Let $ (M, g) $ be a compact Riemannian manifold, and let $ A $ be a smooth symmetric, positive definite (1, 1)-tensor on $ M $. Denote by $ \nabla $ and $ \Delta $ the gradient operator and the Lapacian of $ M $, respectively. Define the operator $ L_A $ as follows:

$ \begin{equation*} L_A=\rm div \it A\nabla. \end{equation*} $

It is easy to see that $ L_A $ is an elliptic operator, and if $ A $ is the $ (1, 1) $-tensor associated to the tensor $ g $, $ L_A=\Delta $.

For the eigenvalue problem

$ \begin{equation} L_A(f)=-\lambda^A f, \ on \ M, \end{equation} $

(1.1)

when $ L_A=\Delta $, Lichnerowicz [1] proved that if $ M $ is an $ n $-dimensional compact Riemannian manifold with Ricci curvature bounded below by $ (n-1)K, K>0 $, then the first nonzero eigenvalue $ \lambda_1 $ of the problem (1.1) satisfies $ \lambda_1\geq nK^2 $. Moreover, one can get the fact that the above equality holds if and only if $ M $ is isometric to a sphere from the proof of Obata Theorem(see[2]). This is the so-called Lichnerowicz-Obata type estimate of Laplacian eigenvalue problem.

In recent years, there are also many other Lichnerowicz-Obata type results about the first nonzero eigenvalue of the problem (1.1). When $ M $ is an $ n $-dimensional compact immersed hypersurfaces of a space form and $ T_1 $ is the first Newton transformation associated to the shape operator of the immersion, [3] obtained a Lichnerowicz-Obata type estimate of the eigenvalue problem (1.1) with $ A=T_1 $. The operator $ L_{T_1} $ plays a key role in the study of stability for hypersurfaces with constant high order curvature (cf.[4, 5, 6]). Later, we [7] obtained more Lichnerowicz-Obata type estimates of the problem (1.1) with $ A=T_r $($ T_r $ is the $ r $th Newton transformation, $ 2\leq r\leq n-1 $). In [3], they also considered the case $ A=S $ (here $ S $ is the Schouten operator of an $ n(n\geq 4) $-dimensional compact Riemannian manifold which has harmonic Weyl tensr), and proved a Lichnerowicz-Obata type result for $ L_S $.

On the basis of the above researches, our aim in this paper is to establish Lichnerowicz-Obata type estimate for the eigenvalue problem (1.1) with $ A=E $. $ E $ is the Einstein operator defined by $ E=\frac{1}{2}RId-Ric $, where $ R $ is the scalar curvature and $ Ric $ is the linear operator associated with the Ricci tensor. This kind of $ L_E $ operator has very important application value both in general relativity and fuzzy mathematics(cf.[8]).

For this sake, we prove the following result:

Theorem 1.1 Let $ (M, g) $ be an $ n $-dimensional compact Riemannian manifold and $ E $ be the Einstein operator on $ M $. Suppose that Einstein operator $ E $ satisfies

$ \begin{equation*} 0<aId\leq E\leq bId, \end{equation*} $

where $ a, b $ are positive constants. Then, for the first nonzero eigenvalue $ \lambda^E_1 $ of the problem (1.1) with $ A=E $, we have

$ \begin{equation*} \lambda^E_1\geq\frac{nb}{2(nb-a)}[R_0a-2b^2-\frac{1}{2}c+d] \end{equation*} $

where $ R_0 $ is the lower bound of the scalar curvature of $ M $, $ c $ is the supremum of laplacian of the scalar curvature $ R $ and $ d $ is the infimum of laplacian of all eigenvalue functions of $ Ric $.

Furthermore, the equalities hold if and only if $ M $ is a sphere.

Let $ \{\omega_1, \ldots, \omega_n\} $ be a locally orthonormal coframe field on the $ n $-dimensional Riemannian manifold $ (M, g) $. Let $ \phi=\Sigma_{i, j=1}^n \phi_{ij}\omega_i\otimes\omega_j $ be a symmetric $ (0, 2) $-tensor on $ M $. Associated to tensor $ \phi $ we have the $ (1, 1) $-tensor, still denoted by $ \phi $, defined by

$ \begin{equation*} \langle\phi(X), Y\rangle=\phi(X, Y), \forall X, Y \in T(M), \end{equation*} $

and vice versa.

Then, we denote by $ Ric $ the Ricci tensor of $ M $. Namely

$ \begin{equation*} Ric(X, Y)=\sum\limits_{i=1}^{n}\langle Rm(X, e_i)Y, e_i\rangle, \end{equation*} $

where $ Rm(X, Y)Z=\nabla_Y\nabla_XZ-\nabla_X\nabla_YZ+\nabla_{[X, Y]}Z $ is the curvature tensor of $ M $ and $ \{e_1, \ldots, e_n\} $ is an orthonormal frame. We will also denote by $ Ric $ the linear operator associated with the Ricci tensor, (i.e., $ Ric(X, Y)=\langle Ric(X), Y\rangle $), as well as its coordinates will be denoted by $ Ric_{ij} $.

In [9], Cheng and Yau introduced an operator $ \square $ associated to $ \phi $ by

$ \begin{equation} \square l=\sum\limits_{i, j=1}^{n}\phi_{ij}l_{ij}, \end{equation} $

(2.1)

where $ l $ is any smooth function on $ M $.

Now, let us review two following basic properties of the operator $ \square $:

1. It follows from Cheng and Yau (Proposition 1 in [9]) that $ \square l=\rm div (\it \phi (\nabla l))- $ $ \sum_{i=1}^n(\sum_{j=1}^n \phi_{ijj})l_i. $

2. One says that $ \phi $ is divergence free if div$ \phi=0 $ or, equivalently, $ \sum_{j=1}^{n}\phi_{ijj}=0 $, for all $ 1\leq i\leq n $.

Remark If $ M $ is compact, it is not hard to check that $ \square $ is self-adjoint if and only if $ \phi $ is divergence free from [9]. Of course, by the above properties, we know that $ \square f=\rm div (\it \phi (\nabla f)) $ when $ \phi $ is divergence free. If $ \phi $ is symmetric and positive definite, then $ \square $ is strictly elliptic. therefore, we can assert that $ \square $ is strictly elliptic and self-adjoint when $ \phi $ is divergence free, symmetric and positive definite. Furthermore, the spectrum of $ \square $ is discrete and it makes sense to consider the eigenvalue problem.

To prove the main theorem, we also need the following lemmas.

Lemma 2.1 Let $ (M, g) $ be a Riemannian manifold and $ E $ is the Einstein operator(Einstein tensor) on $ M $. Then we have $ \rm div \it E=\rm0 $.

Proof It is well known that (see[10], P39) $ \rm div \it Ric=\rm\frac{1}{2}\it dR. $ Then, we get

$ \begin{equation*} \rm div \it E=\rm div \it Ric-\rm\frac{1}{2}div\it (RId)=\rm\frac{1}{2}\it dR-\rm\frac{1}{2}\it dR=\rm0. \end{equation*} $

Lemma 2.2 (Bochner type formula[7]) Let $ (M, g) $ be an $ n $-dimensional Riemannian manifold and $ \phi=\Sigma_{i, j=1}^n \phi_{ij}\omega_i\otimes\omega_j $ be a divergence free, symmetric tensor defined on $ M $. Then, for any smooth function $ l:M\rightarrow \mathbb{R} $, we have,

$ \begin{equation} \begin{aligned} \frac{1}{2} \rm div \it(\phi(\nabla|\nabla l|^{\rm 2}))=\rm \frac{1}{2}\square(|\nabla \it l|^{\rm 2})=&\langle\nabla l, \nabla (\rm div (\it \phi (\nabla l)))\rangle+\langle\phi(\nabla l), \nabla (\rm\Delta \it l)\rangle\\ &+\rm 2 \it\sum\limits_{i, j, k=1}^{n}\phi_{ij}l_{ik}l_{kj}+\rm 2 \it\sum\limits_{i, j, k, m=1}^{n}l_i l_j \phi_{im}R_{mkjk}\\ &-\sum\limits_{i, j=1}^{n}l_i l_j \Delta\phi_{ij}+\sum\limits_{k=1}^{n}(\sum\limits_{i, j=1}^{n}l_i l_j(\phi_{jik}-\phi_{jki}))_k\\ &-\sum\limits_{k=1}^{n}(\sum\limits_{i, j=1}^{n} l_{ik} \phi_{ij}l_j)_k. \end{aligned} \end{equation} $

(2.2)

For the proof details of lemma 2.2, one can refer to the lemma 2.1 in [7].

Lemma 2.3 (Generalized Newton inequality[3]) Let $ P $ and $ Q $ be two $ n\times n $ symmetric matrices. If $ Q $ is positive definite, then

$ \begin{equation} tr(P^2Q)\geq\frac{[tr(PQ)]^2}{trQ} \end{equation} $

(2.3)

and the equality holds if and only if $ P=\alpha I $ for some $ \alpha \in \mathbb{R} $.

Proof Let $ B $ be a positive definite matrix. By the fact $ tr[(PQ)^2]\leq tr(P^2Q^2) $, which holds for symmetric matrices, and using the Cauchy-Schwarz inequality with $ P\sqrt{B} $ and $ (\sqrt{B})^{-1}Q $, one can obtian

$ \begin{equation*} [tr(PQ)]^2=tr(P\sqrt{B}(\sqrt{B})^{-1}Q)^2\leq tr(P^2B)tr(Q^2B^{-1}). \end{equation*} $

In particular, since $ Q $ is positive definite, we can choose $ B=Q $ to obtain

$ \begin{equation*} [tr(PQ)]^2\leq tr(P^2Q)trQ, \end{equation*} $

The equality holds if and only if

$ \begin{equation*} \begin{aligned} P\sqrt{Q}=&\alpha(\sqrt{Q})^{-1}Q\Leftrightarrow (P\sqrt{Q})\sqrt{Q}\\ =&\alpha(\sqrt{Q})^{-1}Q\sqrt{Q}\Leftrightarrow PQ=\alpha Q\Leftrightarrow P=\alpha I. \end{aligned} \end{equation*} $

Proof of Theorem 1.1 Let $ f $ be an eigenfunction of $ \lambda^E_1 $, i.e. $ \it L_Ef=-\lambda^E_1f $. From Lemma 2.1, we know that $ E $ is divergence free. Now, by applying the Bochner type formula in Lemma 2.2 to tensor $ E $ and $ f $, we obtain

$ \begin{equation} \begin{aligned} \frac{1}{2} \rm div \it(E(\nabla|\nabla f|^{\rm 2}))=&\langle\nabla f, \nabla (\rm div (\it E (\nabla f)))\rangle+\langle E(\nabla f), \nabla (\rm\Delta \it f)\rangle\\ &+\rm 2 \it\sum\limits_{i, j, k=1}^{n}(E)_{ij}f_{ik}f_{kj}+\rm 2 \it\sum\limits_{i, j, k, m=1}^{n}f_i f_j (E)_{im}R_{mkjk}\\ &-\sum\limits_{i, j=1}^{n}f_i f_j \Delta(E)_{ij}+\sum\limits_{k=1}^{n}(\sum\limits_{i, j=1}^{n}f_i f_j((E)_{jik}-(E)_{jki}))_k\\ &-\sum\limits_{k=1}^{n}(\sum\limits_{i, j=1}^{n} f_{ik} (E)_{ij}f_j)_k. \end{aligned} \end{equation} $

(3.1)

By integrating both sides of (3.1) and using the divergence theorem, we have

$ \begin{equation} \begin{aligned} 0=&\int_M\langle\nabla f, \nabla (L_Ef))\rangle+\int_M\langle E(\nabla f), \nabla (\Delta f)\rangle+ 2\int_M\sum\limits_{i, j, k=1}^{n}(E)_{ij}f_{ik}f_{kj}\\ &+2\int_M\sum\limits_{i, j, k, m=1}^{n}f_i f_j (E)_{im}R_{mkjk}-\int_M\sum\limits_{i, j=1}^{n}f_i f_j \Delta E_{ij} \end{aligned} \end{equation} $

(3.2)

Then, with the fact $ L_Ef=-\lambda^E_1f $, one have

$ \begin{equation} \int_M\langle\nabla f, \nabla (L_Ef))\rangle=-\lambda^E_1\int_M |\nabla f|^2. \end{equation} $

(3.3)

We also have

$ \begin{equation} \begin{aligned} \rm div(\Delta \it f \ E\nabla f))=&\rm\Delta \it f \rm div( \it E(\nabla f))+ \langle E(\nabla f), \nabla (\rm\Delta \it f)\rangle\\ =&\rm\Delta \it f \cdot L_E(\nabla f)+ \langle E(\nabla f), \nabla (\rm\Delta \it f)\rangle. \end{aligned} \end{equation} $

(3.4)

Hence we get

$ \begin{equation} \begin{aligned} \int_M\langle E(\nabla f), \nabla (\rm\Delta \it f)\rangle =&-\int_M\rm\Delta \it f \cdot L_E f\\ =&\lambda^E_1\int_M f\rm\Delta \it f \\ =&-\lambda^E_1\int_M|\nabla f|^2. \end{aligned} \end{equation} $

(3.5)

Then we estimate other parts in (3.2). For convenience, we choose an orthonormal frame $ \{e_1, \ldots, e_n\} $ such that $ Ric $ is diagonalized in a neighborhood of any point $ p\in M^{n} $, i.e. $ Ric_{ij}=\mu_i\delta_{ij} $, where $ \mu_i $ is eigenvalue of the Ricci tensor at point $ p $.

Thus, for Einstein tensor $ E=\frac{1}{2}RId-Ric $, at point $ p $, we have

$ \begin{equation} (E)_{ij}=\frac{1}{2}R\delta_{ij}-\mu_i\delta_{ij}. \end{equation} $

(3.6)

From Lemma 2.3 and the fact $ E $ is positive definite, divergence free, we can obtain

$ \begin{equation} \begin{aligned} 2\int_M\sum_{i, j, k=1}^{n}(E)_{ij}f_{ik}f_{kj} \geq&2\int_M\frac{(\sum\limits_{i, j=1}^{n}(E)_{ij}f_{ij})^2}{tr(E)}\\ =&2\int_M\frac{(L_Ef)^2}{tr(E)}\\ =&2\int_M\frac{(\lambda^E_1f)^2}{tr(E)}. \end{aligned} \end{equation} $

(3.7)

By appling divergence Theorem and the fact $ L_E (f^2)=2f L_E f+2\langle E(\nabla f), \nabla f\rangle $, we have

$ \begin{equation} \begin{aligned} \int_M\langle E(\nabla f), \nabla f\rangle =-\int_M f \cdot \lambda^E_1 f =\lambda^E_1\int_M f^2. \end{aligned} \end{equation} $

(3.8)

With the condition $ 0<aId\leq E\leq bId $, it is easy to check that

$ \begin{equation} \begin{aligned} a|\nabla f|^2\leq\langle E(\nabla f), \nabla f\rangle\leq b|\nabla f|^2. \end{aligned} \end{equation} $

(3.9)

Then, from (3.7), (3.8) and (3.9), we obtain

$ \begin{equation} \begin{aligned} 2\int_M\sum\limits_{i, j, k=1}^{n}(E)_{ij}f_{ik}f_{kj} \geq\frac{2\lambda^E_1 a}{n b}\int_M|\nabla f|^2. \end{aligned} \end{equation} $

(3.10)

We can also obtain

$ \begin{equation} \begin{aligned} 2\int_M\sum\limits_{i, j, k, m=1}^{n}f_i f_j (E)_{im}R_{mkjk}=&2\int_M Ric(\nabla f, E(\nabla f))\\ =&2\int_M \langle Ric(\nabla f), E(\nabla f)\rangle\\ =&2\int_M \langle(\frac{1}{2}RId-E) \nabla f, E(\nabla f)\rangle\\ =&\int_M [R\langle \nabla f, E(\nabla f)\rangle-2\langle \nabla f, E^2(\nabla f)\rangle]\\ \geq&(R_0a-2b^2)\int_M|\nabla f|^2, \end{aligned} \end{equation} $

(3.11)

where $ R_0 $ is the lower bound of the scalar curvature $ R $.

For any function $ h\in C^2(M) $, by the Hopf maximum principle, it is not hard to find that $ (\Delta h)_{min}\leq0 $ and $ (\Delta h)_{max} \geq0 $ on $ M $. Then, let $ d $ be the infimum of laplacian of all eigenvalue functions of $ Ric $. Hence we have $ c\triangleq(\Delta R)_{max}\geq0 $ and $ d\leq0 $.

Under the above frame, we have

$ \begin{equation} \begin{aligned} -\sum\limits_{i, j=1}^{n}f_i f_j \Delta E_{ij}=&-\sum\limits_{i, j, p=1}^{n}f_i f_j E_{ijpp}\\ =&-\Delta R\sum\limits_{i=1}^{n}f_i^2+\sum\limits_{i}^{n}(\Delta \mu_{i})f_i^2\\ \geq&(-c+d)|\nabla f|^2. \end{aligned} \end{equation} $

(3.12)

Hence

$ \begin{equation} -\int_M\sum\limits_{i, j=1}^{n}f_i f_j \Delta E_{ij}\geq(-c+d)\int_M|\nabla f|^2. \end{equation} $

(3.13)

Finally, taking (3.3), (3.5), (3.10), (3.11) and (3.13) back into (3.2), we obtain

$ \begin{equation} \begin{aligned} 0\geq[-2\lambda^E_1+\frac{2\lambda^E_1 a}{n b}+R_0a-2b^2-c+d]\int_M|\nabla f|^2. \end{aligned} \end{equation} $

(3.14)

Therefore

$ \begin{equation} \begin{aligned} \lambda^E_1\geq\frac{nb}{2(nb-a)}[R_0a-2b^2-\frac{1}{2}c+d]. \end{aligned} \end{equation} $

(3.15)

Now, we consider the equality case. If we suppose $ M=\mathbb{S}^{n}(1) $, we have $ R_0=n(n-1) $, $ E=\frac{(n-1)(n-2)}{2}Id $, $ L_Ef=\frac{(n-1)(n-2)}{2}\Delta f $ and $ \lambda^E_1=\frac{n(n-1)(n-2)}{2} $. In this case, the estimate becomes euqality with the assumption that $ a=b=\frac{(n-1)(n-2)}{2} $. On the other hand, if the equality holds, the equality case of Lemma 2.3, implies that $ f_{ij}=pg_{ij} $, for some real constant $ p $, and following the proof of Obata Theorem, cf [2], we can obtain that $ M $ is a sphere.

Remark There is still much to be studied about the Lichnerowicz-Obata type estimate of this kind of problem (1.1). Especially, when the ambient space of $ M $ is an Einstein manifold, the first nonzero eigenvalue $ \lambda^T_1 $ is of great significance to the study of variational problem that characterizes hypersurfaces with constant $ 2 $-mean curvature in Einstein manifolds (cf.[11]). To the author's knowledge, the Lichnerowicz-Obata type estimate of $ L_{T_1} $ in this kind of ambient space is still wide open.