product $\langle \cdot ,\cdot \rangle$ on E defined by $\langle \alpha + f e,\beta +h e\rangle :=\langle \alpha ,\beta \rangle + fh$ for all $\alpha ,\beta \in \Gamma (T^*M)$ and $f,h\in C^\infty (M)$. Put

$$\begin{aligned} S_i:=d f_i +f_i e\in \Gamma (E) \end{aligned}$$

for each i, and

$$\begin{aligned} \alpha :=S_1\wedge \cdots \wedge S_{n-p+1}\in \Gamma (\bigwedge ^{n-p+1} E). \end{aligned}$$

Then, we have $\alpha \wedge \omega =e\wedge V$, and so

$$\begin{aligned} |\alpha \wedge \omega |=|V|. \end{aligned}$$

(124)

For each $k=1,\ldots ,n-p+1$, we have

$$\begin{aligned} \begin{aligned}&\Big \Vert \big \langle S_k\wedge \cdots \wedge S_{n-p+1}\wedge \omega , \left( \iota (S_{k-1})\cdots \iota (S_1)\alpha \right) \wedge \omega \big \rangle \\&\qquad -\big \langle S_{k+1} \wedge \cdots \wedge S_{n-p+1}\wedge \omega , \left( \iota (S_k)\cdots \iota (S_1)\alpha \right) \wedge \omega \big \rangle \Big \Vert _1\\&\quad =\left\| \big \langle S_{k+1} \wedge \cdots \wedge S_{n-p+1}\wedge \omega , \left( \iota (S_{k-1})\cdots \iota (S_1)\alpha \right) \wedge \iota (d f_k)\omega \big \rangle \right\| _1\\&\quad \le C\Vert \iota (d f_k)\omega \Vert _2\le C\delta ^{1/4} \end{aligned} \end{aligned}$$

by Lemma 4.10 (i). By induction, we get

$$\begin{aligned} \Vert |\alpha \wedge \omega |^2-|\alpha |^2|\omega |^2\Vert _1\le C\delta ^{1/4}. \end{aligned}$$

(125)

In particular, we have

$$\begin{aligned} \left| \Vert \alpha \wedge \omega \Vert _2^2-\Vert |\alpha |^2|\omega |^2\Vert _1\right| \le C\delta ^{1/4}. \end{aligned}$$

(126)

Since we have $\left| \langle S_i(x), S_j(x)\rangle -\delta _{i j}\right| \le \delta ^{1/1600n}$ for all $x\in G=G(f_1,\ldots ,f_{n-p+1})$ and i, j by Lemma 4.49 (ii), we get $||\alpha |^2(x)-1|\le C\delta ^{1/1600n}$ for all $x\in G$. Thus, we get

$$\begin{aligned}&\left| \frac{1}{{\mathrm {Vol}}(M)}\int _M(|\alpha |^2|\omega |^2-1) \,d\mu _g \right| \nonumber \\&\quad =\Bigg | \frac{1}{{\mathrm {Vol}}(M)}\int _G(|\alpha |^2-1)|\omega |^2 \,d\mu _g\nonumber \\&\qquad +\frac{1}{{\mathrm {Vol}}(M)}\int _{M\setminus G}(|\alpha |^2-1)|\omega |^2 \,d\mu _g+\frac{1}{{\mathrm {Vol}}(M)}\int _M(|\omega |^2-1) \,d\mu _g \Bigg |\nonumber \\&\quad \le C\delta ^{1/1600n} \end{aligned}$$

(127)

by Lemmas 3.5 and 4.49 (i). By (124), (126) and (127), we get

$$\begin{aligned} |\Vert V\Vert _2^2-1|\le C\delta ^{1/1600n}. \end{aligned}$$

(128)

We next estimate $\Vert \nabla V\Vert _2^2$. We have

$$\begin{aligned} \begin{aligned}&\nabla V = \sum _{i=1}^{n-p+1} (-1)^{i-1} d f_i\otimes d f_1\wedge \cdots \wedge \widehat{d f_i}\wedge \cdots \wedge d f_{n-p+1}\wedge \omega \\&\qquad +\sum _{j<i}\sum _{k=1}^n (-1)^{i-1}(-1)^{j-1} f_i e^k\otimes (\nabla _{e_k} d f_j)\wedge d f_1\wedge \cdots \wedge \widehat{d f_j} \wedge \cdots \wedge \widehat{d f_i}\wedge \cdots \wedge d f_{n-p+1}\wedge \omega \\&\qquad +\sum _{i<j}\sum _{k=1}^n (-1)^{i-1}(-1)^{j} f_i e^k\otimes (\nabla _{e_k} d f_j)\wedge d f_1\wedge \cdots \wedge \widehat{d f_i} \wedge \cdots \wedge \widehat{d f_j}\wedge \cdots \wedge d f_{n-p+1}\wedge \omega \\&\qquad +\sum _{i=1}^{n-p+1} \sum _{k=1}^n (-1)^{i-1} f_i e^k \otimes d f_1\wedge \cdots \wedge \widehat{d f_i}\wedge \cdots \wedge d f_{n-p+1}\wedge \nabla _{e_k}\omega . \end{aligned} \end{aligned}$$

Thus, we get

$$\begin{aligned}&\left\| \nabla V - \sum _{i=1}^{n-p+1} (-1)^{i-1} d f_i\otimes d f_1\wedge \cdots \wedge \widehat{d f_i}\wedge \cdots \wedge d f_{n-p+1}\wedge \omega \right\| _2\nonumber \\&\quad \le \Bigg \Vert \sum _{j<i}\sum _{k=1}^n (-1)^{i-1}(-1)^{j-1} f_i f_j e^k\otimes e^k\wedge d f_1\wedge \cdots \wedge \widehat{d f_j} \wedge \cdots \wedge \widehat{d f_i}\wedge \cdots \wedge d f_{n-p+1}\wedge \omega \nonumber \\&\qquad +\sum _{i<j}\sum _{k=1}^n (-1)^{i-1}(-1)^{j} f_i f_j e^k\otimes e^k\wedge d f_1\wedge \cdots \wedge \widehat{d f_i} \wedge \cdots \wedge \widehat{d f_j}\wedge \cdots \wedge d f_{n-p+1}\wedge \omega \Bigg \Vert _2\nonumber \\&\qquad +C\sum _{i=1}^{n-p+1}\left\| \sum _{k=1}^n e^k\otimes (\nabla _{e_k}d f_i+f_i e^k)\wedge \omega \right\| _2 + C\Vert \nabla \omega \Vert _2\nonumber \\&\quad \le C\delta ^{1/8} \end{aligned}$$

(129)

by Lemma 4.48.

Similarly to (125), we have

$$\begin{aligned} \begin{aligned}&\Bigg \Vert \left| \sum _{i=1}^{n-p+1} (-1)^{i-1} d f_i\otimes d f_1\wedge \cdots \wedge \widehat{d f_i}\wedge \cdots \wedge d f_{n-p+1}\wedge \omega \right| ^2\\&\qquad -\left| \sum _{i=1}^{n-p+1} (-1)^{i-1} d f_i\otimes d f_1\wedge \cdots \wedge \widehat{d f_i}\wedge \cdots \wedge d f_{n-p+1}\right| ^2|\omega |^2\Bigg \Vert _1\\&\quad \le C\delta ^{1/4}. \end{aligned} \end{aligned}$$

(130)

Since we have $d f_1\wedge \cdots \wedge d f_{n-p+1}\wedge \omega =0,$ we get

$$\begin{aligned} \begin{aligned}&\Vert |d f_1\wedge \cdots \wedge d f_{n-p+1}|^2|\omega |^2 \Vert _1\\&\quad = \Vert |d f_1\wedge \cdots \wedge d f_{n-p+1}|^2|\omega |^2- |d f_1\wedge \cdots \wedge d f_{n-p+1}\wedge \omega |^2 \Vert _1 \le C\delta ^{1/4} \end{aligned}\nonumber \\ \end{aligned}$$

(131)

similarly to (125). By (6), we get

$$\begin{aligned} \begin{aligned}&\left| \sum _{i=1}^{n-p+1}(-1)^{i-1}d f_i\otimes d f_1\wedge \cdots \wedge \widehat{d f_i}\wedge \cdots \wedge d f_{n-p+1} \right| ^2\\&\quad =(n-p+1) |d f_1\wedge \cdots \wedge d f_{n-p+1}|^2. \end{aligned} \end{aligned}$$

(132)

By (131) and (132), we get

$$\begin{aligned} \left\| \left| \sum _{i=1}^{n-p+1}(-1)^{i-1}d f_i\otimes d f_1\wedge \cdots \wedge \widehat{d f_i}\wedge \cdots \wedge d f_{n-p+1}\right| ^2|\omega |^2 \right\| _1\le C\delta ^{1/4}. \end{aligned}$$

(133)

By (130) and (133), we have

$$\begin{aligned} \left\| \sum _{i=1}^{n-p+1}(-1)^{i-1}d f_i\otimes d f_1\wedge \cdots \wedge \widehat{d f_i}\wedge \cdots \wedge d f_{n-p+1}\wedge \omega \right\| _2^2 \le C\delta ^{1/4}. \end{aligned}$$

(134)

By (129) and (134), we get

$$\begin{aligned} \Vert \nabla V\Vert _2\le C\delta ^{1/8}. \end{aligned}$$

(135)

By (128) and (135), we get $\lambda _1(\Delta _{C,n})\le C\delta ^{1/4},$ and so we get the theorem by Claim 4.51. $\square$

Combining Theorems 4.47 and 4.50, we get Main Theorem 2.

4.8 Almost Parallel $(n-p)$-form II

In this subsection, we show that the assumption “$\lambda _{n-p}(g)$ is close to $n-p$” implies the condition “$\lambda _{n-p+1}(g)$ is close to $n-p$” under the assumption $\lambda _1(\Delta _{C,n-p})\le \delta$.

Lemma 4.52

Suppose that Assumption 4.1 for $k=n-p$ and Assumption 4.3 hold. Put $F:= \langle d f_1\wedge \ldots \wedge d f_{n-p}, \xi \rangle \in C^\infty (M).$ Then, we have

$$\begin{aligned} \left| \Vert F\Vert _2^2-\frac{1}{n-p+1}\right| \le C\delta ^{1/1600n},\quad \left| \Vert \nabla F\Vert _2^2-\frac{n-p}{n-p+1}\right| \le C\delta ^{1/1600n} \end{aligned}$$

and

$$\begin{aligned} \left| \frac{1}{{\mathrm {Vol}}(M)}\int _M f_i F\,d\mu _g\right| \le C\delta ^{1/2} \end{aligned}$$

for all $i=1,\ldots , n-p$.

Proof

If M is not orientable, we take the two-sheeted oriented Riemannian covering $\pi :({\widetilde{M}},{\tilde{g}})\rightarrow (M,g)$, and put ${\widetilde{F}}:=F\circ \pi$ and ${\tilde{f}}_i:=f_i\circ \pi .$ Then, we have $\Vert F\Vert _2=\Vert {\widetilde{F}}\Vert _2$, $\Vert \nabla F\Vert _2=\Vert \nabla {\widetilde{F}}\Vert _2,$

$$\begin{aligned} \frac{1}{{\mathrm {Vol}}({\widetilde{M}})}\int _{{\widetilde{M}}} {\tilde{f}}_i {\widetilde{F}} \,d\mu _{{\tilde{g}}}= \frac{1}{{\mathrm {Vol}}(M)}\int _M f_i F \,d\mu _g \end{aligned}$$

and ${\widetilde{F}}=\langle d {\tilde{f}}_1\wedge \ldots \wedge d {\tilde{f}}_{n-p}, \pi ^*\xi \rangle .$ Thus, it is enough to consider the case when M is orientable. In the following, we assume that M is orientable, and we fix an orientation of M.

Put $\omega :=*\xi \in \Gamma (\bigwedge ^p T^*M).$ Let $V_g\in \Gamma (\bigwedge ^n T^*M)$ be the volume form of (M, g). Then, we have

$$\begin{aligned} F V_g= d f_1\wedge \cdots \wedge d f_{n-p}\wedge \omega . \end{aligned}$$

(136)

Define a vector bundle $E:=T^*M\oplus {\mathbb {R}}e$ and an inner product $\langle ,\rangle$ on it as in the proof of Theorem 4.50. Put

$$\begin{aligned} S_i:=d f_i +f_i e\in \Gamma (E) \end{aligned}$$

for each i, and

$$\begin{aligned} \beta :=S_1\wedge \cdots \wedge S_{n-p}\in \Gamma (\bigwedge ^{n-p} E). \end{aligned}$$

Since we have $|F|=|F V_g|$, we get $\Vert |F|^2-|d f_1\wedge \cdots \wedge d f_{n-p} |^2|\omega |^2\Vert _1\le C\delta ^{1/4}$ similarly to (125) by (136), and so

$$\begin{aligned} \left| \Vert F\Vert _2^2-\left\| |d f_1\wedge \cdots \wedge d f_{n-p} |^2|\omega |^2\right\| _1\right| \le C\delta ^{1/4} \end{aligned}$$

(137)

By Lemma 4.48 and (136), we have

$$\begin{aligned}&\left\| \nabla (F V_g)+\sum _{i=1}^{n-p} \sum _{k=1}^n(-1)^{i-1} f_i e^k\otimes e^k\wedge d f_1\wedge \cdots \wedge \widehat{d f_i}\wedge \cdots \wedge d f_{n-p}\wedge \omega \right\| _2\\&\quad \le C\delta ^{1/8}. \end{aligned}$$

Since $|\nabla (F V_g)|=|\nabla F|$, we get

$$\begin{aligned}&\left| \Vert \nabla F\Vert _2^2-\left\| \left| \sum _{i=1}^{n-p} \sum _{k=1}^n(-1)^{i-1} f_i e^k\otimes e^k\wedge d f_1\wedge \cdots \wedge \widehat{d f_i}\wedge \cdots \wedge d f_{n-p}\wedge \omega \right| ^2\right\| _1\right| \nonumber \\&\quad \le C\delta ^{1/8}. \end{aligned}$$

(138)

We have

$$\begin{aligned} \begin{aligned}&\left| \sum _{i=1}^{n-p} \sum _{k=1}^n(-1)^{i-1} f_i e^k\otimes e^k\wedge d f_1\wedge \cdots \wedge \widehat{d f_i}\wedge \cdots \wedge d f_{n-p}\wedge \omega \right| ^2\\&\quad =\left| \sum _{i=1}^{n-p} (-1)^{i-1} f_i d f_1\wedge \cdots \wedge \widehat{d f_i}\wedge \cdots \wedge d f_{n-p}\wedge \omega \right| ^2. \end{aligned} \end{aligned}$$

(139)

Similarly to (125), we have

$$\begin{aligned} \begin{aligned}&\Bigg \Vert \left| \sum _{i=1}^{n-p} (-1)^{i-1} f_i d f_1\wedge \cdots \wedge \widehat{d f_i}\wedge \cdots \wedge d f_{n-p}\wedge \omega \right| ^2\\&\quad -\left| \sum _{i=1}^{n-p} (-1)^{i-1} f_i d f_1\wedge \cdots \wedge \widehat{d f_i}\wedge \cdots \wedge d f_{n-p}\right| ^2|\omega |^2\Bigg \Vert _1\le C\delta ^{1/4}. \end{aligned} \end{aligned}$$

Since we have

$$\begin{aligned} \iota (e)\beta =\sum _{i=1}^{n-p} (-1)^{i-1} f_i d f_1\wedge \cdots \wedge \widehat{d f_i}\wedge \cdots \wedge d f_{n-p}, \end{aligned}$$

we get

$$\begin{aligned} \left\| \left| \sum _{i=1}^{n-p} (-1)^{i-1} f_i d f_1\wedge \cdots \wedge \widehat{d f_i}\wedge \cdots \wedge d f_{n-p}\wedge \omega \right| ^2 -|\iota (e)\beta |^2|\omega |^2\right\| _1\le C\delta ^{1/4}.\nonumber \\ \end{aligned}$$

(140)

By (138), (139) and (140), we get

$$\begin{aligned} \left| \Vert \nabla F\Vert _2^2-\left\| |\iota (e)\beta |^2|\omega |^2\right\| _1\right| \le C\delta ^{1/8}. \end{aligned}$$

(141)

We have

$$\begin{aligned} |\beta |^2=|d f_1\wedge \cdots \wedge d f_{n-p}|^2+|\iota (e)\beta |^2. \end{aligned}$$

(142)

We calculate $\sum _{k=1}^n\left| e^k\wedge \beta \right| ^2$ in two ways. We have

$$\begin{aligned} \begin{aligned} \sum _{k=1}^n|e^k\wedge \beta |^2&=(p+1)|\beta |^2-|e\wedge \beta |^2\\&=(p+1)|\beta |^2-|d f_1\wedge \cdots \wedge d f_{n-p}|^2= p|\beta |^2+|\iota (e)\beta |^2 \end{aligned} \end{aligned}$$

(143)

by (142). For all $\eta \in \Gamma (T^*M)$, we have

$$\begin{aligned}&|\eta \wedge \beta |^2\\&\quad =|\eta |^2|\beta |^2-\langle \iota (\eta )\beta ,\iota (\eta )\beta \rangle \\&\quad =|\eta |^2|\beta |^2-\sum _{i,j=1}^{n-p}(-1)^{i+j}\langle \eta , d f_i\rangle \langle \eta , d f_j\rangle \\&\qquad \langle S_1\wedge \cdots \wedge \widehat{S_i}\wedge \cdots \wedge S_{n-p},S_1\wedge \cdots \wedge \widehat{S_j}\wedge \cdots \wedge S_{n-p} \rangle , \end{aligned}$$

and so we get

$$\begin{aligned} \begin{aligned}&\sum _{k=1}^n|e^k\wedge \beta |^2\\&\quad =n|\beta |^2-\sum _{i,j=1}^{n-p}(-1)^{i+j}\langle d f_i,d f_j\rangle \\&\qquad \langle S_1\wedge \cdots \wedge \widehat{S_i}\wedge \cdots \wedge S_{n-p},S_1\wedge \cdots \wedge \widehat{S_j}\wedge \cdots \wedge S_{n-p}\rangle . \end{aligned} \end{aligned}$$

(144)

By (143) and (144), we get

$$\begin{aligned} \begin{aligned}&|\iota (e)\beta |^2\\&\quad =(n-p)|\beta |^2-\sum _{i,j=1}^{n-p}(-1)^{i+j}\langle d f_i,d f_j\rangle \\&\qquad \langle S_1\wedge \cdots \wedge \widehat{S_i}\wedge \cdots \wedge S_{n-p},S_1\wedge \cdots \wedge \widehat{S_j}\wedge \cdots \wedge S_{n-p}\rangle \end{aligned} \end{aligned}$$

(145)

Since we have $|\langle S_i,S_j\rangle (x)-\delta _{i j}|\le C\delta ^{1/1600n}$ for all $x\in G=G(f_1,\ldots , f_{n-p})$ by Lemma 4.49 (ii), we have

$$\begin{aligned} \begin{aligned}&\Bigg \Vert \sum _{i=1}^{n-p}|d f_i|^2\\&\quad -\sum _{i,j=1}^{n-p}(-1)^{i+j}\langle d f_i,d f_j\rangle \\&\qquad \langle S_1\wedge \cdots \wedge \widehat{S_i}\wedge \cdots \wedge S_{n-p},S_1\wedge \cdots \wedge \widehat{S_j}\wedge \cdots \wedge S_{n-p}\rangle |\omega |^2 \Bigg \Vert _1 \le C\delta ^{1/1600n} \end{aligned} \end{aligned}$$

(146)

and

$$\begin{aligned} \left| \left\| |\beta |^2|\omega |^2\right\| _1-1\right| \le C\delta ^{1/1600n} \end{aligned}$$

(147)

by Lemmas 3.5 and 4.49 (i). By the assumption, we have

$$\begin{aligned} \left| \sum _{i=1}^{n-p}\Vert d f_i\Vert _2^2-\frac{(n-p)^2}{n-p+1}\right| \le C\delta ^{1/2}. \end{aligned}$$

(148)

By (145), (146), (147) and (148), we get

$$\begin{aligned} \left| \left\| |\iota (e)\beta |^2|\omega |^2\right\| _1-\frac{n-p}{n-p+1}\right| \le C\delta ^{1/1600n}, \end{aligned}$$

(149)

and so

$$\begin{aligned} \left| \left\| |d f_1\wedge \cdots \wedge d f_{n-p}|^2|\omega |^2\right\| _1-\frac{1}{n-p+1}\right| \le C\delta ^{1/1600n} \end{aligned}$$

(150)

by (142) and (147). By (137) and (150), we get

$$\begin{aligned} \left| \Vert F\Vert _2^2-\frac{1}{n-p+1}\right| \le C\delta ^{1/1600n}. \end{aligned}$$

By (141) and (149), we get

$$\begin{aligned} \left| \Vert \nabla F\Vert _2^2- \frac{n-p}{n-p+1}\right| \le C\delta ^{1/1600n}. \end{aligned}$$

Let us show the remaining assertion. Since we have

$$\begin{aligned} f_i F V_g&=\frac{1}{2}(-1)^{i-1} d \left( f_i^2 d f_1\wedge \cdots \wedge \widehat{d f_i}\wedge \cdots \wedge d f_{n-p}\wedge \omega \right) \\&\quad -\frac{1}{2}(-1)^{i-1} (-1)^{n-p-1}f_i^2 d f_1\wedge \cdots \wedge \widehat{d f_i}\wedge \cdots \wedge d f_{n-p}\wedge d \omega , \end{aligned}$$

we get

$$\begin{aligned} \left| \frac{1}{{\mathrm {Vol}}(M)}\int _M f_i F\,d\mu _g\right| \le C\Vert \nabla \omega \Vert _2 \le C\delta ^{1/2} \end{aligned}$$

by the Stokes theorem. $\square$

By applying the min-max principle

$$\begin{aligned}&\lambda _{n-p+1}(g)\\&\quad =\inf \left\{ \sup _{f\in V\setminus \{0\}}\frac{\Vert \nabla f\Vert _2^2}{\Vert f\Vert _2^2}: V\text { is an }(n-p+1)\text {-dimensional subspace of } C^\infty (M) \right\} \end{aligned}$$

to the subspace ${\mathrm {Span}}_{{\mathbb {R}}}\{f_1,\ldots , f_{n-p}, F\}$, we immediately get the following corollary:

Corollary 4.53

If Assumption 4.1 for $k=n-p$ and Assumption 4.3 hold, then we have $\lambda _{n-p+1}(g)\le n-p+C\delta ^{1/1600n}.$

Combining Theorem 4.47 and Corollary 4.53, we get Main Theorem 4.

Finally, we investigate the Gromov-Hausdorff limit of the sequence of the Riemannian manifolds that satisfy our pinching condition.

Theorem 4.54

Take $n\ge 5$ and $2\le p < n/2$. Let $\{(M_i,g_i)\}_{i\in {\mathbb {N}}}$ be a sequence of n-dimensional closed Riemannian manifolds with ${\mathrm {Ric}}_{g_i}\ge (n-p-1)g_i$ that satisfies one of the following:

• (i) $\lim _{i\rightarrow \infty }\lambda _{n-p+1}(g_i)=n-p$ and $\lim _{i\rightarrow \infty }\lambda _1(\Delta _{C,p},g_i)=0$,

• (ii) $\lim _{i\rightarrow \infty }\lambda _{n-p}(g_i)=n-p$ and $\lim _{i\rightarrow \infty }\lambda _1(\Delta _{C,n-p},g_i)=0$.

If $\{(M_i,g_i)\}_{i\in {\mathbb {N}}}$ converges to a geodesic space X, then there exists a geodesic space Y such that X is isometric to $S^{n-p}\times Y$.

Proof

By Main Theorems 2 and 4, we get that there exist a sequence of positive real numbers $\{\epsilon _i\}$ and compact metric spaces $\{Y_i\}$ such that $\lim _{i\rightarrow \infty }\epsilon _i=0$ and $d_{GH}(M_i,S^{n-p}\times Y_i)\le \epsilon _i$. Then, $\{S^{n-p}\times Y_i\}$ converges to X in the Gromov-Hausdorff topology, and so $\{Y_i\}$ is pre-compact in the Gromov-Hausdorff topology by [20, Theorem 11.1.10]. Thus, there exists a subsequence that converges to some compact metric space Y. Therefore, we get that X is isometric to $S^{n-p}\times Y$. Since X is a geodesic space, Y is also a geodesic space. $\square$

Appendices

Appendix A. Limit Spaces and Unorientability

In this appendix, we show the stability of unorientability under the non-collapsing Gromov-Hausdorff convergence assuming the two-sided bound on the Ricci curvature. Similarly to Claim 4.51, we have the following.

Lemma A.1

For any n-dimensional unorientable closed Riemannian manifold (M, g) with ${\mathrm {Ric}}\ge -K g$ and ${\mathrm {diam}}(M)\le D$ $(K,D>0)$ we have $\lambda _1(\Delta _{C,n},g)\ge C_1(n,K,2D),$ where $C_1(n,K,D)$ is defined by

$$\begin{aligned} C_1(n,K,D):=\frac{1}{(n-1)D^2\exp \left( 1+\sqrt{1+4(n-1)KD^2}\right) }. \end{aligned}$$

Note that we have $\lambda _1(g_1)\ge C_1(n,K,D)$ for any n-dimensional closed Riemannian manifold $(N_1,g_1)$ with ${\mathrm {Ric}}_{g_1}\ge -K g_1$ and ${\mathrm {diam}}(N_1)\le D$ by the Li-Yau estimate [22, p.116].

We immediately get the following corollary.

Corollary A.2

Let (M, g) be an n-dimensional closed Riemannian manifold with ${\mathrm {Ric}}\ge -K g$ and ${\mathrm {diam}}(M)\le D$ $(K,D>0)$. If $\lambda _1(\Delta _{C,n},g)< C_1(n,K,2D),$ then M is orientable.

The following theorem is the main result of this section.

Theorem A.3

Take real numbers $K_1,K_2\in {\mathbb {R}}$ and positive real numbers $D>0$ and $v>0$. Let $\{(M_i,g_i)\}$ be a sequence of n-dimensional unorientable closed Riemannian manifolds with $K_1 g_i\le {\mathrm {Ric}}_{g_i}\le K_2 g_i$, ${\mathrm {diam}}(M)\le D$ and ${\mathrm {Vol}}(M)\ge v$. Suppose that $\{(M_i,g_i)\}$ converges to a limit space X in the Gromov-Hausdorff sense. Then, X is not orientable in the sense of Honda [16] (see also the definition below).

Note that Honda [16, Theorem 1.3] showed the stability of orientability without assuming the upper bound on the Ricci curvature.

Before proving Theorem A.3, we fix our notation and recall definitions about limit spaces.

Notation A.4

Take real numbers $K_1,K_2\in {\mathbb {R}}$ and positive real numbers $D>0$ and $v>0$. Let ${\mathcal {M}}={\mathcal {M}}(n,K_1,K_2,D,v)$ be the set of isometry classes of n-dimensional closed Riemannian manifolds (M, g) with $K_1g\le {\mathrm {Ric}}_g \le K_2 g$, ${\mathrm {diam}}(M)\le D$ and ${\mathrm {Vol}}(M)\ge v$. Let $\overline{{\mathcal {M}}}=\overline{{\mathcal {M}}}(n,K_1,K_2,D,v)$ be the closure of ${\mathcal {M}}$ in the Gromov-Hausdorff topology.

If $X_i\in \overline{{\mathcal {M}}}$ ($i\in {\mathbb {N}}$) converges to $X\in \overline{{\mathcal {M}}}$ in the Gromov-Hausdorff topology, then there exist a sequence of positive real numbers $\{\epsilon _i\}_{i\in {\mathbb {N}}}$ with $\lim _{i\rightarrow \infty }\epsilon _i=0$, and a sequence of $\epsilon _i$-Hausdorff approximation maps $\phi _i :X_i\rightarrow X$. Fix such a sequence. We say a sequence $x_i\in X_i$ converges to $x\in X$ if $\lim _{i\rightarrow \infty }\phi _i(x_i)=x$ (denote it by $x_i{\mathop {\rightarrow }\limits ^{GH}} x$). By the volume convergence theorem [8, Theorem 5.9], $(X_i,H^n)$ converges to $(X,H^n)$ in the measured Gromov-Hausdorff sense, i.e., for all $r>0$ and all sequence $x_i\in X_i$ that converges to $x\in X$, we have $\lim _{i\rightarrow \infty }H^n(B_r (x_i))=H^n(B_r(x))$, where $H^n$ denotes the n-dimensional Hausdorff measure.

For all $X\in \overline{{\mathcal {M}}}$, we can consider the cotangent bundle $\pi :T^*X \rightarrow X$ with a canonical inner product by [5] and [9] (see also [15, Section 2] for a short review). We have $H^n(X\setminus \pi (T^*X))=0$ and $T^*_x X:=\pi ^{-1}(x)$ is an n-dimensional vector space for all $x\in \pi (T^*X)$. For all Lipschitz function f on X, we can define $d f(x)\in T_x^*X$ for almost all $x\in X$, and we have $d f\in L^\infty (T^*X)$.

Let us recall definitions of functional spaces on limit spaces briefly. Note that we can define such functional spaces on more general spaces than our assumption. Some of the following functional spaces are first introduced by Gigli [11].

Definition A.5

Let $X\in \overline{{\mathcal {M}}}$.

• (i) Let ${\mathrm {LIP}}(X)$ be the set of the Lipschitz functions on X. For all $f\in {\mathrm {LIP}}(X)$, we define $\Vert f\Vert _{H^{1,2}}^2=\Vert f\Vert _2^2+\Vert d f\Vert _2^2$. Let $H^{1,2}(X)$ be the completion of ${\mathrm {LIP}}(X)$ with respect to this norm.

• (ii) Define

  $$\begin{aligned} \begin{aligned}&{\mathcal {D}}^2(\Delta ,X):=\Big \{f\in H^{1,2}(X) : \text {there exists }F\in L^2(X)\text { such that}\\&\quad \int _X \langle df, dh \rangle \,d H^n=\int _X F h\,d H^n \text { for all }h\in H^{1,2}(X) \Big \}. \end{aligned} \end{aligned}$$

  For any $f\in {\mathcal {D}}^2(\Delta ,X)$, the function $F\in L^2(X)$ is uniquely determined. Thus, we define $\Delta f:=F$.

• (iii) Define

  $$\begin{aligned} \begin{aligned} {\mathrm {Test}}F(X)&:=\left\{ f\in {\mathcal {D}}^2(\Delta ,X)\cap {\mathrm {LIP}}(X):\Delta f\in H^{1,2}(X)\right\} ,\\ {\mathrm {TestForm}}_p(X)&:=\left\{ \sum _{i=1}^N f_{0,i} d f_{1,i}\wedge \ldots \wedge d f_{p,i}: N\in {\mathbb {N}},\, f_{j,i}\in {\mathrm {Test}}F(X)\right\} \end{aligned} \end{aligned}$$

  for all $p\in \{1,\ldots ,n\}$.

• (vi) The operator $\nabla :{\mathrm {TestForm}}_p(X)\rightarrow L^2(T^*X \otimes \bigwedge ^p T^*X)$ is defined by

  $$\begin{aligned}&\nabla \sum _{i=1}^N f_{0,i} d f_{1,i}\wedge \ldots \wedge d f_{p,i}\\&\quad :=\sum _{i=1}^N \left( d f_{0,i}\otimes d f_{1,i}\wedge \ldots \wedge d f_{p,i}\right. \\&\qquad \left. + \sum _{j=1}^p f_{0,i} d f_{1,i}\wedge \ldots \wedge \nabla ^2 f_{j,i}\wedge \ldots \wedge d f_{p,i}\right) , \end{aligned}$$

  where $\nabla ^2$ denotes the Hessian ${\mathrm {Hess}}$ defined in [11, Definition 3.3.1] or [14].

• (v) For any $\omega \in {\mathrm {TestForm}}_p(X)$, we define $\Vert \omega \Vert _{H_C^{1,2}}^2:=\Vert \omega \Vert _2^2+\Vert \nabla \omega \Vert _2^2$. Let $H^{1,2}_C(\bigwedge ^p T^*X)$ be the completion of ${\mathrm {TestForm}}_p (X)$ with respect to this norm.

• (vi) Define

  $$\begin{aligned} \begin{aligned}&{\mathcal {D}}^2(\Delta _{C,p},X):=\Big \{\omega \in H^{1,2}_C(\bigwedge ^p T^*X) : \text {there exists }{\hat{\omega }}\in L^2(\bigwedge ^p T^*X)\text { such that}\\&\quad \int _X \langle \nabla \omega , \nabla \eta \rangle \,d H^n=\int _X \langle {\hat{\omega }}, \eta \rangle \,d H^n \text { for all }\eta \in H_C^{1,2}(\bigwedge ^p T^*X)\Big \}. \end{aligned} \end{aligned}$$

  For any $\omega \in {\mathcal {D}}^2(\Delta _{C,p},X)$, the form ${\hat{\omega }}\in L^2(\bigwedge ^p T^*X)$ is uniquely determined. Thus, we put $\Delta _{C,p} \omega :={\hat{\omega }}$.

• (viii) For all $k\in {\mathbb {Z}}_{>0}$, we define

  $$\begin{aligned} \begin{aligned} \lambda _k(\Delta _{C,p},X):=\inf \left\{ \sup _{\omega \in {\mathcal {E}}_k\setminus \{0\}}\frac{\Vert \nabla \omega \Vert ^2_2}{\Vert \omega \Vert ^2_2}: {\mathcal {E}}_k\subset H^{1,2}_C(\bigwedge ^p T^*X)\text { is a }k\text {-dimensional subspace}\right\} . \end{aligned} \end{aligned}$$

Similarly to the smooth case, there exists a complete orthonormal system of eigenforms of the connection Laplacian $\Delta _{C,p}$ in $L^2(\bigwedge ^p T^*M)$, and each eigenform is an element of ${\mathcal {D}}^2(\Delta _{C,p},X)$ (see [17, Theorem 4.17]).

Honda [17] showed the following theorem:

Theorem A.6

([17]) Let $\{X_i\}_{i\in {\mathbb {N}}}$ be a sequence in $\overline{{\mathcal {M}}}$ and let $X\in \overline{{\mathcal {M}}}$ be its Gromov-Hausdorff limit. Then, we have $\lim _{i\rightarrow \infty }\lambda _k(\Delta _{C,p},X_i)=\lambda _k(\Delta _{C,p},X)$ for all $p\in \{0,\ldots ,n\}$ and $k\in {\mathbb {Z}}_{>0}$.

Definition A.7

(Orientation [16]) Let $X\in \overline{{\mathcal {M}}}$. We say that X is orientable if there exists $\omega \in L^\infty (\bigwedge ^n T^*X)$ such that $|\omega |(z)=1$ for almost all $z\in X$ and that $\langle \omega ,\eta \rangle \in H^{1,2}(X)$ for any $\eta \in {\mathrm {TestForm}}_n(X)$. We call $\omega$ an orientation of X.

Lemma A.8

Let $X\in \overline{{\mathcal {M}}}$. Then, X is orientable if and only if $\lambda _1(\Delta _{C,n},X)=0$.

Proof

We first suppose that X is orientable and show $\lambda _1(\Delta _{C,n},X)=0$. Let $\omega \in L^\infty (\bigwedge ^n T^*X)$ be the orientation of X. By [16, Proposition 6.5], for almost all $z\in X$, $\omega$ is differentiable at z and $\nabla ^{g_X}\omega (z)=0$, where $\nabla ^{g_X}$ denotes the Levi-Civita connection defined in [14]. By Proposition 4.5 and Remark 4.7 in [17], we have $\omega \in H^{1,2}_C(\bigwedge ^p T^*X)$. By [18, Corollary 7.10], we have $\nabla \omega (z)=\nabla ^{g_X}\omega (z)=0$ for almost all $z\in X$. Thus, we get $\lambda _1(\Delta _{C,n},X)=0$ by the definition of $\lambda _1(\Delta _{C,n},X)$.

We next suppose $\lambda _1(\Delta _{C,n},X)=0$ and show that X is orientable. Let $\{(M_i,g_i)\}_{i\in {\mathbb {N}}}$ be a sequence in ${\mathcal {M}}$ that converges to X in the Gromov-Hausdorff topology. Then, we have $\lim _{i\rightarrow \infty }\lambda _1(\Delta _{C,n},g_i)=0$ by Theorem A.6. Thus, by Corollary A.2, we get that $M_i$ is orientable for sufficiently large i, and so X is orientable by the stability of orientability [16, Theorem 1.3]. $\square$

Proof of Theorem A.3

Let $\{(M_i,g_i)\}_{i\in {\mathbb {N}}}$ be a sequence in ${\mathcal {M}}$ and let X be its Gromov-Hausdorff limit. Suppose that each $M_i$ is not orientable. Then, we have $\lambda _1(\Delta _{C,n},g_i)\ge C_1(n,K_1,2D)$ by Lemma A.1. By Theorem A.6, we get $\lambda _1(\Delta _{C,n},X)\ge C_1(n,K_1,2D).$ Thus, by Lemma A.8, we get the theorem. $\square$

Theorem A.9

Let $X\in \overline{{\mathcal {M}}}$. If X is not orientable, then we have $\lambda _1(\Delta _{C,n},X)\ge C_1(n,K_1,2D).$

Proof

Let $\{(M_i,g_i)\}_{i\in {\mathbb {N}}}$ be a sequence in ${\mathcal {M}}$ that converges to X in the Gromov-Hausdorff topology. By Lemma A.8, we have $\lambda _1(\Delta _{C,n},X)>0$, and so we get $\lambda _1(\Delta _{C,n},g_i)>0$ for sufficiently large i by Theorem A.6. Thus, $M_i$ is not orientable and $\lambda _1(\Delta _{C,n},g_i)\ge C_1(n,K_1,2D)$ for sufficiently large i by Lemma A.1. By Theorem A.6, we get the theorem. $\square$

We immediately get the following corollaries:

Corollary A.10

Let $\{X_i\}_{i\in {\mathbb {N}}}$ be a sequence in $\overline{{\mathcal {M}}}$ and let $X\in \overline{{\mathcal {M}}}$ be its Gromov-Hausdorff limit. If $X_i$ is not orientable for each i, then X is not orientable.

Corollary A.11

Let $\{X_i\}_{i\in {\mathbb {N}}}$ be a sequence in $\overline{{\mathcal {M}}}$ and let $X\in \overline{{\mathcal {M}}}$ be its Gromov-Hausdorff limit. Then, the following two conditions are mutually equivalent.

• (i) $X_i$ is orientable for sufficiently large i.

• (ii) X is orientable.

By Corollary A.11, we have that if $X_1\in \overline{{\mathcal {M}}}$ is orientable and $X_2\in \overline{{\mathcal {M}}}$ is unorientable, then $X_1$ and $X_2$ belong to different connected components in $\overline{{\mathcal {M}}}$ with respect to the Gromov-Hausdorff topology.

Appendix B. Eigenvalue Estimate for $L^2$ Almost Kähler Manifolds

In this section, we consider $L^2$ almost Kähler manifolds, i.e., we assume that there exists a 2-form $\omega$ which satisfies that $\Vert \nabla \omega \Vert _2$ and $\Vert J_\omega ^2+{\mathrm {Id}}\Vert _1$ are small, where $J_\omega \in \Gamma (T^*M\otimes T M)$ is defined so that $\omega =g(J_\omega \cdot ,\cdot )$. The main goal is to give the almost version of (1).

Notation B.1

Let (M, g) be a Riemannian manifold. For each 2-form $\omega \in \Gamma (\bigwedge ^2 T^*M)$, let $J_\omega \in \Gamma (T^*M\otimes T M)$ denotes the anti-symmetric tensor that satisfies $\omega =g(J_\omega \cdot ,\cdot )$.

We first show the following easy lemmas.

Lemma B.2

Let (M, g) be an n-dimensional closed Riemannian manifold. If there exists a 2-form $\omega$ such that $\Vert J_\omega ^2+{\mathrm {Id}}\Vert _1<1$ holds, then n is an even integer.

Proof

There exists a point $x\in M$ such that $|J_\omega ^2(x)+{\mathrm {Id}}_{T_x M}|<1$. For any $v\in T_x M$ with $|v|=1$, we have $|J_\omega ^2(x)(v)+v|<1$, and so $|J_\omega ^2(x)(v)|>0$. Thus, $J_\omega (x)$ is non-degenerate. Therefore, $(T_x M,\omega _x)$ is a symplectic vector space. This implies the lemma. $\square$

Lemma B.3

Given integers $n\ge 2$, $1\le p\le n-1$, and positive real numbers $K>0$, $D>0$, there exists $\delta _0(n,p,K,D)>0$ such that if (M, g) is an n-dimensional closed Riemannian manifold with ${\mathrm {Ric}}\ge -K g$ and ${\mathrm {diam}}(M)\le D$, then we have $\lambda _{\alpha (n,p)+1}(\Delta _{C,p})\ge \delta _0(n,p,K,D),$ where we defined $\alpha (n,p):=n!/(p!(n-p)!).$

Proof

Put $\delta :=\lambda _{\alpha (n,p)+1}(\Delta _{C,p})$. If $\delta \ge 1$, we get the lemma. Thus, we assume that $\delta <1$. Let $\omega _i\in \Gamma (\bigwedge ^p T^*M)$ denotes the i-th eigenform of the connection Laplacian $\Delta _{C,p}$ acting on p-forms with $\Vert \omega _i\Vert _2=1$.

We have

$$\begin{aligned} \Vert \langle \omega _i,\omega _j\rangle \Vert _2^2\le \frac{1}{\lambda _1(g)}\Vert \nabla \langle \omega _i,\omega _j\rangle \Vert _2^2\le C(n,p,K,D)\delta \end{aligned}$$

(151)

for each $i,j=1,\ldots , \alpha (n,p)+1$ with $i\ne j$ by the Li-Yau estimate [22, p.116] and Lemma 3.7. By Lemma 3.5 and (151), we have

$$\begin{aligned} \Vert \langle \omega _i,\omega _j\rangle \Vert _1&\le C(n,p,K,D)\delta ^{1/2} \quad (i,j=1,\ldots , \alpha (n,p)+1 \text { with } i\ne j),\\ \Vert |\omega _i|^2-1\Vert _1&\le C(n,p,K,D)\delta ^{1/2} \quad (i=1,\ldots , \alpha (n,p)+1). \end{aligned}$$

Put

$$\begin{aligned}&G:=\Big \{x\in M: ||\omega _i|^2-1|(x)\le \delta ^{1/4}\text { for all }i=1,\ldots , \alpha (n,p)+1,\text { and}\\&\quad \left| \langle \omega _i,\omega _j\rangle \right| (x)\le \delta ^{1/4}\text { for all }i,j=1,\ldots , \alpha (n,p)+1\text { with }i\ne j \Big \}. \end{aligned}$$

Then, we have ${\mathrm {Vol}}(M\backslash G)\le C_1(n,p,K,D)\delta ^{1/4}{\mathrm {Vol}}(M)$ for some positive constant $C_1(n,p,K,D)$ depending only on n, p, K and D similarly to Lemma 4.49.

Let us show $\delta \ge \min \left\{ 1/C_1(n,p,K,D)^4,1/(\alpha (n,p)+1)^4\right\}$ by contradiction. Suppose that that $\delta < \min \left\{ 1/C_1(n,p,K,D)^4,1/(\alpha (n,p)+1)^4\right\}$. Then, we have $G\ne \emptyset$, and so we can take a point $x_0\in G$. We show that $\omega _{1}(x_0),\ldots , \omega _{\alpha (n,p)+1}(x_0)\in \bigwedge ^p T_{x_0}^*M$ are linearly independent. Take arbitrary $a_1,\ldots , a_{\alpha (n,p)+1}\in {\mathbb {R}}$ with $a_1 \omega _1(x_0)+\cdots +a_{\alpha (n,p)+1} \omega _{\alpha (n,p)+1}(x_0)=0$. Take i with $|a_i|=\max \{|a_1|,\ldots ,|a_{\alpha (n,p)+1}|\}$. Since we have $\langle a_1 \omega _1(x_0)+\cdots +a_k \omega _{\alpha (n,p)+1}(x_0),\omega _i(x_0)\rangle =0$, we get

$$\begin{aligned} 0\ge |a_i||\omega _i(x_0)|^2-\sum _{i\ne j}\left| a_j\langle \omega _i(x_0), \omega _j(x_0)\rangle \right| \ge |a_i|\left( 1-(\alpha (n,p)+1)\delta ^{1/4}\right) . \end{aligned}$$

Thus, $|a_i|=0$, and so $a_1=\cdots =a_k=0$. This implies the linearly independence of $\omega _{1}(x_0),\ldots , \omega _{\alpha (n,p)+1}(x_0)$. This contradicts to $\dim \left( \bigwedge ^p T^*_{x_0} M\right) =\alpha (n,p)$. Thus, we get $\lambda _{\alpha (n,p)+1}(\Delta _{C,p})=\delta \ge \min \left\{ 1/C_1(n,p,K,D)^4,1/(\alpha (n,p)+1)^4\right\}$. $\square$

Lemma B.4

Let (M, g) be an n-dimensional closed Riemannian manifold. Suppose that a 2-form $\omega$ satisfies

• (i) $\Vert \nabla \omega \Vert _2^2\le \delta \Vert \omega \Vert _2^2$,

• (ii) $\Vert J_\omega ^2+{\mathrm {Id}}\Vert _1\le \delta ^{1/4}\Vert \omega \Vert _2^2$

for some $0<\delta \le 1/4$. Let $\omega _\alpha$ be its image of the orthogonal projection

$$\begin{aligned} P_{\delta }:L^2\left( \bigwedge ^2 T^*M \right) \rightarrow \bigoplus _{\lambda _i(\Delta _{C,2})\le \delta ^{1/2}} {\mathbb {R}}\omega _i, \end{aligned}$$

where $\omega _i$ denotes the i-th eigenform of the connection Laplacian $\Delta _{C,2}$ with $\Vert \omega _i\Vert _2=1$ $(\omega _\alpha :=P_{\delta } (\omega ))$. Then, we have

• $\Vert \nabla \omega _\alpha \Vert _2^2\le 2\delta \Vert \omega _\alpha \Vert _2^2$,

• $\Vert J_{\omega _\alpha }^2+{\mathrm {Id}}\Vert _1\le 10\delta ^{1/4}\Vert \omega _\alpha \Vert _2^2$.

Proof

Put $\omega _\beta :=\omega -\omega _\alpha$. Then, we have $\Vert \omega \Vert ^2_2=\Vert \omega _\alpha \Vert ^2_2+\Vert \omega _\beta \Vert ^2_2$. By the assumption (i), we have

$$\begin{aligned} \delta \Vert \omega \Vert _2^2 \ge \Vert \nabla \omega \Vert _2^2 =\Vert \nabla \omega _\alpha \Vert _2^2+\Vert \nabla \omega _\beta \Vert _2^2 \ge \Vert \nabla \omega _\alpha \Vert _2^2+\delta ^{1/2}\Vert \omega _\beta \Vert _2^2. \end{aligned}$$

Thus, we get

$$\begin{aligned} \Vert \nabla \omega _\alpha \Vert _2^2&\le \delta \Vert \omega \Vert _2^2, \end{aligned}$$

(152)

$$\begin{aligned} \Vert \omega _\beta \Vert _2^2&\le \delta ^{1/2}\Vert \omega \Vert _2^2, \end{aligned}$$

(153)

and so

$$\begin{aligned} \Vert \omega _\alpha \Vert _2^2=\Vert \omega \Vert ^2_2 -\Vert \omega _\beta \Vert ^2_2\ge (1-\delta ^{1/2})\Vert \omega \Vert _2^2\ge \frac{1}{2}\Vert \omega \Vert _2^2. \end{aligned}$$

(154)

By the definitions of the norms, we have $|J_\omega |^2=2|\omega |^2$ and $|J_{\omega _\alpha }|^2=2|\omega _\alpha |^2$. Since we have $J_\omega ^2-J_{\omega _\alpha }^2 =J_\omega (J_\omega -J_{\omega _\alpha })+(J_\omega -J_{\omega _\alpha })J_{\omega _\alpha },$ we get $|J_\omega ^2-J_{\omega _\alpha }^2| \le 2(|\omega |+|\omega _\alpha |)|\omega _\beta |.$ Therefore, we have

$$\begin{aligned} \Vert J_\omega ^2-J_{\omega _\alpha }^2\Vert _1 \le 4\Vert \omega \Vert _2\Vert \omega _\beta \Vert _2\le 4\delta ^{1/4}\Vert \omega \Vert _2^2 \end{aligned}$$

by (153), and so

$$\begin{aligned} \Vert J_{\omega _\alpha }^2+{\mathrm {Id}}\Vert _1 \le \Vert J_{\omega }^2+{\mathrm {Id}}\Vert _1+\Vert J_\omega ^2-J_{\omega _\alpha }^2\Vert _1\le 5 \delta ^{1/4}\Vert \omega \Vert _2^2\le 10 \delta ^{1/4}\Vert \omega _\alpha \Vert _2^2 \end{aligned}$$

(155)

by (154). By (152) and (155), we get the lemma. $\square$

Let us show the orientability for $L^2$ almost Kähler manifolds.

Proposition B.5

For any integer $n\ge 2$ and positive real numbers $K>0$, $D>0$, there exists a constant $\delta _1(n,K,D)>0$ such that the following property holds. Let (M, g) be an n-dimensional closed Riemannian manifold with ${\mathrm {Ric}}\ge -K g$ and ${\mathrm {diam}}(M)\le D$. If there exists a 2-form $\omega$ such that

• (i) $\Vert \nabla \omega \Vert _2^2\le \delta _1\Vert \omega \Vert _2^2$,

• (ii) $\Vert J_\omega ^2+{\mathrm {Id}}\Vert _1\le \delta _1^{1/4}\Vert \omega \Vert _2^2$,

then M is orientable.

Proof

By Lemma B.2, we have that $n=2m$ is an even integer. We first assume that $\delta _1< \min \{1/4m^2,\delta _0(n,2,K,D)^2\}$. Since $J_{\omega }$ is anti-symmetric, we have $|J_{\omega }|^2\le \sqrt{2m}|J_{\omega }^2|$. Thus, we get

$$\begin{aligned} \sqrt{\frac{2}{m}}\Vert \omega \Vert _2^2= \frac{1}{\sqrt{2m}}\Vert J_{\omega }\Vert _2^2\le \sqrt{2m}+\delta _1^{1/4} \Vert \omega \Vert _2^2 \end{aligned}$$

by $|{\mathrm {Id}}|=\sqrt{2m}$. This and $\delta _1^{1/4}\le \frac{1}{2}\sqrt{\frac{2}{m}}$ imply that $\Vert \omega \Vert _2\le \sqrt{2m}$. Put $\omega _\alpha :=P_{\delta _1}(\omega )$. Note that we have that $\Vert \omega _\alpha \Vert _2\le \Vert \omega \Vert _2\le \sqrt{2m}$ and that $\Vert \omega _\alpha \Vert _\infty \le C(n,K,D)$ by Lemmas 3.7 and B.3.

We first fix $x\in M$, and consider the ${\mathbb {C}}$-linear map

$$\begin{aligned} J_{\omega _\alpha }(x):T_x M\otimes _{\mathbb {R}} {\mathbb {C}}\rightarrow T_x M\otimes _{\mathbb {R}} {\mathbb {C}}. \end{aligned}$$

Let us extend the Riemannian metric $\langle \cdot ,\cdot \rangle$ to $T_x M\otimes _{\mathbb {R}} {\mathbb {C}}$ so that

$$\begin{aligned} \langle u_1 + i v_1, u_2+iv_2 \rangle =(\langle u_1,u_2\rangle +\langle v_1,v_2\rangle )+i(\langle v_1, u_2\rangle -\langle u_1, v_2\rangle ) \end{aligned}$$

for all $u_1,u_2,v_1,v_2\in T_x M$. Since $J_{\omega _\alpha }(x)$ is anti-symmetric, there exist eigenvalues $\{\lambda _1,\overline{\lambda _1},\ldots ,\lambda _m,\overline{\lambda _m}\}$ of $J_{\omega _\alpha }(x)$ and an orthogonal basis $\{E_1,\overline{E_1},\ldots ,E_m,\overline{E_m}\}$ of $T_{x} M\otimes _{\mathbb {R}} {\mathbb {C}}$ such that $J_{\omega _\alpha }(x) E_i=\lambda _i E_i$, where the overline denotes the complex conjugate. Note that each $\lambda _i$ is a pure imaginary number. Let $\{E^1,\overline{E^1},\ldots ,E^m,\overline{E^m}\}\subset T_x^*M\otimes _{{\mathbb {R}}} {\mathbb {C}}\cong (T_x M\otimes _{{\mathbb {R}}} {\mathbb {C}} )^*$ be the dual basis of $\{E_1,\overline{E_1},\ldots ,E_m,\overline{E_m}\}$. If we extend $\omega _\alpha (x)$ to a complex bilinear form, then we have $\omega _\alpha (x)=\sum _{i=1}^m \lambda _i E^i\wedge \overline{E^i}.$ Thus, we get $\omega _{\alpha }^m(x)=m! \lambda _1\cdots \lambda _m E^1\wedge \overline{E^1}\wedge E^m\wedge \overline{E^m},$ and so $|\omega _{\alpha }^m(x)|=m!|\lambda _1|\cdots |\lambda _m|.$ Since we have $|\lambda _i|^2 =|(J_{\omega _\alpha }^2+{\mathrm {Id}})E_i-E_i|,$ we get $\left| 1-|\lambda _i|^2\right| \le |J_{\omega _\alpha }^2+{\mathrm {Id}}|(x)$ and $|\lambda _i|\le C(n,K,D)$. Therefore, we get

$$\begin{aligned} \left| |\omega _\alpha ^m|^2-(m!)^2\right| \le C|J_{\omega _\alpha }^2+{\mathrm {Id}}|, \end{aligned}$$

and so $\left| \Vert \omega _\alpha ^m\Vert _2^2-(m!)^2\right| \le C\delta _1^{1/4}$ by Lemma B.4. Since we have $\Vert \nabla (\omega _\alpha ^m)\Vert _2^2\le C\delta _1$ by Lemma B.4, we get the proposition taking $\delta _1$ sufficiently small by Corollary A.2 (ii). $\square$

The following theorem is the goal of this section.

Theorem B.6

For any integer $n\ge 2$, there exists a constant $C(n)>0$ such that the following property holds. Let (M, g) be an n-dimensional closed Riemannian manifold with ${\mathrm {Ric}}\ge (n-1) g$. If there exists a 2-form $\omega$ such that

• (i) $\Vert \nabla \omega \Vert _2^2\le \delta \Vert \omega \Vert _2^2$,

• (ii) $\Vert J_\omega ^2+{\mathrm {Id}}\Vert _1\le \delta ^{1/4}\Vert \omega \Vert _2^2$,

for some $\delta >0$, then we have $\lambda _1(g)\ge 2(n-1)- C(n)\delta ^{1/2}.$

Remark B.1

It is enough to prove the theorem when $\delta$ is small. Thus, we can assume that $n=2m$ is an even integer by Lemma B.2. If $n=2$, then $\lambda _1(g)\ge 2(n-1)$ is the original Lichnerowicz estimate. If $n=4$, the conclusion of the theorem can also be deduced from Main Theorem 1.

Proof

We first assume that $\delta < \min \{1/4m^2,\delta _0(n,2,K,D)^2\}$. Put $\omega _\alpha :=P_{\delta }(\omega )=\sum _{i=1}^k a_i \omega _i$. Here, $\omega _i$ is the i-th eigenform of the connection Laplacian $\Delta _{C,2}$ with $\Vert \omega _i\Vert _2=1$ corresponding to the eigenvalue $\lambda _i(\Delta _{C,2})\le \delta ^{1/2}$ for each $i=1,\ldots , k$. Similarly to Proposition B.5, we have $\Vert \omega _\alpha \Vert _\infty \le C$.

Let $f\in C^\infty (M)$ be the first eigenfunction of the Laplacian with $\Vert f\Vert _2=1$. If $\lambda _1(g)\ge 2(n-1)+1$, we get the theorem. Thus, we assume that $\lambda _1(g)\le 2(n-1)+1$. Then, we have $\Vert f\Vert _\infty \le C$ and $\Vert \nabla f\Vert _\infty \le C$ by Lemma 3.7. By Lemma 3.6 (i) and (iii), we have

$$\begin{aligned} \begin{aligned}&\frac{1}{{\mathrm {Vol}}(M)}\left| \int _M \langle \Delta (\iota (\nabla f)\omega _\alpha ),\iota (\nabla f)\omega _\alpha \rangle -\lambda _1(g)\langle \iota (\nabla f)\omega _\alpha ,\iota (\nabla f)\omega _\alpha \rangle \,d\mu _g \right| \\&\quad \le C\delta ^{1/2}\Vert \omega _\alpha \Vert _2^2 \end{aligned} \end{aligned}$$

(156)

and

$$\begin{aligned} \Vert d^*(\iota (\nabla f)\omega _\alpha )\Vert _2^2\le C\delta \Vert \omega _\alpha \Vert _2^2. \end{aligned}$$

(157)

By (4), (156), (157) and the Bochner formula, we get

$$\begin{aligned} \begin{aligned}&\frac{n-1}{{\mathrm {Vol}}(M)}\int _M \langle J_{\omega _\alpha }\nabla f,J_{\omega _\alpha }\nabla f\rangle \,d\mu _g\\&\quad \le \frac{1}{{\mathrm {Vol}}(M)}\int _M{\mathrm {Ric}}(J_{\omega _\alpha }\nabla f,J_{\omega _\alpha }\nabla f) \,d\mu _g\\&\quad =\frac{1}{{\mathrm {Vol}}(M)}\int _M \langle \Delta (\iota (\nabla f)\omega _\alpha ),\iota (\nabla f)\omega _\alpha \rangle \,d\mu _g-\frac{1}{{\mathrm {Vol}}(M)}\int _M |\nabla (\iota (\nabla f)\omega _\alpha )|^2\,d\mu _g\\&\quad \le \frac{\lambda _1(g)}{2{\mathrm {Vol}}(M)}\int _M \langle \iota (\nabla f)\omega _\alpha ,\iota (\nabla f)\omega _\alpha \rangle \,d\mu _g+C\delta ^{1/2}\\&\quad =\frac{\lambda _1(g)}{2{\mathrm {Vol}}(M)}\int _M \langle J_{\omega _\alpha }\nabla f,J_{\omega _\alpha }\nabla f \rangle \,d\mu _g+C\delta ^{1/2}. \end{aligned}\nonumber \\ \end{aligned}$$

(158)

Since $J_{\omega _\alpha }$ is anti-symmetric, we have

$$\begin{aligned} \begin{aligned}&\frac{1}{{\mathrm {Vol}}(M)}\left| \int _M \langle J_{\omega _\alpha }\nabla f,J_{\omega _\alpha }\nabla f\rangle \,d\mu _g- \int _M \langle \nabla f,\nabla f\rangle \,d\mu _g\right| \\&\quad \le \frac{1}{{\mathrm {Vol}}(M)}\left| \int _M \langle (J_{\omega _\alpha }^2+{\mathrm {Id}})\nabla f,\nabla f\rangle \,d\mu _g\right| \\&\quad \le C\Vert J_{\omega _\alpha }^2+{\mathrm {Id}}\Vert _1\le C\delta ^{1/4} \end{aligned} \end{aligned}$$

by Lemma B.4. Thus, taking $\delta$ sufficiently small, we get

$$\begin{aligned} \Vert J_{\omega _\alpha }\nabla f\Vert _2^2 \ge \Vert \nabla f\Vert _2^2-C\delta ^{1/4}= \lambda _1(g) -C\delta ^{1/4}\ge n-C\delta ^{1/4}\ge \frac{n}{2} \end{aligned}$$

(159)

by the Lichnerowicz estimate. By (158) and (159), we get the theorem. $\square$