Within this paper, we have formulated the first Chen inequality for bi-warped product sub-manifolds within Riemannian space forms. This inequality intricately involved extrinsic invariants such as mean curvature and the lengths of the warping functions, while also incorporating intrinsic invariants like sectional curvature and δ-invariants. Furthermore, we extensively explored and analyzed the scenarios where equality conditions were met within the context of this inequality.
Citation: Biswabismita Bag, Meraj Ali Khan, Tanumoy Pal, Shyamal Kumar Hui. First chen inequality for biwarped product submanifold of a Riemannian space forms[J]. AIMS Mathematics, 2025, 10(4): 9917-9932. doi: 10.3934/math.2025454
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Within this paper, we have formulated the first Chen inequality for bi-warped product sub-manifolds within Riemannian space forms. This inequality intricately involved extrinsic invariants such as mean curvature and the lengths of the warping functions, while also incorporating intrinsic invariants like sectional curvature and δ-invariants. Furthermore, we extensively explored and analyzed the scenarios where equality conditions were met within the context of this inequality.
The theory of submanifolds has evolved naturally from the classical study of curves and surfaces in Euclidean space, employing the tools of differential calculus. The extensive applications of extrinsic and intrinsic Riemannian invariants of some sub-manifold, span across various scientific disciplines, particularly within the realm of general relativity. The main motivation for such a study lies in establishing the relationships between these invariants.
In the work [1], Chen demonstrated an inequality that involves these invariants for sub-manifolds within Riemannian space forms. Building upon Nash's theorem, researchers have derived geometric constraints, known as intrinsic and extrinsic invariants, for warped products in diverse space forms (see [2,3,4]). Of particular interest is Chen's first invariant, a significant intrinsic measure in our inequality, which was defined in [2].
δMm(x)=τ(TxMm)−inf{K(π):π⊂TxMm,x∈Mm,dim π=2}, | (1.1) |
where K(π) is the sectional curvature of the plane section π, τ(TxMm) is the scalar curvature of TxMm.
Bishop and O'Neill [5] introduced the warped product of Riemannian manifolds to generate a broad class of complete manifolds characterized by negative curvature.
In [6], Chen and Dillen introduced the notion of multiply warped product manifolds and sub-manifolds as follows: Let M1,…,Mk be Riemannian manifolds and let
M=M1×⋯×Mk |
be the Cartesian product of M1,…,Mk. For each i, denote by
πi:M→Mi, |
the canonical projection of M onto Mi. When there is no confusion, we identify Mi with a horizontal lift of Mi in M via πi. If
f2,…,fk:M1→R+ |
are positive real-valued functions, then
g(L1,L2):=g(π1∗L1,π1∗L2)+k∑i=2(fi∘π1)2g(πi∗L1,πi∗L2) |
defines a Riemannian metric g on M, called a multiply warped product metric, for any vector fields L1,L2 on M and πi∗ denotes the push tangential map. The product manifold M endowed with this metric is denoted by
M1×f2M2×⋯×fkMk. |
In this case, the warping functions f2,…,fk are nonconstant functions on M1. It is clear that if all f1,…,fk are constant except one function fj such that 2≤j≤n, then M is a single warped product manifold. Also, if any two functions are not constant and all others are constant, then multiply warped product reduces to bi-warped product, which we discuss in this paper. Multiply warped products play crucial roles in both physics and differential geometry, particularly in the realm of relativity theory. Standard spacetime models such as Robertson-Walker and Schwarzschild are examples of warped products. Furthermore, elementary models describing the regions around stars and black holes often align with the warped product framework [7]. Additionally, many solutions to Einstein's field equations find expression in terms of warped products.[8].
In a study by [9], an enhanced form of the initial Chen inequality for Legendrian sub-manifolds within Sasakian space forms was introduced. Furthermore, the authors in [10] presented the first Chen inequality for general warped product sub-manifolds in a Riemannian space form. Similarly, in [11], Alghamdi et al. established a comparable inequality for warped product Legendrian sub-manifolds within Kenmotsu space forms. Additionally, in [12], Li et al. set forth a similar inequality for warped product sub-manifolds within Qmϵ×R. Some applications of inequalities in other fields can be found on [13].
Building upon the research mentioned above, this paper introduces the first Chen inequality for bi-warped product sub-manifolds within Riemannian space forms. Additionally, we delve into the examination of the equality case and provide insights into the applications of these inequalities within the context of this study.
In this section, we present the foundational mathematical framework required to establish the first Chen inequality for bi-warped product sub-manifolds in Riemannian space forms.
Consider the n-dimensional sub-manifold Nn of a Riemannian manifold (Mm,g) of dimension m. Let ˉ∇ denote the Levi-Civita connection on Mm, and let ∇ be the induced connection on Nn. The Gauss and Weingarten formulas, which connect the geometry of the sub-manifold to that of the ambient manifold, are given as follows:
ˉ∇L1L2=∇L1L2+h(L1,L2) | (2.1) |
and
ˉ∇L1υ=−AVL1+∇⊥L1υ, | (2.2) |
where L1,L2∈TNn,υ∈T⊥Nn, and h,∇⊥,Aυ denotes the second fundamental form, normal connection, and the shape operator, respectively.
The well-known equation of Gauss is given by
R(L1,L2,L3,L4)=ˉR(L1,L2,L3,L4)+g(h(L1,L4),h(L2,L3))−g(h(L1,L3),h(L2,L4)), | (2.3) |
for any L1,L2,L3,L4∈Γ(TMm), where ˉR and R are the curvature tensors of Mm and Nn, respectively.
If we select two linearly independent tangent vectors L1,L2∈TM, the sectional curvature of the 2-plane π spanned by L1 and L2 can be expressed in terms of the Riemannian curvature tensor ˉR as follows:
ˉK(π)=ˉK(L1∧L2)=g(ˉR(L1,L2)L2,L1)g(L1,L1)g(L2,L2)−(g(L1,L2))2. | (2.4) |
If the 2-plane π is spanned by orthogonal unit vectors L1 and L2 from the tangent space TxMm, where x∈Mm, the previous definition can be expressed as:
ˉK(π)=ˉKMm(L1∧L2)=g(ˉR(L1,L2)L2,L1). | (2.5) |
It is important to note that the sectional curvature is independent of the choice of orthonormal basis for π and fully characterizes the Riemannian curvature tensor ˉR. Furthermore, if ˉK(π) is constant for all planes π in TxMm and for all points x∈Mm, specifically
ˉK(π)=c, |
we refer to Mm(c) as a real space form.
Here, then we have:
ˉR(L1,L2)L3=c(g(L2,L3)L1−g(L1,L3)L2), | (2.6) |
for any L1,L2,L3∈Γ(TMm(c)).
Scalar curvature for Mm is defined in terms of the sectional curvature as:
ˉτ(TxMm) =∑1≤i<j≤mˉKij⟹2ˉτ(TxMn)=∑1≤i≠j≤mˉKij. | (2.7) |
An interesting invariant for some manifold is the Chen's first invariant, which is defined as:
ˉδMm(x)=ˉτ(TxMm)−inf{ˉK(π):π⊂TxMm,x∈Mm,dim π=2}. | (2.8) |
Whenever we consider the above geometric objects such as the sectional curvature, scalar curvature, Chen's first invariant, etc., for the sub-manifold Nn, we simply denote them as K,τ and δ, respectively.
In particular, the scalar curvature τ(x) of Nn at x is identical with the scalar curvature of the tangent space TxNn of Nn at x, that is,
τ(x)=τ(TxNn). |
Bi-warped product manifolds are special classes of manifolds. Let us consider,
Nn=N1×f1N2×f2N3 |
as the bi-warped product sub-manifold of the Riemannian space form Mm(c). We choose an orthonormal basis
{ϰ1,⋯,ϰn1,ϰn1+1,⋯,ϰn1+n2,ϰn1+n2+1,⋯,ϰn} |
of TxNn, where
{ϰ1,⋯,ϰn1},{ϰn1+1,⋯,ϰn1+n2},{ϰn1+n2+1,⋯,ϰn} |
are orthonormal bases of N1,N2, and N3, respectively. Let {ϰn+2,⋯,ϰm} be an orthonormal basis of T⊥xNn.
The coefficients of the second fundamental form h of Nn with respect to the above local frame are denoted as
hrij=g(h(ϰi,ϰj),ϰr), | (2.9) |
where i,j∈{1,⋯,n} and r∈{n+1,⋯,m}. The mean curvature vector H is defined with respect to the same local frame above, as
H=1nn∑i=1h(ϰi,ϰi). | (2.10) |
We say that Nn is a minimal sub-manifold of Mm if H vanishes identically.
If f is a smooth function on Mm, then its gradient ∇f and Laplacian Δf are defined as
g(∇f,X)=Xf,Δf=m∑i=1((∇ϰiϰi)f−ϰiϰif). | (2.11) |
So from Eqs (2.3), (2.4), and (2.9), we get
K(ϰi∧ϰj) =ˉK(ϰi∧ϰj)+m∑r=n+1(g(hriiϰr,hrjjϰr)−g(hrijϰr,hrijϰr))⟹K(ϰi∧ϰj)=ˉK(ϰi∧ϰj)+m∑r=n+1(hriihrjj−(hrij)2), | (2.12) |
where ˉK(ϰi∧ϰj) denotes the sectional curvature of the 2-plane spanned by ϰi and ϰj at x in the ambient manifold Mm(c).
Taking the summation over the orthonormal frame of the tangent space of Nn in (2.12), we have
2τ(TxNn)=2ˉτ(TxNn)+n2||H||2−||h||2. | (2.13) |
The sectional curvature and warping functions are related by the following formulas [14]:
(ⅰ) ∑n1a=1∑n1+n2A=n1+1KaA=n2Δf3f3;
(ⅱ) ∑n1a=1∑nb=n1+n2+1Kab=n3Δf4f4;
(ⅲ) ∑n1+n2A=n1+1∑nb=n1+n2+1KAb=−n2n3f3f4g3(∇f3,∇f4).
So, we have
τ(TxNn)=∑1≤i<j≤nKij=n1∑a=1n1+n2∑A=n1+1KaA+n1∑a=1n∑b=n1+n2+1Kab+n1+n2∑A=n1+1n∑b=n1+n2+1KAb+∑1≤a<a′≤n1Kaa′+∑n1+1≤A<A′≤n1+n2KAA′+∑n1+n2+1≤b<b′≤nKbb′⟹τ(TxNn)=n2Δ1f1f1+n3Δ1f2f2−n2n3g(∇1(lnf1),∇1(lnf2))+τ(TxN1)+τ(TxN2)+τ(TxN3). | (2.14) |
The following lemma will also be needed to prove our main result.
Lemma 2.1. [10] Let α1,α2,⋯,αn,β be (n+1)(n≥2) real numbers such that
(n∑i=1αi)2=(n−1)(n∑i=1α2i+β), |
then 2α1α2≥β, with equality holds if, and only if,
α1+α2=α3=⋯=αn. |
Theorem 3.1. Let
ϕ:Nn=N1×f1N2×f2N3→Mm(c) |
be an isometric immersion of a bi-warped product sub-manifold Nn into a Riemannian space form Mm(c). Then, for each point x∈Nn and each plane section
πi⊂TxNnii, ni=dim Ni≥2 |
for i=1,2,3, we have:
(1) If π1⊂TxN1, then
δN1n1≤n22||H||2−n2Δf1f1−n3Δf2f2+n2n3g(∇(lnf1),∇(lnf2))+n12(n1+2n2+2n3−1)c+n2n3c−c. |
(2) If π2⊂TxN2, then
δN2n2≤n22||H||2−n2Δf1f1−n3Δf2f2+n2n3g(∇(lnf1),∇(lnf2))+n22(n2+2n1+2n3−1)c+n1n3c−c. |
(3) If π3⊂TxN3, then
δN3n3≤n22||H||2−n2Δf1f1−n3Δf2f2+n2n3g(∇(lnf1),∇(lnf2))+n32(n3+2n1+2n2−1)c+n1n2c−c. |
the equality holds at x∈Nn if, and only if, there exist an orthonormal basis {ϰ1,⋯,ϰn} of TxNn and an orthonormal basis {ϰn+1,⋯,ϰn} of T⊥xNn such that:
(a) π1=span{ϰ1.ϰ2},π2=span{ϰn1+1,ϰn1+2},π3=span{ϰn1+n2+1,n1+n2+2}.
(b) The shape operator takes the following forms, where Om×n, denoting the zero matrix of order m×n:
[i] If π1⊂TxN1, then for r=n+1, we have
Aϰn+1=(BLOCK(I)On1×n2On1×n3On2×n1BLOCK(II)On2×n3On3×n1On3×n2BLOCK(III)), |
where BLOCK(I),BLOCK(II), and BLOCK(III) are given, respectively, as follows:
(μ1hn+1120⋯01n1hn+121μ20⋯02n100μ⋯⋮⋮⋮⋮⋱⋮0(n1)(1)⋯⋯⋯μ),(hn+1(n1+1)(n1+1)⋯⋯hn+1(n1+1)(n1+n2)⋮⋯⋮⋮⋯⋮⋮⋯⋮hn+1(n1+n2)(n1+1)⋯hn+1(n1+n2)(n1+n2)),(hn+1(n1+n2+1)(n1+n2+1)⋯⋯hn+1(n1+n2+1)(n)⋮⋯⋮⋮⋯⋮⋮⋯⋮hn+1(n)(n1+n2+1)⋯hn+1(n)(n)), |
where
μ=μ1+μ2=hn+111+hn+122. |
Also, for r∈{n+2,⋯,m},
Aϰr=(BLOCK(I)On1×n2On1×n3On2×n1BLOCK(II)On2×n3On3×n1On3×n2BLOCK(III)), |
where BLOCK(I),BLOCK(II) and BLOCK(III) are given, respectively, as
(hr11hr120⋯01n1hr21−hr110⋯02n100033⋯⋮⋮⋮⋮⋱⋮0(n1)(1)⋯⋯⋯0n1n1),(hr(n1+1)(n1+1)⋯⋯hr(n1+1)(n1+n2)⋮⋯⋮⋮⋯⋮⋮⋯⋮h(n1+n2)(n1+1)cr⋯hr(n1+n2)(n1+n2)),(hr(n1+n2+1)(n1+n2+1)⋯⋯hr(n1+n2+1)(n)⋮⋯⋮⋮⋯⋮⋮⋯⋮hr(n)(n1+n2+1)⋯hr(n)(n)). |
[ii] If π2⊂TxN2, then for r=n+1, we have
Aϰn+1=(BLOCK(I)On1×n2On1×n3On2×n1BLOCK(II)On2×n3On3×n1On3×n2BLOCK(III)), |
where BLOCK(I),BLOCK(II), and BLOCK(III) are given, respectively, as follows:
(hn+111⋯⋯hn+1(1)(n1)⋮⋯⋮⋮⋯⋮⋮⋯⋮hn+1(n1)(1)⋯hn+1(n1)(n1)),(μ1hn+1(n1+1)(n1+2)0⋯0(n1+1)(n1+n2)hn+1(n1+2)(n1+1)μ20⋯0(n1+2)(n1+n2)00μ⋯⋮⋮⋮⋮⋱⋮0(n1+n2)(n1+1)⋯⋯⋯μ),(hn+1(n1+n2+1)(n1+n2+1)⋯⋯hn+1(n1+n2+1)(n)⋮⋯⋮⋮⋯⋮⋮⋯⋮hn+1(n)(n1+n2+1)⋯hn+1(n)(n)), |
where
μ=μ1+μ2=hn+1(n1+1)(n1+1)+hn+1(n1+2)(n1+2). |
Also for r∈{n+2,⋯,m},
Aϰr=(BLOCK(I)On1×n2On1×n3On2×n1BLOCK(II)On2×n3On3×n1On3×n2BLOCK(III)), |
where BLOCK(I),BLOCK(II), and BLOCK(III) are given, respectively, as
(hr11⋯⋯hr(1)(n1)⋮⋯⋮⋮⋯⋮⋮⋯⋮hr(n1)(1)⋯hr(n1)(n1)),(hr(n1+1)(n1+1)hr(n1+1)(n1+2)0⋯0(n1+1)(n1+n2)hr(n1+2)(n1+1)−hr(n1+1)(n1+1)0⋯0(n1+2)(n1+n2)000(n1+3)(n1+3)⋯⋮⋮⋮⋮⋱⋮0(n1+n2)(n1+1)⋯⋯⋯0(n1+n2)(n1+n2)),(hr(n1+n2+1)(n1+n2+1)⋯⋯hr(n1+n2+1)(n)⋮⋯⋮⋮⋯⋮⋮⋯⋮hr(n)(n1+n2+1)⋯hr(n)(n)). |
[iii] If π3⊂TxN3, then for r=n+1, we have
Aϰn+1=(BLOCK(I)On1×n2On1×n3On2×n1BLOCK(II)On2×n3On3×n1On3×n2BLOCK(III)), |
where BLOCK(I),BLOCK(II), and BLOCK(III) are given, respectively, as follows:
(hn+111⋯⋯hn+1(1)(n1)⋮⋯⋮⋮⋯⋮⋮⋯⋮hn+1(n1)(1)⋯hn+1(n1)(n1)),(hn+1(n1+1)(n1+1)⋯⋯hn+1(n1+1)(n1+n2)⋮⋯⋮⋮⋯⋮⋮⋯⋮hn+1(n1+n2)(n1+1)⋯hn+1(n1+n2)(n1+n2)),(μ1hn+1(n1+n2+1)(n1+n2+2)0⋯0(n1+n2+1)(n)hn+1(n1+n2+2)(n1+n2+1)μ20⋯0(n1+n2+2)(n)00μ⋯⋮⋮⋮⋮⋱⋮0(n)(n1+n2+1)⋯⋯⋯μ), |
where
μ=μ1+μ2=hn+1(n1+n2+1)(n1+n2+1)+hn+1(n1+n2+2)(n1+n2+2). |
For r∈{n+2,⋯,m},
Aϰr=(BLOCK(I)On1×n2On1×n3On2×n1BLOCK(II)On2×n3On3×n1On3×n2BLOCK(III)), |
where BLOCK(I),BLOCK(II), and BLOCK(III) are given, respectively, as
(hr11⋯⋯hr(1)(n1)⋮⋯⋮⋮⋯⋮⋮⋯⋮hr(n1)(1)⋯hr(n1)(n1)),(hr(n1+1)(n1+1)⋯⋯hr(n1+1)(n1+n2)⋮⋯⋮⋮⋯⋮⋮⋯⋮h(n1+n2)(n1+1)cr⋯hr(n1+n2)(n1+n2)),(hr(n1+n2+1)(n1+n2+1)hr(n1+n2+1)(n1+n2+2)0⋯0(n1+n2+1)(n)hr(n1+n2+2)(n1+n2+1)−hr(n1+n2+1)(n1+n2+1)0⋯0(n1+n2+2)(n)000(n1+n2+3)(n1+n2+3)⋯⋮⋮⋮⋮⋱⋮0(n)(n1+n2+1)⋯⋯⋯0nn). |
(4) If equality of (i) or (ii) or (iii) holds, then
N1×f1N2×f2N3 |
is mixed totally geodesic in Mm(c). Moreover,
N1×f1N2×f2N3 |
is both D1,D2, and D3-minimal. Thus,
N1×f1N2×f2N3 |
is a minimal bi-warped product sub-manifold of Mm(c).
Proof. For simplicity we are providing the complete proof of (1). The other two parts viz., (2) and (3) can be proved in a similar way together with the equality cases.
We start our proof by considering a point x∈Nn and let π1⊂TxN1 be a 2-plane. We choose an orthonormal basis
{ϰ1,⋯,ϰn1,ϰn1+1,⋯,ϰn1+n2,ϰn1+n2+1,⋯,ϰn} |
of TxNn, where
{ϰ1,⋯,ϰn1},{ϰn1+1,⋯,ϰn1+n2},{ϰn1+n2+1,⋯,ϰn} |
are orthonormal bases of N1,N2, and N3, respectively, and {ϰn+1,⋯,ϰm} is an orthonormal basis of T⊥xNn. First, put
π1=span{ϰ1,ϰ2} |
such that the normal vector ϰn+1 is in the direction of the mean curvature vector H. We have from (2.3) and (2.6)
n2||H||2=2τ(TxMn)+||h||2−n(n−1)c, |
which gives
(n1∑a=1hn+1aa)2=2τ(TxMn)+||h||2−n(n−1)c−(n1+n2∑A=n1+1hn+1AA)2−(n∑b=n1+n2+1hn+1bb)2−2n1∑a=1n1+n2∑A=n1+1hn+1aahn+1AA−2n1∑a=1n∑b=n1+n2+1hn+1aahn+1bb2n1+n2∑A=n1+1n∑b=n1+n2+1hn+1AAhn+1bb. |
Assume
τ1=2τ(TxMn)−n1−2n1−1(n1∑a=1hn+1aa)2−(n1+n2∑A=n1+1hn+1AA)2−(n∑b=n1+n2+1hn+1bb)2−2∑a∑Ahn+1aahn+1AA−2∑a∑bhn+1aahn+1bb−2∑A∑bhn+1AAhn+1bb−n(n−1)c. |
Thus,
(n1−1)τ1 =(n1−1)[2τ(TxMn)−(n1−1)−1n1−1(n1∑a=1hn+1aa)2−(n1+n2∑A=n1+1hn+1AA)2−(n∑b=n1+n2+1hn+1bb)2−2∑a∑Ahn+1aahn+1AA−2∑a∑bhn+1aahn+1bb−2∑A∑bhn+1AAhn+1bb−n(n−1)c]⟹(n1−1)τ1=(n1−1)[2τ(TxMn)−(n1∑a=1hn+1aa)2−(n1+n2∑A=n1+1hn+1AA)2−(n∑b=n1+n2+1hn+1bb)2−2∑a∑Ahn+1aahn+1AA−2∑a∑bhn+1aahn+1bb−2∑A∑bhn+1AAhn+1bb−n(n−1)c]−(n1∑a=1hn+1aa)2⟹(n1−1)τ1=(n1−1)[−||h||2]−(n1∑a=1hn+1aa)2⟹(n1∑a=1hn+1aa)2=(n1−1)[τ1+||h||2]⟹(n1∑a=1hn+1aa)2=(n1−1)[τ1+n∑i,j=1(hn+1ij)2+m∑r=n+2n∑i,j=1(hrij)2]⟹(n1∑a=1hn+1aa)2=(n1−1)[τ1+n1∑a=1(hn+1aa)2+n1+n2∑A=n1+1(hn+1AA)2+n∑b=n1+n2+1(hn+1bb)2+n∑i,j=1i≠j(hn+1ij)2+m∑r=n+2n∑i,j=1(hrij)2]. |
From Lemma 2.1 with
αa=hn+1aa,β=[τ1+n1∑a=1(hn+1aa)2+n1+n2∑A=n1+1(hn+1AA)2+n∑b=n1+n2+1(hn+1bb)2+n∑i,j=1i≠j(hn+1ij)2+m∑r=n+2n∑i,j=1(hrij)2], |
we have
hn+111hn+122≥12[τ1+n1∑a=1(hn+1aa)2+n1+n2∑A=n1+1(hn+1AA)2+n∑b=n1+n2+1(hn+1bb)2+n∑i,j=1i≠j(hn+1ij)2+m∑r=n+2n∑i,j=1(hrij)2]. |
Assume
π1=<ϰ1,ϰ2>. |
Hence,
K(π1) =c+m∑r=n+1(hr11hr22−(hr12)2)⟹K(π1)=c+hn+111hn+122+m∑r=n+2hr11hr22−m∑r=n+1(hr12)2⟹K(π1)≥c+12[τ1+n1+n2∑A=n1+1(hn+1AA)2+n∑b=n1+n2+1(hn+1bb)2+n∑i,j=1i≠j(hn+1ij)2+m∑r=n+2n∑i,j=1(hrij)2]−m∑r=n+1(hr12)2+m∑r=n+2hr11hr22. |
Arranging the terms of the right hand side in the last inequality, we derive
K(π1)≥c+12τ1+12n1+n2∑A=n1+1(hn+1AA)2+12n∑b=n1+n2+1(hn+1bb)2+m∑r=n+2hr11hr22−m∑r=n+1(hr12)2+n∑i,j=1i≠j(hn+1ij)2+m∑r=n+2n∑i,j=1(hrij)2. |
Applying [10, Lemma 2] with the last four terms of the above inequality, we find that
K(π1)≥c+12τ1+12n1+n2∑A=n1+1(hn+1AA)2+12n∑b=n1+n2+1(hn+1bb)2+12n∑i,j=3i≠j(hn+1ij)2+12m∑r=n+2n∑i,j=3(hrij)2+12(hr11+hr22)2+m∑r=n+1n∑j+3((hr1j)2+(hr2j)2)⟹K(π1)≥c+12τ1+12n1+n2∑A=n1+1(hn+1AA)2+12n∑b=n1+n2+1(hn+1bb)2+12n∑i,j=3i≠j(hn+1ij)2+12m∑r=n+2n∑i,j=3(hrij)2, |
and substituting
12τ1=τ(TxMn)−(n1∑a=1hn+1aa)2+12(n1−1)(n1∑a=1hn+1aa)2−12(n1+n2∑A=n1+1hn+1AA)2−12(n∑b=n1+n2+1hn+1bb)2−∑a∑Ahn+1aahn+1AA−∑a∑bhn+1aahn+1bb−∑A∑bhn+1AAhn+1bb−n(n−1)c, |
we get
K(π1) ≥c+τ(TxMn)−(n1∑a=1hn+1aa)2+12(n1−1)(n1∑a=1hn+1aa)2−12(n1+n2∑A=n1+1hn+1AA)2−12(n∑b=n1+n2+1hn+1bb)2−∑a∑Ahn+1aahn+1AA−∑a∑bhn+1aahn+1bb−∑A∑bhn+1AAhn+1bb−n(n−1)c+12n1+n2∑A=n1+1(hn+1AA)2+12n∑b=n1+n2+1(hn+1bb)2+12n∑i,j=3i≠j(hn+1ij)2+12m∑r=n+2n∑i,j=3(hrij)2⟹K(π1)≥c+τ(TxMn)+12(n1−1)(n1∑a=1hn+1aa)2−n(n−1)c+12n1+n2∑A=n1+1(hn+1AA)2+12n∑b=n1+n2+1(hn+1bb)2+12n∑i,j=3i≠j(hn+1ij)2+12m∑r=n+2n∑i,j=3(hrij)2−n22||H||2. |
Using (2.14), we have
K(π1)≥c+n2Δ1f1f1+n3Δ1f2f2−n2n3g(∇1(lnf1),∇1(lnf2))+τ(TxN1)+τ(TxN2)+τ(TxN3)+12(n1−1)(n1∑a=1hn+1aa)2−n(n−1)c+12n1+n2∑A=n1+1(hn+1AA)2+12n∑b=n1+n2+1(hn+1bb)2+12n∑i,j=3i≠j(hn+1ij)2+12m∑r=n+2n∑i,j=3(hrij)2−n22||H||2⟹τ(TxN1)−K(π1)≤n22||H||2−[c+n2Δ1f1f1+n3Δ1f2f2−n2n3g(∇1(lnf1),∇1(lnf2))+τ(TxN2)+τ(TxN3)+12(n1−1)(n1∑a=1hn+1aa)2−n(n−1)c+12n1+n2∑A=n1+1(hn+1AA)2+12n∑b=n1+n2+1(hn+1bb)2+12n∑i,j=3i≠j(hn+1ij)2+12m∑r=n+2n∑i,j=3(hrij)2]. |
From (2.3), we have
−2τ2(TxN2)=−2ˉτ2(TxN2)+m∑r=n+1n1+n2∑A,A′=n1+1(hrAA′)2−m∑r=n+1(hr(n1+1)(n1+1)+⋯+hr(n1+n2)(n1+n2))2 |
and
−2τ3(TxN3)=−2ˉτ3(TxN3)+m∑r=n+1n∑b,b′=n1+n2+1(hrbb′)2−m∑r=n+1(hr(n1+n2+1)(n1+n2+1)+⋯+hrnn)2. |
So,
τ(TxN1)−K(π1) ≤n22||H||2−n2Δ1f1f1−n3Δ1f2f2+n2n3g(∇1(lnf1),∇1(lnf2))+(n22−n2−1)c−ˉτ2(TxN2)−ˉτ3(TxN3)−12n∑i,j=3(hn+2ij)2−12m∑r=n+2n∑i,j=3(hrij)2−12∑A(hn+1AA)2−12∑b(hn+1bb)2+12m∑r=n+1n1+n2∑A,A′=n1+1(hrAA′)2+12m∑r=n+1n∑b,b′=n1+n2+1(hrbb′)2⟹τ(TxN1)−K(π1)≤n22||H||2−n2Δ1f1f1−n3Δ1f2f2+n2n3g(∇1(lnf2),∇1(lnf2))+(n22−n2−1)c−ˉτ2(TxN2)−ˉτ3(TxN3)−12n1∑a,a′=3a≠a′(hn+1aa′)2−12×2n1∑a=3n1+n2∑A=n1+1(hn+1aA)2−12×2n1∑a=3n∑b=n1+n2+1(hn+1ab)2−12×2n1+n2∑A=n1+1n∑b=n1+n2+1(hn+1Ab)2−12m∑r=n+2n1∑a,a′=3(hraa′)2−12×2m∑r=n+2∑a∑A(hraA)2−12×2m∑r=n+2∑a∑b(hrab)2−12×2m∑r=n+2∑A∑b(hrAb)2. |
Putting
n=(n1+n2+n3), ˉτ2(TxN2)=n2(n2−1)2 |
and
ˉτ3(TxN3)=n3(n3−1)2, |
we have our inequality. Clearly, equalities hold if, and only if:
(ⅰ)
α1+α2=α3=⋯=αn, |
which implies
hn+111+hn+122=hn+133=⋯=hn+1n1n1,μ=μ1+μ2=hn+111+hn+122. |
(ⅱ)
m∑r=n+2(hr11+hr22)2+m∑r=n+1n∑j=3((hr1j)2+(hr2j)2)=0. |
(ⅲ)
(n1∑a=1hn+1aa)2=m∑r=n+1(hr(n1+1)(n1+1)+⋯+hr(n1+n2)(n1+n2))2=m∑r=n+1(hr(n1+n2+1)(n1+n2+1)+⋯+hrnn)2. |
(ⅳ)
n1∑a,a′=3a≠a′(hn+1aa′)2+n1∑a=3n1+n2∑A=n1+1(hn+1aA)2+n1∑a=3n∑b=n1+n2+1(hn+1ab)2+n1+n2∑A=n1+1n∑b=n1+n2+1(hn+1Ab)2+m∑r=n+2n1∑a,a′=3(hraa′)2+m∑r=n+2n1∑a=3n1+n2∑A=n1+1(hraA)2+m∑r=n+2n1∑a=3n∑b=n1+n2+1(hrab)2+m∑r=n+2n1+n2∑A=n1+1n∑b=n1+n2+1(hrAb)2=0. |
That is,
Aϰn+1=(BLOCK(I)On1×n2On1×n3On2×n1BLOCK(II)On2×n3On3×n1On3×n2BLOCK(III)), |
where BLOCK(I),BLOCK(II), and BLOCK(III) are given, respectively, as follows:
(μ1hn+1120⋯01n1hn+121μ20⋯02n100μ⋯⋮⋮⋮⋮⋱⋮0(n1)(1)⋯⋯⋯μ),(hn+1(n1+1)(n1+1)⋯⋯hn+1(n1+1)(n1+n2)⋮⋯⋮⋮⋯⋮⋮⋯⋮hn+1(n1+n2)(n1+1)⋯hn+1(n1+n2)(n1+n2)),(hn+1(n1+n2+1)(n1+n2+1)⋯⋯hn+1(n1+n2+1)(n)⋮⋯⋮⋮⋯⋮⋮⋯⋮hn+1(n)(n1+n2+1)⋯hn+1(n)(n)). |
For r∈{n+2,⋯,m}, since
hr11+hr22=n∑j=3hr1j=n∑j=3hr2j=n1∑a,a′=3hraa′=∑a∑AhraA=∑a∑bhrab=∑A∑bhrAb=0, |
Aϰr=(BLOCK(I)On1×n2On1×n3On2×n1BLOCK(II)On2×n3On3×n1On3×n2BLOCK(III)), |
where BLOCK(I),BLOCK(II), and BLOCK(III) are given, respectively, as
(hr11hr120⋯01n1hr21−hr110⋯02n100033⋯⋮⋮⋮⋮⋱⋮0(n1)(1)⋯⋯⋯0n1n1),(hr(n1+1)(n1+1)⋯⋯hr(n1+1)(n1+n2)⋮⋯⋮⋮⋯⋮⋮⋯⋮h(n1+n2)(n1+1)cr⋯hr(n1+n2)(n1+n2)),(hr(n1+n2+1)(n1+n2+1)⋯⋯hr(n1+n2+1)(n)⋮⋯⋮⋮⋯⋮⋮⋯⋮hr(n)(n1+n2+1)⋯hr(n)(n)). |
This completes the inequality part of (1).
If equality of (ⅰ) or (ⅱ) or (ⅲ) holds, Nn is mixed totally geodesic in Mm(c). Moreover, Nn is N1-minimal, N2-minimal, and N3-minimal. Thus, Nn is a minimal warped product sub-manifold in the Riemannian space form Mm(c).
The above theorem provides partial answer to the Chen problem of finding a necessary condition for the bi-warped product sub-manifold of a Riemannian space form to be minimal. The conditions are stated in the following corollaries.
Corollary 3.1. Let
ϕ:Nn=N1×f1N2×f2N3 |
be an isometric immersion of a bi-warped product sub-manifold Nn into a Riemannian space form Mm(c). Then, for each point x∈Nn,
δN1n1+n2Δf1f1+n3Δf2f2≤n2n3g(∇(lnf1),∇(lnf2))+n12(n1+2n2+2n3−1)c+n2n3c−c, |
and if the equality holds, then the immersion ϕ is minimal.
Corollary 3.2. Let
ϕ:Nn=N1×f1N2×f2N3 |
be an isometric immersion of a bi-warped product sub-manifold Nn into a Riemannian space form Mm(c). Then, for each point x∈Nn,
δN2n2+n2Δf1f1+n3Δf2f2≤n2n3g(∇(lnf1),∇(lnf2))+n22(n2+2n1+2n3−1)c+n1n3c−c, |
and if the equality holds, then the immersion ϕ is minimal.
Corollary 3.3. Let
ϕ:Nn=N1×f1N2×f2N3 |
be an isometric immersion of a bi-warped product sub-manifold Nn into a Riemannian space form Mm(c). Then, for each point x∈Nn,
δN3n3+n2Δf1f1+n3Δf2f2≤n2n3g(∇(lnf1),∇(lnf2))+n32(n3+2n1+2n2−1)c+n1n2c−c, |
and if the equality holds, then the immersion ϕ is minimal.
If the warping function is f1=1, then the first warped product becomes an ordinary product and a whole Nn becomes a simply warped product sub-manifold of Riemannian space form Mm(c). So, Theorem 3.1 gives the results of [10]. The main limitation for our article is that we only investigated on Riemannian space forms without any additional structures on it. It should be noted that imposing additional structures like Sasakian, Kenmotsu, etc. may extend our results.
The Chen delta invariant has applications in physics, particularly in the study of topological field theories. In algebraic topology it measures the extent to which a loop in space fails to be the boundary of a surface. If a loop is the boundary of a surface, then the Chen delta invariant is zero [11]. We investigated the effect of warping functions on the Chen's delta invariant for bi-warped product sub-manifolds on Riemannian space forms. So finding the Chen inequality for various space forms with additional structures will be an interesting area to discover.
B. Bag: conceptualization, methodology, formal analysis, investigation, writing---original draft preparation, writing---review and editing; M. Ali Khan: validation, writing---original draft preparation, writing---review and editing, project administration, funding acquisition; T. Pal: methodology, validation, formal analysis; S. K. Hui: conceptualization, validation, investigation, supervision, project administration. All authors have read and agreed to the published version of the manuscript.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
This work was supported and funded by the Deanship of Scientific Research at Imam Mohammad Ibn Saud Islamic University (IMSIU) (grant number IMSIU-DDRSP2502).
The authors declare no conflicts of interest.
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