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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 16, Issue 1

# Chen’s inequalities for submanifolds in (κ, μ)-contact space form with a semi-symmetric metric connection

/ He Guoqing
/ Tang Wanxiao
/ Zhao Peibiao
Published Online: 2018-04-18 | DOI: https://doi.org/10.1515/math-2018-0034

## Abstract

In this paper, we obtain Chen’s inequalities for submanifolds in (κ, μ)-contact space form endowed with a semi-symmetric metric connection.

MSC 2010: 53C40; 53B05

## 1 Introduction

Since Chen [1, 2] proposed the so-called Chen invariants, and chen’s inequalities, the study of Chen invariants and inequalities has been an active field over the past two decades. There are many works studying Chen’s inequalities for different submanifolds in various ambient spaces (one can see [3, 4, 5, 6, 7, 8, 9, 10, 11, 12] for details).

In 1924, Friedmann and Schoutenn [13] first introduced the notion of a semi-symmetric linear connection on a differentiable manifold. Later many scholars studied the geometric and analytic problems on a manifold associated with a semi-symmetric metric connection. For instance, H. A. Hayden [14] studied the geometry of subspaces with a semi-symmetric metric connection. K. Yano [15] studied some geometric properties of a Riemannian manifold endowed with a semi-symmetric metric connection. In [16], Z. Nakao studied submanifolds of a Riemannian manifold with a semi-symmetric metric connection.

Recently, T. Koufogiorgos [17] introduced the notation of (κ, μ)-contact space form, in particular, when κ = 1, (κ, μ)-contact space form is the well known class of Sasakian space form, and A. Mihai and C. Özgür [18, 19] arrived at Chen’s inequalities for submanifolds of Sasakian space forms, real(complex) space forms with a semi-symmetric metric connection. On the other hand, P. Zhang et al [20] also established Chen’s inequalities for submanifolds of a Riemannian manifold of quasi-constant curvature with a semi-symmetric metric connection.

Motivated by the celebrated works above, a natural question is: can we establish the Chen’s inequalities for submanifolds in a (κ, μ)-contact space form endowed with a semi-symmetric metric connection?

In this paper, we will consider and derive Chen’s first inequality and Chen-Ricci inequalities for submanifolds of a (κ, μ)-contact space form with a semi-symmetric metric connection.

## 2 Preliminaries

Let Nn+p be an (n+p)-dimensional Riemannian manifold with a Riemannian metric g and ▽ be a linear connection on Nn+p. If the torsion tensor T of ▽ satisfies $T¯(X¯,Y¯)=ϕ(Y¯)X¯−ϕ(X¯)Y¯$

for a 1-form ϕ, then the connection ▽ is called a semi-symmetric connection. Furthermore, if ▽ satisfies ▽g = 0, then ▽ is called a semi-symmetric metric connection.

Let ▽ denote the Levi-Civita connection associated with the Riemannian metric g. In [15] K.Yano obtained a relation between the semi-symmetric metric connection ▽ and the Levi-Civita connection ▽ which is given by $▽¯X¯Y¯=▽¯′X¯Y¯+ϕ(Y¯)X¯−g(X¯,Y¯)U,∀X¯,Y¯∈X(Nn+p)$

where U is a vector field given by g(U, X) = ϕ(X).

Let Mn be an n-dimensional submanifold of Nn+p with the semi-symmetric metric connection ▽ and the Levi-Civita connection ▽. On Mn we consider the induced semi-symmetric metric connection denoted by ▽ and the induced Levi-Civita connection denoted by ▽. The Gauss formula with respect to ▽ and ▽ can be written as $▽¯XY=▽XY+σ(X,Y),▽¯X′Y=▽X′Y+σ′(X,Y),∀X,Y∈X(Mn),$

where σ is the second fundamental form of Mn and σ is a (0, 2)-tensor on Mn. According to [16] σ is also symmetric.

Let R and R denote the curvature tensors with respect to ▽ and ▽ respectively. We also denote the curvature tensor associated with ▽ and ▽ by R and R, repectively. From [16], we know $R¯(X,Y,Z,W)=R¯′(X,Y,Z,W)−α(Y,Z)g(X,W)+α(X,Z)g(Y,W)−α(X,W)g(Y,Z)+α(Y,W)g(X,Z),$(1)

for all X, Y, Z, W ∈ 𝓧(Mn), where α is a (0, 2)-tensor field defined by $α(X,Y)=(▽¯X′ϕ)Y−ϕ(X)ϕ(Y)+12ϕ(U)g(X,Y).$

Denote by λ the trace of α.

In [16] the Gauss equation for Mn with respect to the semi-symmetric metric connection is given by $R¯(X,Y,Z,W)=R(X,Y,Z,W)+g(σ(X,Z),σ(Y,W))−g(σ(X,W),σ(Y,Z))$(2)

for all X, Y, Z, W ∈ 𝓧(Mn).

In Nn+p we can choose a local orthonormal frame {e1, e2,…, en, en+1, …, en+p} such that {e1, e2, …, en} are tangent to Mn. Setting $\begin{array}{}{\sigma }_{ij}^{r}\end{array}$ = g(σ(ei, ej), er), then the squared length of σ is given by $∥σ∥2=∑i,j=1ng(σ(ei,ej),σ(ei,ej))=∑r=nn+p∑i,j=1n(σijr)2.$

The mean curvature vector of Mn associated to ▽ is $\begin{array}{}{H}^{\mathrm{\prime }}=\frac{1}{n}\sum _{i=1}^{n}{\sigma }^{\mathrm{\prime }}\left({e}_{i},{e}_{i}\right).\end{array}$ The mean curvature vector of Mn associated to ▽ is defined by $\begin{array}{}H=\frac{1}{n}\sum _{i=1}^{n}\sigma \left({e}_{i},{e}_{i}\right).\end{array}$

According to the formula (7) in [16], we have the following result.

#### Lemma 2.1

([16]). If U is a tangent vector field on Mn, then H = H and σ = σ.

Let πTpMn be a 2-plane section for any pMn and K(π) be the sectional curvature of Mn associated to the semi-symmetric metric connection ▽. The scalar curvature τ associated to the semi-symmetric metric connection ▽ at p is defined by $τ(p)=∑1⩽i(3)

Let Lk be a k-plane section of TpMn and {e1, e2,…, ek} be any orthonormal basis of Lk. The scalar curvature τ(Lk) of Lk associated to the semi-symmetric metric connection ▽ is given by $τ(Lk)=∑1⩽i(4)

We denote by (inf K)(p) = inf{K(π)∣πTpMn, dimπ = 2}. In [1] B.-Y. Chen introduced the first Chen invariant δM(p) = τ(p) − (inf K)(p), which is certainly an intrinsic character of Mn.

Suppose L is a k-plane section of TpM and X is a unit vector in L, we choose an orthonormal basis {e1, e2,…, ek} of L, such that e1 = X. The Ricci curvature Ricp of L at X is given by $RicL(X)=K12+K13+...+K1k$(5)

where Kij = K(eiej). We call RicL(X) a k-Ricci curvature.

For each integer k, 2 ⩽ kn, the Riemannian invariant Θk [2] on Mn is defined by $Θk(p)=1k−1infL,X{RicL(X)},p∈Mn$(6)

where L is a k-plane section in TpMn and X is a unit vector in L.

Recently, T. Koufogiorgos [17] introduced the notion of (κ, μ)-contact space form, which contains the well known class of Sasakian space forms for κ = 1. Thus it is worthwhile to study relationships between intrinsic and extrinsic invariants of submanifolds in a (κ, μ) -contact space form with a semi-symmetric metric connection ▽.

A (2m+1)-dimensional differentiable manifold is called an almost contact metric manifold if there is an almost contact metric structure (φ, ξ, η, g) consisting of a (1, 1) tensor field φ, a vector field ξ, a 1-form η and a compatible Riemannian metric g satisfying $φ2=−I+η⊗ξ,η(ξ)=1,φξ=0,η∘φ=0,$ $g(X,φY)=−g(φX,Y),g(X,ξ)=η(X),∀X,Y∈X(M^).$(7)

An almost contact metric structure becomes a contact metric structure if = Φ, where Φ(X, Y) = g(X, φY) is the fundamental 2-form of .

In a contact metric manifold , the (1, 1)-tensor field h defined by 2h = 𝓛ξφ is symmetric and satisfies $hξ=0,hφ+φh=0,▽¯′ξ=−φ−φh,trace(h)=trace(φh)=0$

where ▽ is a Levi-civita connection associated with the Riemannian metric g.

The (κ, μ)-nullity distribution of a contact metric manifold is a distribution $N(κ,μ):p→Np(κ,μ)={Z∈TpM^∣R^(X,Y)Z=κ[g(Y,Z)X−g(X,Z)Y]+μ[g(Y,Z)hX−g(X,Z)hY]}$

where κ and μ are constants. If ξN(κ, μ), is called a (κ, μ)-contact metric manifold. Since in a (κ, μ)-contact metric manifold one has h2 = (κ − 1)φ2, therefore κ ⩽ 1 and if κ = 1 then the structure is Sasakian.

The sectional curvature (X, φX) of a plane section spanned by a unit vector orthogonal to ξ is called a φ-sectional curvature. If the (κ, μ)-contact metric manifold has constant φ-sectional curvature C, then it is called a (κ, μ)-contact space form and it is denoted by (C). The curvature tensor of (C) is given by [17]. $R¯′(X,Y)Z=c+34{g(Y,Z)X−g(X,Z)Y}+c+3−4κ4{η(X)η(Z)Y−η(Y)η(Z)X+g(X,Z)η(Y)ξ−g(Y,Z)η(X)ξ)}+c−14{2g(X,φY)φZ+g(X,φZ)φY−g(Y,φZ)φX}+12{g(hY,Z)hX−g(hX,Z)hY+g(φhX,Z)φhY−g(φhY,Z)φhX}−g(X,Z)hY+g(Y,Z)hX+η(X)η(Z)hY−η(Y)η(Z)hX−g(hX,Z)Y+g(hY,Z)X−g(hY,Z)η(X)ξ+g(hX,Z)η(Y)ξ+μ{η(Y)η(Z)hX−η(X)η(Z)hY+g(hY,Z)η(X)ξ−g(hX,Z)η(Y)ξ}$(8)

for all X, Y, Z ∈ 𝓧(), where c+2κ = −1 = κμ if κ < 1.

For a vector field X on a submanifold M of a (κ, μ)-contact form (c), let PX be the tangential part of φX. Thus, P is an endomorphism of the tangent bundle of M and satisfies g(X, PY) = −g(PX, Y) for all X, Y ∈ 𝓧(M). (φh)TX and hTX are the tangential parts of φhX and hX respectively. Let {e1, e2, …, en} be an orthonormal basis of TpM. We set $‖ϑ‖2=∑i,j=1ng(ei,ϑej)2,ϑ∈{P,(φh)T,hT}.$

Let πTpM be a 2-plane section spanned by an orthonormal basis {e1, e2}. Then, β(π) is given by $β(π)=〈e1,Pe2〉2$

is a real number in [0, 1], which is independent of the choice of orthonormal basis {e1, e2}. Put $γ(π)=(η(e1))2+(η(e2))2,θ(π)=η(e1)2g(hTe2,e2)+η(e2)2g(hTe1,e1)−2η(e1)η(e2)g(hTe1,e2).$

Then γ(π) and θ(π) are also real numbers and do not depend on the choice of orthonormal basis {e1, e2}, of course, γ(π) ∈ [0, 1]

## 3 Chen first inequality

For submanifolds of a (κ, μ) -contact space form endowed with a semi-symmetric metric connection ▽, we establish the following optimal inequality relating the scalar curvature and the squared mean curvature associated with the semi-symmetric metric connection, which is called Chen first inequality [1]. At first we recall the following lemma.

#### Lemma 3.1

([20]). Let f(x1, x2, …, xn) (n ≥ 3) be a function in Rn defined by $f(x1,x2,...,xn)=(x1+x2)∑i=3nxi+∑3⩽i

If x1+x2+⋯+xn = (n − 1)ε, then we have $f(x1,x2,...,xn)⩽(n−1)(n−2)2ε2$

with the equality holding if and only if x1+x2 = x3 = ⋯ = xn = ε.

#### Theorem 3.2

Let M ba an n-dimensional (n ≥ 3) submanifold of a (2m+1)-dimensional (κ, μ)-contact form M̂(c) endowed with a semi-symmetric metric connectionsuch that ξTM. For each 2-plane section πTpM, we have $τ(p)−K(π)⩽n2(n−2)2(n−1)∥H∥2+18n(n−3)(c+3)+(n−1)κ+3(c−1)8{∥P∥2−2β(π)}+14(c+3−4κ)γ(π)−(μ−1)θ(π)−12{2trace(h|π)+det(h|π)−det((φh)|π)}+(μ+n−2)trace(hT)+14{∥(φh)T∥2−∥hT∥2−(trace(φh)T)2+(trace(hT))2}−(n−1)λ+trace(α|π).$(9)

If U is a tangent vector field to M, then the equality in (9) holds at pM if and only if there exists an orthonormal basis {e1, e2, …, en} of TpM and an orthonormal basis {en+1, …, e2m+1} of $\begin{array}{}{T}_{p}^{\perp }M\end{array}$ such that $π=Span{e1,e2}$

and the forms of shape operators ArAer (r = n+1,…, 2m+1) become $An+1=σ11n+1000σ22n+1000(σ11n+1+σ22n+1)In−2,Ar=σ11rσ12r0σ12r−σ11r0000n−2,r=n+2,⋯,2m+1.$

#### Proof

Let πTpM be a 2-plane section. We choose an orthonormal basis {e1, e2,…, en} for TpM and {en+1,…, e2m+1} for $\begin{array}{}{T}_{p}^{\perp }M\end{array}$ such that π = Span{e1, e2}.

Setting X = W = ei, Y = Z = ej, ij, i, j = 1,…, n, and using (1), (2), (8), we have $Rijji=c+34+c+3−4κ4{−η(ei)2−η(ej)2}+c−14{3g(ei,φej)2}+12{g(ei,φhej)2−g(ei,hej)2+g(ei,hei)g(ej,hej)−g(ei,φhei)g(ej,φhej)}+g(ei,hei)+2η(ei)η(ej)g(ei,hej)−g(hei,ei)η(ej)2−g(hej,ej)η(ei)2+g(hej,ej)+μ{g(hei,ei)η(ej)2+g(hej,ej)η(ei)2−2η(ei)η(ej)g(ei,hej)}+g(σ(ei,ei),σ(ej,ej))−g(σ(ei,ej),σ(ei,ej))−α(ei,ei)−α(ej,ej).$(10)

From (10) we get $τ=18{n(n−1)(c+3)+3(c−1)∥P∥2−2n(n−1)(c+3−4κ)}+14{∥(φh)T∥2−∥hT∥2−(trace(φhT))2+(trace(hT))2}+(μ+n−2)trace(hT)−(n−1)λ+∑r=12m+1∑1⩽i(11)

On the other hand, using (10) we have $R1221=14{c+3+3(c−1)β(π)−(c+3−4κ)γ(π)+4(μ−1)θ(π)}+12{det(h|π)−det((φh)|π)+2trace(h|π)}−trace(α|π)+∑r=n+12m+1[σ11rσ22r−(σ12r)2].$(12)

From (11) and (12) it follows that $τ−K(π)=18n(n−3)(c+3)+(n−1)κ+3(c−1)8[∥P∥2−2β(π)]+14(c+3−4κ)γ(π)−(μ−1)θ(π)−12{2trace(h|π)+det(h|π)−det(φh|π)}+(μ+n−2)trace(hT)+14{∥(φh)T∥2−∥hT∥2−(trace(φh)T)2+(trace(hT))2}−(n−1)λ+trace(α|π)+∑r=n+12m+1[(σ11r+σ22r)∑3⩽i⩽nσiir+∑3⩽i(13)

Let us consider the following problem: $max{fr(σ11r,...,σnnr)=(σ11r+σ22r)∑3⩽i⩽nσiir+∑3⩽i

where kr is a real constant. From Lemma 3.1, we know $fr⩽n−22(n−1)(kr)2,$(14)

with the equality holding if and only if $σ11r+σ22r=σiir=krn−1,i=3,...,n.$(15)

From (13) and (14), we have $τ−K(π)≤n2(n−2)2(n−1)∥H∥2+18n(n−3)(c+3)+(n−1)κ+3(c−1)8[∥P∥2−2β(π)]+14(c+3−4κ)γ(π)−(μ−1)θ(π)−12[2trace(h|π)+det(h|π)−det((φh)|π)]+(μ+n−2)trace(hT)+14[∥(φh)T∥2−∥hT∥2−(trace(φh)T)2+(trace(hT))2]−(n−1)λ+trace(α|π).$

If the equality in (9) holds, then the inequalities given by (13) and (14) become equalities. In this case we have $∑3⩽i⩽n(σ1ir)2=0,∑2⩽i

Since U is a tangent vector field to M, we know σ = σ. By choosing a suitable orthonormal basis, the shape operators take the desired forms.

The converse is easy to show. □

For a Sasakian space form (c), we have κ = 1 and h = 0. So using Theorem 3.2, we have the following corollary.

#### Corollary 3.3

Let M be an n-dimensional (n ≥ 3) submanifold in a Sasakian space form M̂(c) endowed with a semi-symmetric metric connection such that ξTM. Then, for each point pM and each plane section πTpM, we have $τ−K(π)⩽n2(n−2)2(n−1)∥H∥2+18n(n−3)(c+3)+(n−1)+3(c−1)8[∥P∥2−2β(π)]+c−14γ(π)−(n−1)λ+trace(α|π).$(16)

If U is a tangent vector field to M, then the equality in (16) holds at pM if and only if there exists an orthonormal basis {e1, e2, …, en} of TpM and orthonormal basis {en+1, …, e2m+1} of $\begin{array}{}{T}_{p}^{\perp }M\end{array}$ such that $π=Span{e1,e2}$

and the forms of shape operators ArAer (r = n+1,…, 2m+1) become $An+1=σ11n+1000σ22n+1000(σ11r+σ22r)In−2,Ar=σ11rσ12r0σ12r−σ11r0000n−2,r=n+2,⋯,2m+1.$

Since in case of non-Sasakian (κ, μ)-contact space form, we have κ < 1, thus c = −2κ − 1 and μ = κ+1. Putting these values in (9), we can have a direct corollary to Theorem 3.2.

#### Corollary 3.4

Let Let M be an n-dimensional (n ≥ 3) submanifold in a non-Sasakian (κ, μ)-contact space form M̂(c) with a semi-symmetric metric connection such that ξTM. Then, for each point pM and each plane section πTpM, we have $τ−K(π)⩽n2(n−2)2(n−1)∥H∥2−14n(n−3)(κ−1)+(n−1)κ−34(κ+1)∥P∥2+12[3(κ+1)β(π)−(3κ−1)γ(π)−2κθ(π)]−12[2trace(h|π)+det(h|π)−det((φh)|π)]+(κ+n−1)trace(hT)+14[∥(φh)T∥2−∥hT∥2−(trace(φh)T)2+(trace(hT))2]−(n−1)λ+trace(α|π).$(17)

If U is a tangent vector field to M, then the equality in (17) holds at pM if and only if there exists an orthonormal basis {e1, e2, …, en} of TpM and orthonormal basis {en+1,…, e2m+1} of $\begin{array}{}{T}_{p}^{\perp }M\end{array}$ such that $π=Span{e1,e2}$

and the forms of shape operators ArAer(r = n+1,…, 2m+1) become $An+1=σ11n+1000σ22n+1000(σ11r+σ22r)In−2,Ar=σ11rσ12r0σ12r−σ11r0000n−2,r=n+2,⋯,2m+1.$

## 4 Ricci and k-Ricci curvature

In this section, we establish inequalities between Ricci curvature and the squared mean curvature for submanifolds in a (κ, μ)-contact space form with a semi-symmetric metric connection. These inequalities are called Chen-Ricci inequalities [2].

First we give a lemma as follows.

#### Lemma 4.1

([20]). Let f(x1, x2, …, xn) be a function in Rn defined by $f(x1,x2,...,xn)=x1∑i=2nxi.$

If x1+x2+ …+xn = 2ε, then we have $f(x1,x2,...,xn)⩽ε2,$

with the equality holding if and only if x1 = x2+ … +xn = ε.

#### Theorem 4.2

Let M be an n-dimensional (n ⩾ 2) submanifold of a (2m+1)-dimensional (κ, μ)-contact space form M̂(c) endowed with a semi-symmetric metric connection such that ξTM. Then for each point pM,

1. For each unit vector X in TpM, we have $Ric(X)⩽n24∥H∥2+(n−1)(c+3)4+3(c−1)4∥PX∥2−c+3−4κ4[1+(n−2)η(X)2]+12[∥(φhX)T∥2−∥(hX)T∥2−g(φhX,X)trace((φh)T)+g(hX,X)trace(hT)]+(μ+n−3)g(hX,X)+[1+(μ−1)η(X)2]trace(hT)−(n−2)α(X,X)−λ.$(18)

2. If H(p) = 0, a unit tangent vector XTpM satisfies the equality case of (18) if and only if XN(p) = {X ∈ {TpM|σ(X, Y) = 0, ∀YTpM}.

3. The equality of (18) holds identically for all unit tangent vectors if and only if either

1. n ≠ 2, $\begin{array}{}{\sigma }_{ij}^{r}\end{array}$ = 0, i, j = 1, 2, …, n; r = n+1, …, 2m+1;

or

2. n = 2, $\begin{array}{}{\sigma }_{11}^{r}={\sigma }_{22}^{r},\phantom{\rule{thinmathspace}{0ex}}{\sigma }_{12}^{r}=0,\phantom{\rule{thinmathspace}{0ex}}r=3,...,2m+1..\end{array}$

#### Proof

1. Let XTpM be an unit vector. We choose an orthonormal basis e1, …, en, en+1 …, e2m+1 such that e1, …, en are tangential to M at p with e1 = X.

Using (10), we have $Ric(X)=(n−1)(c+3)4−c+3−4κ4[1+(n−2)η(X)2]+3(c−1)4∥PX∥2+12[∥(φhX)T∥2−∥(hX)T∥2+g(hX,X)trace(hT)−g(φhX,X)trace((φh)T)]+(μ+n−3)g(X,hX)+(1−η(X)2+μη(X)2)trace(hT)−(n−2)α(X,X)−λ+∑r=n+12m+1∑i=2n[σ11rσiir−(σ1ir)2]⩽(n−1)(c+3)4+3(c−1)4∥PX∥2−c+3−4κ4[1+(n−2)η(X)2]+12[∥(φhX)T∥2−∥(hX)T∥2−g(φhX,X)trace(φh)T+g(hX,X)trace(hT)]+(μ+n−3)g(hX,X)+[1−η(X)2+μη(X)2]trace(hT)−(n−2)α(X,X)−λ+∑r=n+12m+1∑i=2nσ11rσiir.$(19)

Let us consider the function fr: RnR, defined by $fr(σ11r,σ22r,...,σnnr)=∑i=2nσ11rσiir.$

We consider the problem $max{fr|σ11r+...+σnnr=kr},$

where kr is a real constant. From Lemma 4.1, we have $fr⩽(kr)24,$(20)

with equality holding if and only if $σ11r=∑i=2nσiir=kr2.$(21)

From (19) and (20) we get $Ric(X)⩽n24∥H∥2+(n−1)(c+3)4+3(c−1)4∥PX∥2−c+3−4κ4[1+(n−2)η(X)2]+12[∥(φhX)T∥2−∥(hX)T∥2−g(φhX,X)trace(φh)T+g(hX,X)trace(hT)]+(μ+n−3)g(hX,X)+[1+(μ−1)η(X)2]trace(hT)−(n−2)α(X,X)−λ.$

2. For a unit vector XTpM, if the equality case of (18) holds, from (19), (20) and (21) we have $σ1ir=0,i≠1,∀r.σ11r+σ22r+...+σnnr=2σ11r,∀r.$

Since H(p) = 0, we know $σ11r=0,∀r.$

So we get $σ1jr=0,∀r.$

i.e XN(p)

The converse is trivial.

3. For all unit vectors XTpM, the equality case of (18) holds if and only if $2σiir=σ11r+...+σnnr,i=1,...,n;r=n+1,...,2m+1.σijr=0,i≠j,r=n+1,...,2m+1.$

Thus we have two cases, namely either n ≠ 2 or n = 2.

In the first case we have $σijr=0,i.j=1,...,n;r=n+1,...,2m+1.$

In the second case we have $σ11r=σ22r,σ12r=0,r=3,...,2m+1.$

The converse part is straightforward. □

#### Corollary 4.3

Let M be an n-dimensional (n ⩾ 2) submanifold in a Sasakian space form M̂(c) endowed with a semi-symmetric metric connection such that ξTM. Then for each point pM and for each unit vector XTpM, we have $Ric(X)⩽n24∥H∥2+(n−1)(c+3)4+3(c−1)4∥PX∥2−c−14[1+(n−2)η(X)2]−(n−2)α(X,X)−λ.$

#### Corollary 4.4

Let M be an n-dimensional (n ⩾2) submanifold in a non-Sasakian space form M̂(c) endowed with a semi-symmetric metric connection such that ξTM. Then for each point pM and for each unit vector XTpM, we have $Ric(X)⩽n24∥H∥2+(n−1)(−κ+1)2−3(κ+1)2∥PX∥2−1−3κ2[1+(n−2)η(X)2]+12[∥(φhX)T∥2−∥(hX)T∥2−g(φhX,X)trace(φh)T+g(hX,X)trace(hT)]+(κ+n−2)g(hX,X)+[1+κη(X)2]trace(hT)−(n−2)α(X,X)−λ.$

#### Theorem 4.5

Let M be an n-dimensional (n ⩾ 3) submanifold in a (2m+1)-dimensional (κ, μ)-contact space form M̂(c) endowed with a semi-symmetric connection such that ξTM. Then we have $n(n−1)∥H∥2⩾n(n−1)Θk(p)−14{n(n−1)(c+3)+3(c−1)∥P∥2−2(n−1)(c+3−4κ)}−12{∥(φh)T∥2−∥hT∥2−(trace(φ(h)T))2+(trace(hT))2}−2[μ+(n−2)]trace(hT)+2(n−1)λ.$

#### Proof

Let {e1, …, en} be an orthonormal basis of TpM. We denote by Li1, …, ik the k-plane section spanned by ei1, …, eik. From (4) and (5), it follows that $τ(Li1,...,eik)=12∑i∈{i1,...,ik}RicLi1,...,ik(ei)$(22)

and $τ(p)=1Cn−2k−2∑1⩽i1<...(23)

Combining (6), (22) and (23), we obtain $τ(p)⩾n(n−1)2Θk(p).$(24)

We choose an orthonormal basis {e1, …, en} of TpM such that en+1 is in the direction of the mean curvature vector H(p) and {e1, …, en} diagonalizes the shape operator An+1. Then the shape operators take the following forms: $An+1=a10...00a2...0.....0.....0.....000...an,$(25) $traceAr=0,r=n+2,...,2m+1.$(26)

From (10), we have $2τ=14{n(n−1)(c+3)+3(c−1)∥P∥2−2(n−1)(c+3−4κ)}+12{∥(φh)T∥2−∥hT∥2−(trace(φh)T)2+(trace(hT))2}+2[μ+(n−2)]trace(hT)−2(n−1)λ+n2∥H∥2−∥σ∥2.$(27)

Using (25) and (27), we obtain $n2∥H∥2=2τ+∑i=1nai2+∑r=n+22m+1∑i,j=1n(σijr)2−14{n(n−1)(c+3)+3(c−1)∥P∥2−2(n−1)(c+3−4κ)}−12{∥(φh)T∥2−∥hT∥2−(trace(φh)T)2+(trace(hT))2}−2[μ+(n−2)]trace(hT)+2(n−1)λ.$(28)

On the other hand from (25) and (26), we have $(n∥H∥)2=(∑ai)2⩽n∑i=1nai2.$(29)

From (28) and (29), it follows that $n(n−1)∥H∥2≥2τ−14{n(n−1)(c+3)+3(c−1)∥P∥2−2(n−1)(c+3−4κ)}−12{∥(φh)T∥2−∥hT∥2−(trace(φh)T)2+(trace(hT))2}−2[μ+(n−2)]trace(hT)+2(n−1)λ+∑r=n+22m+1∑i,j=1n(σijr)2⩾2τ−14{n(n−1)(c+3)+3(c−1)∥P∥2−2(n−1)(c+3−4κ)}−12{∥(φh)T∥2−∥hT∥2−(trace(φh)T)2+(trace(hT))2}−2[μ+(n−2)]trace(hT)+2(n−1)λ.$

Using (24), we obtain $n(n−1)∥H∥2⩾n(n−1)Θk(p)−14{n(n−1)(c+3)+3(c−1)∥P∥2−2(n−1)(c+3−4κ)}−12{∥(φh)T∥2−∥hT∥2−(trace(φh)T)2+(trace(hT))2}−2[μ+(n−2)]trace(hT)+2(n−1)λ.$

#### Corollary 4.6

Let M be an n-dimensional (n ⩾ 3) submanifold in a Sasakian space form M̂(c) endowed with a semi-symmetric connection such that ξTM. Then for each point pM and for each unit vector XTpM, we have $n(n−1)∥H∥2⩾n(n−1)Θk(p)−14{n(n−1)(c+3)+3(c−1)∥P∥2−2(n−1)(c−1)}+2(n−1)λ.$

#### Corollary 4.7

Let M be an n-dimensional (n ⩾ 3) submanifold in a non-Sasakian space form M̂(c) endowed with a semi-symmetric connection such that ξTM. Then for each point pM and for each unit vector XTpM, we have $n(n−1)∥H∥2⩾n(n−1)Θk(p)−14{2n(n−1)(−κ+1)−6(κ+1)∥P∥2+4(n−1)(3κ−1)}−12{∥(φh)T∥2−∥hT∥2−(trace(φh)T)2+(trace(hT))2}−2(κ+n−1)trace(hT)+2(n−1)λ.$

## Acknowledgement

Supported by NNSF of CHINA(11371194).

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Accepted: 2018-02-27

Published Online: 2018-04-18

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 380–391, ISSN (Online) 2391-5455,

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