Generalized Semi-Symmetric Non-Metric Connections of Non-Integrable Distributions

S S symmetry

Article Generalized Semi-Symmetric Non-Metric Connections of Non-Integrable Distributions

Tong Wu and Yong Wang *

School of Mathematics and Statistics, Northeast Normal University, Changchun 130024, China; [email protected] * Correspondence: [email protected]

Abstract: In this work, the cases of non-integrable distributions in a Riemannian manifold with the ﬁrst generalized semi-symmetric non-metric connection and the second generalized semi-symmetric non-metric connection are discussed. We obtain the Gauss, Codazzi, and Ricci equations in both cases. Moreover, Chen’s inequalities are also obtained in both cases. Some new examples based on non-integrable distributions in a Riemannian manifold with generalized semi-symmetric non-metric connections are proposed.

Keywords: non-integrable distributions; semi-symmetric non-metric connections; Chen’s inequalities; Einstein distributions; distributions with constant scalar curvature

MSC: 53C40; 53C42

1. Introduction In [1], the notion of a semi-symmetric metric connection on a Riemannian manifold was introduced by H. A. Hayden. Some properties of a Riemannian manifold endowed with a semi-symmetric metric connection were studied by K. Yano [2]. Later, the properties Citation: Wu, T.; Wang, Y. of the curvature tensor of a semi-symmetric metric connection in a Sasakian manifold Generalized Semi-Symmetric were also investigated by T. Imai [3,4]. Z. Nakao [5] studied the Gauss curvature equation Non-Metric Connections of and the Codazzi–Mainardi equation with respect to a semi-symmetric metric connection Non-Integrable Distributions. Symmetry 2021, 13, 79. https:// on a Riemannian manifold and a submanifold. The idea of studying the tangent bundle doi.org/10.3390/sym13010079 of a hypersurface with semi-symmetric metric connections was presented by Gozutok and Esin [6]. In [7], Demirbag investigated the properties of a weakly Ricci-symmetric Received: 16 December 2020 manifold admitting a semi-symmetric metric connection. N. S. Agashe and M. R. Chafle Accepted: 2 January 2021 showed some properties of submanifolds of a Riemannian manifold with a semi-symmetric Published: 5 January 2021 non-metric connection in [8,9]. In [10,11], the study of non-integrable distributions, as a generalized version of distributions, was initiated by Synge. In [12], a regular distribution Publisher’s Note: MDPI stays neu- was shown in a Riemannian manifold. tral with regard to jurisdictional clai- Besides this, in [13–15], an important inequality was established by B. Y. Chen, called ms in published maps and institutio- Chen inequality. In geometry, by studying different submanifolds in various ambient nal affiliations. spaces, we can obtain similar results. In [16,17], Mihai and O¨ zgu¨ presented the relationships between the mean curvature associated with the semi-symmetric metric connection, scalar, and sectional curvatures and the k-Ricci curvature. In this paper, we obtain the Chen inequalities of non-integrable distributions of real-space forms endowed with the first Copyright: © 2021 by the authors. Li- censee MDPI, Basel, Switzerland. generalized semi-symmetric non-metric connection and the second generalized semi- This article is an open access article symmetric non-metric connection. distributed under the terms and con- In the literature, we find several works that were conducted with Einstein manifolds ditions of the Creative Commons At- and manifolds involving a constant scalar curvature. In [18], Dobarro and Unal studied tribution (CC BY) license (https:// Ricci-flat and Einstein Lorentzian multiply-warped products and constant scalar curvatures creativecommons.org/licenses/by/ for this class of warped products. In [19–21], the authors obtained some results with Einstein 4.0/). warped-product manifolds with a semi-symmetric non-metric connection.

Symmetry 2021, 13, 79. https://doi.org/10.3390/sym13010079 https://www.mdpi.com/journal/symmetry Symmetry 2021, 13, 79 2 of 19

In Section2, we obtain the Gauss, Codazzi, and Ricci equations for non-integrable distributions with the ﬁrst generalized semi-symmetric non-metric connection by establishing the Gauss formula and the Weingarten formula. Meanwhile, the result of the Chen inequality is presented. In Section3, we obtain the Gauss, Codazzi, and Ricci equations for non-integrable distributions by establishing the Gauss formula and the Weingarten formula and the second generalized semi-symmetric non-metric connection. Meanwhile, we obtain the result of the Chen inequality. Finally, in Section4, some examples based on non-integrable distributions in a Riemannian manifold with generalized semi-symmetric non-metric connections are presented.

2. Non-Integrable Distributions with the First Generalized Semi-Symmetric Non-Metric Connection Let (M, g) be a m-dimensional smooth Riemannian manifold, where g is the Rieman- nian metric and ∇ is the Levi–Civita connection on (M, g). For X, Y ∈ Γ(M), denote ∇XY the covariant derivative of Y with respect to X and represent by Γ(M) the C∞(M)-module of vector ﬁelds on M.

Deﬁnition 1. If there are X, Y ∈ Γ(D) such that [X, Y] is not in Γ(D), we say that D is a non-integrable distribution, where D is a sub-bundle of the tangent bundle TM with a constant rank n and Γ(D) is the space of sections of D.

⊥ Let gD be a metric tensor ﬁeld in the distribution D and let gD be a metric tensor ⊥ ﬁeld in the orthogonal distribution to D, such that g = gD ⊕ gD .

⊥ Deﬁnition 2. Let πD : TM → D, πD : TM → D⊥ be the projections associated to the tangent D D D D D⊥ D⊥ bundle TM; then, ∇X Y = π (∇XY) and [X, Y] = π ([X, Y]) and [X, Y] = π ([X, Y]) for any X, Y ∈ Γ(D).

By [12], we obtain

D D D D ∇ f XY = f ∇X Y, ∇X ( f Y) = X( f )Y + f ∇X Y, (1) where X, Y ∈ Γ(D) and f ∈ C∞(M).

D D D D D⊥ ∇X g = 0, T(X, Y) := ∇X Y − ∇Y X − [X, Y] = −[X, Y] , (2)

and D ∇XY = ∇X Y + B(X, Y). (3) D⊥ where B(X, Y) = π ∇XY and B(X, Y) 6= B(Y, X).

Definition 3. For any V ∈ Γ(TM), let ω be a 1-form satisfying ω(V) = g(U, V), here U ∈ ∞ Γ(TM) is a vector field. Let λ1, λ2 ∈ C (M), we give the definition of the first generalized semi-symmetric non-metric connection on M

∇e XY = ∇XY + λ1ω(Y)X − λ2g(X, Y)U. (4)

⊥ ⊥ ⊥ Let UD = πDU and UD = πD U; then, U = UD + UD .

Deﬁnition 4. Let

D D D D⊥ ∇e XY = ∇e X Y + Be(X, Y), ∇e X Y = π ∇e XY, Be(X, Y) = π ∇e XY. (5) Symmetry 2021, 13, 79 3 of 19

Then,

D D D D⊥ ∇e X Y = ∇X Y + λ1ω(Y)X − λ2g(X, Y)U , Be(X, Y) = B(X, Y) − λ2g(X, Y)U , (6)

where Be(X, Y) is called the second fundamental form with the ﬁrst generalized semi-symmetric non-metric connection.

Then, by (2) and (6), we obtain

D D ∇e X (g )(Y, Z) = (λ2 − λ1)[g(X, Z)ω(Y) + g(X, Y)ω(Z)], (7) D D⊥ Te (X, Y) = −[X, Y] + λ1[ω(Y)X − ω(X)Y].

If D = TM, we obtain the following results:

Theorem 1. If a linear connection ∇e D : Γ(D) × Γ(D) → Γ(D) on D satisﬁes Equation (7), then this connection is unique.

1 n We choose {E1, ... , En} as an orthonormal basis of D and let He = n ∑i=1 Be(Ei, Ei) ∈ ⊥ 1 n Γ(D ) be the mean curvature vector associated to ∇e on D. Similarly, let H = n ∑i=1 B(Ei, Ei); D⊥ then, He = H − λ2U . If He = 0, we say that D is minimal with the ﬁrst generalized semi- symmetric non-metric connection ∇e . If ∇e γ˙ γ˙ = 0, we say that curve γ is ∇e -geodesic. If every ∇e -geodesic with an initial condition in D is contained in D, we say that D is totally geodesic with the ﬁrst generalized semi-symmetric non-metric connection ∇e . 1 1 Let h(X, Y) = 2 [B(X, Y) + B(Y, X)] and eh(X, Y) = 2 [Be(X, Y) + Be(Y, X)]; then, according to [12], we obtain the following:

Proposition 1. (1) If D is totally geodesic with respect to the ﬁrst generalized semi-symmetric non-metric connection ∇e , then Be(X, Y) is dissymmetrical. (2) When U ∈ Γ(D),H = H,Ue ∈ Γ(D), and vice versa. (3) If h = HgD (or eh = Hge D), then D is umbilical with respect to ∇ (resp. ∇e ).

Proposition 2. If D is umbilical with respect to ∇, then D is umbilical with respect to ∇e , and vice versa.

D⊥ D⊥ Proof. For X, Y ∈ D, by eh(X, Y) = h(X, Y) − λ2g(X, Y)U and He = H − λ2U , D D D⊥ D then Hge (X, Y) = Hg (X, Y) − λ2U g (X, Y). Therefore, we obtain Proposition 1. Thus, by Deﬁnition 4, we obtain

∇e Xη = ∇Xη + λ1ω(η)X, (8)

where η ∈ Γ(D⊥) and X ∈ Γ(D). We deﬁne

D D⊥ g (Aη X, Y) := g (B(X, Y), η), (9)

⊥ D⊥ where Aη : Γ(D) → Γ(D) is the shape operator with respect to ∇. Let LX η = π ∇Xη; D ⊥ then, ∇Xη = π ∇Xη + LX η, and so we can get the Weingarten formula with respect to ∇

D ⊥ π ∇Xη = −Aη X, ∇Xη = −Aη X + LX η, (10)

⊥ ⊥ ⊥ ⊥ where LX η : Γ(D) × Γ(D ) → Γ(D ) is a metric connection on D along Γ(D). Let Aeη = (Aη − λ1ω(η))I; then, by (8) and (10), we have the Weingarten formula with respect to ∇e

⊥ ∇e Xη = −Aeη X + LX η, (11) Symmetry 2021, 13, 79 4 of 19

Given X1, X2, X3 ∈ Γ(TM), we deﬁne the curvature tensor Re with respect to ∇e

R(X X )X = ∇ ∇ X − ∇ ∇ X − ∇ X e 1, 2 3 : e X1 e X2 3 e X2 e X1 3 e [X1,X2] 3. (12)

D D Given X1, X2, X3 ∈ Γ(D), we deﬁne the curvature tensor Re on D with respect to ∇e

D D D D D D D D⊥ Re (X1, X2)X3 := ∇e ∇e X3 − ∇e ∇e X3 − ∇e D X3 − π [[X1, X2] , X3]. (13) X1 X2 X2 X1 [X1,X2]

D D D⊥ In (13), Re is a tensor ﬁeld created by adding the extra term −π [[X1, X2] , X3]. Given X1, X2, X3, X4 ∈ Γ(D), similarly, we deﬁne the Riemannian curvature tensor Re and ReD

D D Re(X1, X2, X3, X4) = g(Re(X1, X2)X3, X4), Re (X1, X2, X3, X4) = g(Re (X1, X2)X3, X4). (14)

Theorem 2. If X1, X2, X3, X4 ∈ Γ(D), we obtain the Gauss equation for D with respect to ∇e

D Re(X1, X2, X3, X4) = Re (X1, X2, X3, X4) + g(B(X2, X4), B(X1, X3)) − g(B(X1, X4), B(X2, X3)) (15)

+ g(B(X3, X4), [X1, X2]) − λ1ω(Be(X1, X3))g(X2, X4) + λ1ω(Be(X2, X3))g(X1, X4)

− λ2g(X1, X3)ω(B(X2, X4)) + λ2g(X2, X3)ω(B(X1, X4)).

Proof. From (5) and (11), for X1, X2, X3 ∈ Γ(D), we have

D D D ∇e X ∇e X X3 = ∇e ∇e X3 + Be(X , ∇e X3) − A X (16) 1 2 X1 X2 1 X2 Be(X2,X3) 1 + λ ω(B(X , X ))X + L⊥ (B(X , X )), 1 e 2 3 X1 e 2 3

D D D ∇e X ∇e X X3 = ∇e ∇e X3 + Be(X2, ∇e X3) − A X2 (17) 2 1 X2 X1 X1 Be(X1,X3) + λ ω(B(X , X ))Y + L⊥ (B(X , X )). 1 e 1 3 X2 e 1 3

For X1, X2 ∈ Γ(TM), we have

∇ = ∇ + [ ] + ( ) − ( ) e X1 X2 e X2 X1 X1, X2 λ1ω X2 X1 λ1ω X1 X2. (18)

Then, by (11) and (18), we have

⊥ D⊥ D⊥ D⊥ ∇e D⊥ X3 = −A D⊥ X3 + LX ([X1, X2] ) + λ1ω(X3)[X1, X2] + [[X1, X2] , X3]. (19) [X1,X2] [X1,X2] 3

By (19) and (5), we get

∇e [X ,X ]X3 = ∇e [X ,X ]D X3 + ∇e D⊥ X3 (20) 1 2 1 2 [X1,X2] D D = ∇e D X3 + Be([X1, X2] , X3) − A D⊥ X3 [X1,X2] [X1,X2] ⊥ D⊥ D⊥ D⊥ + LZ ([X1, X2] ) + λ1ω(X3)[X1, X2] + [[X1, X2] , X3].

By (12)–(20), we have

⊥ ⊥ R(X , X )X =RD(X , X )X − πD [[X , X ]D , X ] + B(X , ∇D X ) (21) e 1 2 3 e 1 2 3 1 2 3 e 1 e X2 3 D D − Be(X2, ∇e X3) − Be([X , X2] , X3) − A X + A X2 X1 1 Be(X2,X3) 1 Be(X1,X3) + L⊥ (B(X , X )) − L⊥ (B(X , X )) + λ ω(B(X , X ))X − λ ω(B(X , X ))X X1 e 2 3 X2 e 1 3 1 e 2 3 1 1 e 1 3 2 ⊥ D⊥ D⊥ + A D⊥ X3 − LZ ([X1, X2] ) − λ1ω(X3)[X1, X2] . [X1,X2] Symmetry 2021, 13, 79 5 of 19

By the second equality in (6) and (9), (14), (21), we get Theorem 2.

Corollary 1. If U = 0, then ω = 0 and ∇e = ∇, and we have

D R(X1, X2, X3, X4) =R (X1, X2, X3, X4) − g(B(X1, X4), B(X2, X3)) (22)

+ g(B(X2, X4), B(X1, X3)) + g(B(X3, X4), [X1, X2]).

Theorem 3. If X1, X2, X3 ∈ Γ(D), we get the Codazzi equation with respect to ∇e

⊥ (R(X , X )X )D =(L⊥ B)(X , X ) − (L⊥ B)(X , X ) − λ ω(X )B(X , X ) (23) e 1 2 3 X1 e 2 3 X2 e 1 3 1 1 e 2 3 D⊥ D⊥ ⊥ D⊥ + λ1ω(X2)Be(X1, X3) − π [[X1, X2] , X3] − LZ ([X1, X2] ) D⊥ − λ1ω(X3)[X1, X2] ,

where (L⊥ B)(X , X ) = L⊥ (B(X , X )) − B(∇D X , X ) − B(X , ∇D X ). X1 e 2 3 X1 e 2 3 e e X1 2 3 e 2 e X1 3

Proof. By (21), we have

⊥ ⊥ ⊥ (R(X , X )X )D = − πD [[X , X ]D , X ] + B(X , ∇D X ) (24) e 1 2 3 1 2 3 e 1 e X2 3 − B(X , ∇D X ) − B([X , X ]D, X ) + L⊥ (B(X , X )) e 2 e X1 3 e 1 2 3 X1 e 2 3 ⊥ ⊥ − L⊥ (B(X , X )) − L⊥([X , X ]D ) − λ ω(X )[X , X ]D . X2 e 1 3 Z 1 2 1 3 1 2

By (18), (24) and (25), we get

⊥ ⊥ ⊥ (R(X , X )X )D = − πD [[X , X ]D , X ] + B(X , ∇D X ) − B(X , ∇D X ) (25) e 1 2 3 1 2 3 e 1 e X2 3 e 2 e X1 3 − B(∇D X − ∇D X − λ ω(X )X + λ ω(X )X , X ) + L⊥(B(X , X )) e e X1 2 e X2 1 1 2 1 1 1 2 3 X e 2 3 ⊥ ⊥ − L⊥ (B(X , X )) − L⊥([X , X ]D ) − λ ω(X )[X , X ]D . X2 e 1 3 Z 1 2 1 3 1 2

Thus, (23) holds.

Corollary 2. If U = 0, then we have

⊥ (R(X , X )X )D =(L⊥ B)(X , X ) − (L⊥ B)(X , X ) (26) 1 2 3 X1 2 3 X2 1 3 D⊥ D⊥ ⊥ D⊥ − π [[X1, X2] , X3] − LZ ([X1, X2] ).

⊥ Theorem 4. If X1, X2 ∈ Γ(D), η ∈ Γ(D ), we get the Ricci equation for D with respect to ∇e

D⊥ L⊥ (Re(X1, X2)η) = −Be(X1, Aeη X2) + Be(X2, Aeη X1) + Re (X1, X2)η (27)

where

L⊥ ⊥ ⊥ ⊥ ⊥ ⊥ D⊥ Re (X1, X2)η := L L η − L L η − L D η − π ∇e ⊥ η. X1 X2 X2 X1 [X1,X2] [X1,X2]

Proof. From (5) and (11), we have

D ⊥ ⊥ ∇ ∇ η = −∇ (A X ) − B(X , A X ) − A ⊥ X + L L η, (28) e X1 e X2 e X1 eη 2 e 1 eη 2 eL η 1 X1 X2 X2

D ⊥ ⊥ ∇ ∇ η = −∇ (A X ) − B(X , A X ) − A ⊥ X + L L η, (29) e X2 e X1 e X2 eη 1 e 2 eη 1 eL η 2 X2 X1 X1 Symmetry 2021, 13, 79 6 of 19

By ∇e [X ,X ]η = ∇e [X ,X ]D η + ∇e D⊥ η, we have 1 2 1 2 [X1,X2]

D ⊥ D D⊥ ∇e [X ,X ]η = −Aeη([X1, X2] ) + L D η + π ∇e D⊥ η + π ∇e D⊥ η. (30) 1 2 [X1,X2] [X1,X2] [X1,X2]

So (27) holds.

Corollary 3. If U = 0, then we have

D⊥ L⊥ (R(X1, X2)η) = −B(X1, Aη X2) + B(X2, Aη X1) + R (X1, X2)η, (31)

where

L⊥ ⊥ ⊥ ⊥ ⊥ ⊥ D⊥ R (X1, X2)η := L L η − L L η − L D η − π ∇ ⊥ η. X1 X2 X2 X1 [X1,X2] [X1,X2]

Now, we present the proof of the Chen inequality with respect to D and ∇e . By (∇Xω)(Y) = X(ω(Y)) − ω(∇XY), we let

λ α(X , X ) = (∇ ω)(X ) − λ ω(X )ω(X ) + 2 g(X , X )ω(U), 1 2 X1 2 1 1 2 2 1 2 ω(U) β(X , X ) = g(X , X ) + ω(X )ω(X ). 1 2 2 1 2 1 2

where X1, X2 ∈ Γ(TM). In [16], we get

Re(X1, X2, X3, X4) = R(X1, X2, X3, X4) + λ1α(X1, X3)g(X2, X4) − λ1α(X2, X3)g(X1, X4) (32)

− λ2(λ1 − λ2)g(X2, X3)β(X1, X4) + λ2(λ1 − λ2)g(X1, X3)β(X2, X4)

+ λ2g(X1, X3)α(X2, X4) − λ2g(X2, X3)α(X1, X4).

Let {E1, ... , En, En+1, ... , Em} be a local orthonormal frame in M and n n D = span{E1, ... , En}. And let λ = ∑i=1 α(Ei, Ei),µ = ∑i=1 β(Ei, Ei). Let M be an m- dimensional real space form of constant sectional curvature c endowed with the ﬁrst generalized semi-symmetric non-metric connection ∇e . The curvature tensor R with respect to the Levi–Civita connection on M is expressed by

R(X1, X2, X3, X4) = c{g(X1, X4)g(X2, X3) − g(X1, X3)g(X2, X4)}. (33) By (33) and (35), we get

Re(X1, X2, X3, X4) = c{g(X1, X4)g(X2, X3) − g(X1, X3)g(X2, X4) − λ2g(X2, X4)α(X1, X4) (34)

+ λ2(λ1 − λ2)g(X1, X3)β(X2, X4) − λ2(λ1 − λ2)g(X2, X3)β(X1, X4)

− λ1α(X2, X3)g(X1, X4) + λ2g(X1, X3)α(X2, X4).

Let Π ⊂ D, be a two-plane section. Denote by KeD(Π) the sectional curvature of D with the induced connection ∇e D deﬁned by

1 KeD(Π) = [ReD(E , E , E , E ) − ReD(E , E , E , E )], (35) 2 1 2 2 1 1 2 1 2

D where E1, E2 are orthonormal bases of Π and Ke (Π) is independent of the choice of E1, E2. D For any orthonormal basis {E1, ... , En} of D, the scalar curvature τe with respect to D and ∇e D is deﬁned by

D 1 D τe = ∑ Re (Ei, Ej, Ej, Ei). (36) 2 1≤i,j≤n Symmetry 2021, 13, 79 7 of 19

Let E1, E2 be the orthonormal bases of Π ⊂ D such that the following deﬁnitions are independent of the choice of the orthonormal bases:

D 1 A = ∑ g(B(Ej, Ei), [Ej, Ei]), (37) 2 1≤i,j≤n λ + λ 1 ΩΠ = 1 2 [α(E , E ) + α(E , E )] − g(B(E , E ) − B(E , E ), [E , E ]) (38) 2 1 1 2 2 2 1 2 2 1 1 2 λ λ + 2 (λ − λ )[β(E , E ) + β(E , E )] + 2 [ω(B(E , E )) + ω(B(E , E ))] 2 1 2 1 1 2 2 2 1 1 2 2 λ1 + [ω(Be(E , E )) + ω(Be(E , E ))]. 2 1 1 2 2

Theorem 5. Let TM = D ⊕ D⊥, dimD = n ≥ 3, and let M be a manifold with constant sectional curvature c endowed with a connection ∇e ; then, we get the Chen inequality:

(n + 1)(n − 2) λ1 + λ2 λ2 τD − KeD(Π) ≤ c − (n − 1)λ − (λ − λ )(n − 1)µ e 2 2 2 1 2 2 λ2 λ2 n (n − 2) 1 − (n − 1)nω(B) − (n − 1)nω(Be) + AD + ΩΠ + kHk2 + ||B||2. (39) 2 2 2(n − 1) 2

2 n 2 where ||B|| = ∑i,j=1 g(B(Ei, Ej), B(Ei, Ej)) is the squared length of B and ||Be|| = n ∑i,j=1 g(Be(Ei, Ej), Be(Ei, Ej)) is the squared length of B.e

⊥ Proof. We choose the orthonormal bases {E1, ... , En} and {En+1, ... , Em} of D and D , respectively, such that Π ⊂ D = span{E1, E2}. By Theorem 2, (34) and (35), we obtain

m D Π r r r r Ke (Π) =c − Ω + ∑ [h11h22 − h12h21]. (40) r=n+1

Then, we get

D 1 D τe = ∑ Re (Ei, Ej, Ej, Ei) (41) 2 1≤i6=j≤n n(n − 1) λ + λ λ λ = c − 1 2 (n − 1)λ − 2 (λ − λ )(n − 1)µ − 2 (n − 1)nω(B) 2 2 2 1 2 2 m λ1 D r r r r − (n − 1)nω(Be) + A + ∑ ∑ [hiihjj − hijhji]. 2 r=n+1 1≤i

Thus,

(n + 1)(n − 2) λ1 + λ2 λ2 τD − KeD(Π) = c − (n − 1)λ − (λ − λ )(n − 1)µ e 2 2 2 1 2 λ2 λ1 − (n − 1)nω(B) − (n − 1)ω(Be) + AD + ΩΠ 2 2 m r r r r r r r r r + ∑ [(h11 + h22) ∑ hjj + ∑ hiihjj − ∑ hijhji + h12h21]. r=n+1 3≤j≤n 3≤i

By Lemma 2.4 in [22], we get

m n2(n − 2) [(hr + hr ) hr + hr hr ] ≤ kHk2. (42) ∑ 11 22 ∑ jj ∑ ii jj ( − ) r=n+1 3≤j≤n 3≤i

We note that

m r r r r ∑ [− ∑ hijhji + h12h21] (43) r=n+1 1≤i

Remark 1. When U ∈ Γ(D), that is B = B,e we get the following inequality

(n + 1)(n − 2) λ1 + λ2 λ2 τD − KeD(Π) ≤ c − (n − 1)λ − (λ − λ )(n − 1)µ e 2 2 2 1 2 λ + λ n2(n − 2) 1 − 1 2 (n − 1)nω(B) + AD + ΩΠ + kHk2 + ||B||2. (44) 2 2(n − 1) 2

r r Corollary 4. If D is totally geodesic with respect to ∇e and h12 = h21 = 0, then the equality case of (39) holds, and vice versa.

Proof. From the equality case of (42) and the equality case of (43), Corollary 3 holds.

Corollary 5. If D is an integrable distribution—that is if X, Y ∈ Γ(D)—then [X, Y] is in Γ(D). Then,

(n + 1)(n − 2) λ1 + λ2 λ2 τD − KeD(Π) ≤ c − (n − 1)λ − (λ − λ )(n − 1)µ e 2 2 2 1 2 2 λ2 λ2 n (n − 2) 1 − (n − 1)nω(B) − (n − 1)nω(Be) + ΩΠ + kHk2 + ||B||2. (45) 2 2 2(n − 1) 2

where λ + λ λ ΩΠ = 1 2 [α(E , E ) + α(E , E )] + 2 (λ − λ )[β(E , E ) + β(E , E )] (46) 2 1 1 2 2 2 1 2 1 1 2 2 λ2 λ1 + [ω(B(E , E )) + ω(B(E , E ))] + [ω(Be(E , E )) + ω(Be(E , E ))]. 2 1 1 2 2 2 1 1 2 2

We choose the orthonormal basis {E1,..., En} of D and let X = E1. We deﬁne

n n D D D Ricf (X) = ∑ Re (X, Ei, Ei, X); A (X) = ∑ g(B(Ei, X), [Ei, X]); (47) i=2 i=2 n X 2 kB k = ∑[g(B(X, Ei), B(X, Ei)) + g(B(Ei, X), B(Ei, X))]. i=2 Symmetry 2021, 13, 79 9 of 19

Theorem 6. Let TM = D ⊕ D⊥, dimD = n ≥ 2, and let M be a manifold with constant sectional curvature c endowed with a connection ∇e , then

D Ricf (X) ≤ (n − 1)c − λ1λ + λ1α(X, X) + λ2(1 − n)α(X, X) (48)

+ λ2(λ1 − λ2)(n − 1)β(X, X) − λ2(n − 1)ω(B(X, X)) n2 kBXk2 − λ nω(Be) + λ ω(B(X, X)) + kHk2 + + AD(X). 1 1 4 2

Proof. By (34)–(36), we have

D Ricf (X) = (n − 1)c − λ1λ + λ1α(X, X) + λ2(1 − n)α(X, X) (49)

− λ2(λ1 − λ2)(n − 1)β(X, X) − λ2(n − 1)ω(B(X, X)) − λ1nω(Be) m n r r r r D + λ1ω(B(X, X)) + ∑ ∑[h11hjj − h1jhj1] + A (X). r=n+1 j=2

From [22], we get

n+p n 2 r r n 2 ∑ ∑ h11hjj ≤ kHk . (50) r=n+1 j=2 4 We note that

m n m n (hr )2 + (hr )2 X 2 r r 1j j1 kB k − ∑ ∑ h1jhj1 ≤ ∑ ∑ = . (51) r=n+1 j=2 r=n+1 j=2 2 2

Thus, (48) holds.

r r r r r Corollary 6. If h1j = −hj1 for 2 ≤ j ≤ n and h11 − h22 − · · · − hnn = 0, then the equality case of (48) holds, and vice versa.

Corollary 7. If D is an integrable distribution—that is if X, Y ∈ Γ(D)—then [X, Y] is in Γ(D). Then,

D Ricf (X) ≤ (n − 1)c − λ1λ + λ1α(X, X) + λ2(1 − n)α(X, X) + λ2(λ1 − λ2)(n − 1)β(X, X) (52) n2 kBXk2 − λ (n − 1)ω(B(X, X)) − λ nω(Be) + λ ω(B(X, X)) + kHk2 + . 2 1 1 4 2

3. Non-Integrable Distributions with the Second Generalized Semi-Symmetric Non-Metric Connection Definition 5. For any V ∈ Γ(TM), let ω be a one-form satisfying ω(V) = g(U, V); here, ∞ U ∈ Γ(TM) is a vector field. Let f1, f2 ∈ C (M); we give the definition of the second generalized semi-symmetric non-metric connection on M as follows:

∇XY = ∇XY + f1ω(X) + f2ω(Y)X. (53)

Similarly to (2.5), for X, Y ∈ Γ(D)

D D D ∇XY = ∇X Y + B(X, Y), ∇X Y = π ∇XY, (54)

D⊥ where B(X, Y) = π ∇XY, and we call it the second fundamental form with respect to the second generalized semi-symmetric non-metric connection. Therefore, we have

D D ∇X Y = ∇X Y + f1ω(X)Y + f2ω(Y)X, B(X, Y) = B(X, Y), (55) Symmetry 2021, 13, 79 10 of 19

∞ where f1, f2 ∈ C (M). By (3.3), we have

D D D D D ∇X (g )(Y, Z) = −2 f1ω(X)g (Y, Z) − f2ω(Y)g (X, Z) − f2ω(Z)g (X, Y), (56) D D⊥ T (X, Y) = −[X, Y] + ( f2 − f1)[ω(Y)X − ω(X)Y].

If D = TM, we have the following results:

D Theorem 7. If a linear connection ∇ : Γ(D) × Γ(D) → Γ(D) on D satisﬁes the Equation (56), then this connection is the uniqueness.

Proposition 3. D is minimal (or umbilical) with respect to ∇ if and only if D is minimal (or umbilical) with respect to ∇.

Let

⊥ ∇Xη = −Aη X + LX η, (57) D where Aη = (Aη − f2ω(η))I. Then, by the deﬁnition of R and R , we get

Theorem 8. If X1, X2, X3, X4 ∈ Γ(D) and η ∈ Γ(D), we have

D D⊥ D⊥ R(X1, X2, X3, X4) = R (X1, X2, X3, X4) − π [[X1, X2] , X3] − g(B(X1, X3), B(X2, X4)) (58) D⊥ + g(B(X3, X4), [X1, X2]) − f1ω([X1, X2] )g(X3, X4) − f2g(X2, X4)ω(B(X1, X3))

+ g(B(X2, X4), B(X1, X3)) + f2g(X1, X4)ω(B(X2, X3)).

⊥ ⊥ ⊥ (R(X X )X )D =(L B)(X X ) − (L B)(X X ) 1, 2 3 X1 2, 3 X2 1, 3 (59) D⊥ D⊥ + ( f1 − f2)ω(X1)B(V) + ( f2 − f1)ω(X2)B(X1, X3) − π [[X1, X2] , X3] ⊥ ⊥ − L⊥ ([X , X ]D ) + ( f − f )ω(Z)[X , X ]D , X3 1 2 1 2 1 2

⊥ ⊥ D D where (L B)(X X ) = L (B(X X )) − B(∇ X X ) − B(X ∇ X ) X1 2, 3 X1 2, 3 X1 2, 3 2, X1 3 .

⊥ D⊥ L (R(X1, X2)η) = −B(X1, Aη X2) + B(X2, Aη X1) + R (X1, X2)η, (60)

where

⊥ L ⊥ ⊥ ⊥ ⊥ ⊥ D⊥ R (X , X )η := L L η − L L η − L D η − π ∇ ⊥ η. 1 2 X1 X2 X2 X1 [X1,X2] [X1,X2]

D Remark 2. We use the equality ∇XY = ∇X Y + B(X, Y) to prove Theorem 7. We use the equality D ∇e XY = ∇e X Y + Be(X, Y) to prove Theorems 1–3. This is the difference between the two cases. D We may deﬁne K (Π), τD, and for X, Y ∈ Γ(TM), we obtain

α1(X, Y) = (∇Xω)(Y).

n Similarly, let λ1 = ∑i=1 α1(Ei, Ei). In [17], for X1, X2, X3, X4 ∈ Γ(D), we have

R(X1, X2, X3, X4) = R(X1, X2, X3, X4) − f1α1(, X2, X1)g(Z, W) + f1α1(X1, X2)g(X3, X4) (61) 2 − f2α1(X2, X3)g(X1, X4) + f2α1(X1, X3)g(X2, X4) + f2 ω(X2)ω(X4)g(X1, X4) 2 − f2 ω(X1)ω(X4)g(X2, X3). Symmetry 2021, 13, 79 11 of 19

Let M be an m-dimensional real space form of the constant sectional curvature c endowed with the second generalized semi-symmetric non-metric connection ∇. By (33) and (61), we get

R(X1, X2, X3, X4) = c{g(X1, X4)g(X2, X3) − g(X1, X3)g(X2, X4)} − f1α1(X2, X1)g(X3, X4) (62)

+ f1α1(X1, X2)g(X3, X4) − f2α1(X2, X3)g(X1, X4) + f2α1(X1, X3)g(X2, X4) 2 2 + f2 ω(X2)ω(Z)g(X1, X4) − f2 ω(X1)ω(X3)g(X2, X4).

Let

tr(α1|Π) = α1(E1, E1) + α1(E2, E2), tr(B|Π) = B(E1, E1) + B(E2, E2), (63) 1 ΩΠ∗ = − g(B(E , E ) − B(E , E ), [E , E ]), 2 1 2 2 1 1 2 2 2 2 tr(ω |Π) = ω(E1) + ω(E2).

Theorem 9. Let TM = D ⊕ D⊥, dimD = n ≥ 3, and let M be a manifold with constant sectional curvature c endowed with a connection ∇, then

2 D (n + 1)(n − 2) f f f τD − K (Π) ≤ c − 2 (n − 1)λ − 2 n(n − 1)ω(H) + 2 (n − 1)n (64) 2 2 1 2 2 f f f 2 + 2 tr(α | ) + 2 ω(tr(B| )) + AD + ΩΠ∗ + 2 (n − 1)γ 2 1 Π 2 Π 2 f n2(n − 2) 1 − 2 tr(ω |2 ) + kHk2 + ||B||2. 2 Π 2(n − 1) 2

⊥ Proof. We choose orthonormal bases {E1, ··· , En} and {En+1, ··· , Em} of D and D , respectively. Let E1, E2 be the orthonormal bases of Π ⊂ D. By (62), we obtain

2 2 R(E1, E2, E1, E2) = −c + f2α1(E1, E1) − f2 ω(E1) . (65)

By (58), we have

D 2 2 R (E1, E2, E1, E2) = − c + f2α1(E1, E1) − f2 ω(E1) + g(B(E1, E2), B(E2, E1)) (66) − g(B(E1, E1), B(E2, E2)) + f2ω(B(E1, E1)) − g(B(E1, E2), [E1, E2]).

Similarly, we have

D 2 2 R (E1, E2, E2, E1) =c − f2α1(E2, E2) + f2 ω(E2) − g(B(E1, E2), B(E2, E1)) (67) + g(B(E1, E1), B(E2, E2)) − f2ω(B(E2, E2)) − g(B(E2, E1), [E1, E2]).

Thus, we obtain

D f f K (Π) =c − 2 tr(α | ) − 2 g(tr(B| ), U) (68) 2 1 Π 2 Π m Π∗ f2 2 r r r r − Ω + tr(ω |Π) + ∑ [h11h22 − h12h21]. 2 r=n+1

Similarly to (67), we have

D 2 2 R (Ei, Ej, Ej, Ei) =c − f2α1(Ej, Ej) + f2 ω(Ej) − g(B(Ei, Ej), B(Ej, Ei)) (69)

+ g(B(Ei, Ei), B(Ej, Ej)) − f2ω(B(Ej, Ej)) − g(B(Ej, Ei), [Ei, Ej]). Symmetry 2021, 13, 79 12 of 19

Then,

D 1 D τ = ∑ R (Ei, Ej, Ej, Ei) (70) 2 1≤i6=j≤n n(n − 1) f f = c − 2 (n − 1)λ − 2 n(n − 1)ω(H) + AD 2 2 1 2 2 m f2 r r r r + (n − 1)γ + ∑ ∑ [hiihjj − hijhji]. 2 r=n+1 1≤i

n 2 where γ=∑j=1 ω(Ej) . Thus,

D (n + 1)(n − 2) f f τD − K (Π) = c − 2 (n − 1)λ − 2 n(n − 1)ω(H) (71) 2 2 2 2 f f ∗ f + 2 tr(α | ) + 2 g(tr(B| ), U) + AD + ΩΠ + 2 (n − 1)γ 2 1 Π 2 Π 2 m r r r r r r r r + ∑ [ ∑ hiihjj − h11h22 − ∑ hijhji + h12h21]. r=n+1 1≤i

Thus, (64) holds.

r r Corollary 8. If D is totally geodesic with respect to ∇ and h12 = h21 = 0, then the equality case of (3.12) holds, and vice versa.

Corollary 9. If D is an integrable distribution—that is, if X, Y ∈ Γ(D)—then [X, Y] is in Γ(D). Then,

2 D (n + 1)(n − 2) f f f τD − K (Π) ≤ c − 2 (n − 1)λ − 2 n(n − 1)ω(H) + 2 (n − 1)n (72) 2 2 1 2 2 f f f 2 + 2 tr(α | ) + 2 ω(tr(B| )) + 2 (n − 1)γ 2 1 Π 2 Π 2 f n2(n − 2) 1 − 2 tr(ω |2 ) + kHk2 + ||B||2. 2 Π 2(n − 1) 2

Theorem 10. Let TM = D ⊕ D⊥, dimD = n ≥ 2, and let M be a manifold with constant sectional curvature c endowed with a connection ∇; then,

D Ric (X) ≤ (n − 1)c − f2λ1 + f2α1(X, X) − f2nω(H) (73) n2 kBXk2 + f ω(B(X, X)) + f 2γ + f 2ω(X)2 + kHk2 + + AD(X). 2 2 2 4 2

Proof. By (69), we have

D Ric (X) = (n − 1)c − f2λ1 + f2α1(X, X) − f2nω(H) (74) X 2 m n 2 2 2 D kB k r r r r + f2ω(B(X, X)) + f2 γ + f2 ω(X) + A (X) + + ∑ ∑[h11hjj − h1jhj1]. 2 r=n+1 j=2

Thus, (73) holds.

r r r r r Corollary 10. If h1j = −hj1 for 2 ≤ j ≤ n and h11 − h22 − · · · − hnn = 0, then the equality case of (73) holds, and vice versa. Symmetry 2021, 13, 79 13 of 19

Corollary 11. If D is an integrable distribution—that is, if X, Y ∈ Γ(D)—then [X, Y] is in Γ(D). Then,

D Ric (X) ≤ (n − 1)c − f2λ1 + f2α1(X, X) − f2nω(H) (75) n2 kBXk2 + f ω(B(X, X)) + f 2γ + f 2ω(X)2 + kHk2 + . 2 2 2 4 2

4. Examples Example 1. Let S3 be a unit sphere and dimS3 = 3, which we consider as a Riemannian manifold endowed with the metric induced from R4. Denote by TS3 = 3 the tangent space of S3; we choose 3 an orthonormal basis X1, X2, X3 of TS at each point, which satisﬁes

[X1, X2] = 2X3, [X1, X3] = −2X2, [X2, X3] = 2X1. (76)

Let ∇ be the Levi–Civita connection on S3. By (76) and the Koszul formula, we have ∇ = ∇ = − ∇ = ∇ = ∇ = X1 X2 X3, X2 X1 X3,, X1 X1 X2 X2 X3 X3 0, (77) ∇ = − ∇ = ∇ = −∇ = X1 X3 X2, X3 X1 X2, X2 X3 X3 X2 X1.

Consider a non-integrable distribution D1 = span{X1, X2}; then, we can get a metric of D1. Let U = X1 + X3. By (77), we have

∇D1 X = 0, ∀i, j = 1, 2, B(X , X ) = B(X , X ) = 0, (78) Xi j 1 1 2 2

B(X1, X2) = X3, B(X2, X1) = −X3.

By (6), we obtain

D1 D1 ∇e X Y = ∇X Y + λ1g(X1, Y)X − λ2g(X, Y)X1, Be(X, Y) = B(X, Y) − λ2g(X, Y)X3. (79)

specially, let λ1,λ2 be constant. Thus,

D D D ∇e 1 X = (λ − λ )X , ∇e 1 X = 0, ∇e 1 X = λ X , (80) X1 1 1 2 1 X1 2 X2 1 1 2 D ∇e 1 X = −λ X , B(X , X ) = −λ X , B(X , X ) = X , X2 2 2 1 e 1 1 2 3 e 1 2 3

Be(X2, X1) = −X3, Be(X2, X2) = −λ2X3, He = −λ2X3.

By (13), (38), (39) and (80), we have

D1 D1 Re (X1, X2)X1 = [λ1(λ1 − λ2) − 4]X2, Re (X1, X2)X2 = [λ2(λ1 − λ2) + 4]X1, (81) 2 2 1 (λ1 − λ2) (λ1 − λ2) KeD (D ) = 4 + , τD1 = 4 + . 1 2 e 2 By (54), we have

D1 D1 ∇X Y = ∇X Y + f1g(X1, X)Y + f2g(X1, Y)X, B(X, Y) = B(X, Y). (82)

where f1, f2 are constant. Thus,

D D D D ∇ 1 X = ( f + f )X , ∇ 1 X = f X , ∇ 1 X = f X , ∇b 1 X = 0 (83) X1 1 1 2 1 X1 2 1 2 X2 1 2 2 X2 2

B(X1, X1) = 0, B(X1, X2) = X3,

B(X2, X1) = −X3, B(X2, X2) = 0, 1 1 D 2 D R (X1, X2)X1 = (−4 − f2 )X2, R (X1, X2)X2 = 4X1. Symmetry 2021, 13, 79 14 of 19

3 1 3 1 1,⊥ Example 2. Let M = R × S and D = span{X1, X2} and TS = D ⊕ D . Let f (t) ∈ ∞ 3 3 3 C (R) without zero points. Let π1 : R × S → R; (t, x) → t and π2 : R × S → S ; (t, x) → x. Let

M ∗ 2 2 ∗ D1 ∗ D1,⊥ g f = π1 dt ⊕ f π2 g ⊕ π2 g ; (84) ∗ ∗ 1 D ∗ 2 2 ∗ D1 D = π1 (TR) ⊕ π2 D ; g = π1 dt ⊕ f π2 g ,

∗ 2 ∗ D1 ∗ D1,⊥ 2 D1 D1,⊥ ∗ ∗ 1 where π1 dt , π2 g , π2 g denote the pullback metrics of dt , g , g and π1 (TR), π2 D denote the pullback bundles of TR, D1. We call (D, gD) the warped product distribution on M and f M denote ∇ as the Levi–Civita connection on (M, g f ); then, by the Koszul formula and (84), we get

0 0 0 f f f f f f f ∇ ∂t = 0, ∇ X1 = X1, ∇ ∂t = X1, ∇ X2 = X2, (85) ∂t ∂t f X1 f ∂t f 0 f f f f f f 0 ∇ ∂t = X2, ∇ X3 = ∇ ∂t = 0, ∇ X1 = ∇ X2 = − f f ∂t, X2 f ∂t X3 X1 X2 f f f X2 f 1 ∇ X2 = X3, ∇ X1 = −X3, ∇ X3 = − , ∇ X1 = (2 − )X2, X1 X2 X1 f 2 X3 f 2 f X1 f 1 f ∇ X3 = , ∇ X2 = ( − 2)X1, ∇ X3 = 0. X2 f 2 X3 f 2 X3

∂ 0 where ∂t = ∂t and ∂t( f ) = f . Let D = span{∂t, X1, X2}; by (85), we have

0 0 0 D D f D f D f ∇ ∂t = 0, ∇ X = X , ∇ ∂t = X , ∇ X = X , (86) ∂t ∂t 1 f 1 X1 f 1 ∂t 2 f 2 0 D f D D 0 ∇ ∂t = X , ∇ X = ∇ X = − f f ∂t, X2 f 2 X1 1 X2 2 ∇D X = 0, ∇D X = 0. X1 2 X2 1

D For X, Y ∈ Γ(D), let E1, E2, E3 are orthonormal bases of (D, g ), and we deﬁne the Ricci D 3 D D tensor of D by Ric (X, Y) = ∑k=1 g (R (X, Ek)Y, Ek). Then,

2 f 00 RicD(∂ , ∂ ) = , RicD(X , X ) = RicD(X , X ) = f f 00 + ( f 0)2 − 4, (87) t t f 1 1 2 2 D D D D Ric (∂t, X1) = Ric (∂t, X2) = Ric (X1, ∂t) = Ric (X2, ∂t) = 0; D D Ric (X1, X2) = Ric (X2, X1) = 0.

D D D For X, Y ∈ Γ(D), if Ric (X, Y) = c0g (X, Y), we say that (D, g ) is Einstein.

D Theorem 11. (D, g ) is Einstein with the Einstein constant c0 if and only if (1) c0 = 0, f (t) = 2t + c1 or f (t) = −2t + c1, q q c0 c0 2 2 t − 2 t (2) c0 > 0, f (t) = − c c e + c2e , 2 0 q q (3) c < 0, f (t) = c cos( −c0 t) + c sin( −c0 t), c2 + c2 = − 8 . 0 1 2 2 2 1 2 c0 where c1, c2 are constant.

D Proof. By (87), (D, g ) is Einstein with the Einstein constant c0 if and only if c f 00 − 0 f = 0, (88) 2 00 0 2 2 f f + ( f ) − 4 = c0 f . (89) Symmetry 2021, 13, 79 15 of 19

If c0 = 0, by (88), then f = c2x + c1. Using (89), then c2 = 2, or − 2, and so we get case (1). q q c0 c0 2 t − 2 t 0 2 c0 2 If c0 > 0, by (88), then f = c1e + c2e . Using (89), then ( f ) = 4 + 2 f , −2 so c1 = c c and we get case (2). 2 0 q q −c0 −c0 0 2 c0 2 If c0 < 0, by (88), then f = c1cos( 2 t) + c2sin( 2 t). Using ( f ) = 4 + 2 f , we get c2 + c2 = − 8 , and so case (3) holds. 1 2 c0

Let U = ∂t, then

D D ∇e X Y = ∇X Y + λ1g(∂t, Y)X − λ2g(X, Y)∂t, Be(X, Y) = B(X, Y). (90)

where λ1,λ2 are constant. By (90), we get

0 0 0 D D f D f D f ∇e ∂t = (λ − λ )∂t, ∇e X = X , ∇e ∂t = ( + λ )X , ∇e X = X , (91) ∂t 1 2 ∂t 1 f 1 X1 f 1 1 ∂t 2 f 2 0 D f D D 0 2 ∇e ∂t = ( + λ )X , ∇e X = ∇e X = (− f f − λ f )∂t, X2 f 1 2 X1 1 X2 2 2 ∇D X = 0, ∇D X = 0, e X1 2 e X2 1 and 00 0 D f + λ2 f Ricf (∂t, ∂t) = 2[ + λ (λ − λ )], (92) f 1 2 1 D D Ricf (X1, X1) = Ricf (X2, X2), 00 0 2 2 0 2 0 = f f + 2λ1 f f + 2λ1λ2 f − λ2 f + ( f ) + λ2 f f − 4, D D Ricf (∂t, X1) = Ricf (∂t, X2) = 0, D D Ricf (X1, ∂t) = Ricf (X2, ∂t) = 0; D D Ricf (X1, X2) = Ricf (X2, X1) = 0.

So (D, gD, ∇e D) is mixed Ricci ﬂat. By (55) and (86), we have

0 0 0 D D f D f D f ∇ ∂t = ( f + f )∂t, ∇ X = ( + f )X , ∇ ∂t = ( + f )X , ∇ X = ( + f )X , (93) ∂t 1 2 ∂t 1 f 1 1 X1 f 2 1 ∂t 2 f 1 2 0 D f D D 0 ∇ ∂t = ( + f )X , ∇ X = ∇ X = − f f ∂t, X2 f 2 2 X1 1 X2 2 D D ∇ X = ∇ X = X1 2 0, X2 1 0.

D According to the computation of ∇e D, we can obtain the Ricci tensor of ∇ .

Example 3. Let (H3, gH3 ) be the Heisenberg group H3 endowed with the Riemannian metric g;

we choose an orthonormal basis {e1.e2.e3} of (H3, gH3 ) which satisﬁes the commutation relations

[e1, e2] = e3, [e1, e3] = 0, [e2, e3] = 0. (94) Symmetry 2021, 13, 79 16 of 19

By the Koszul formula, we can get the Levi–Civita connection ∇ of H3:

1 1 ∇ e = 0, 1 ≤ j ≤ 3, ∇ e = e , ∇ e = − e , (95) ej j e1 2 2 3 e2 1 2 3 1 1 ∇ e = ∇ e = − e , ∇ e = ∇ e = e . e1 3 e3 1 2 2 e2 3 e3 2 2 1

Let D = span{e e }, by (95), then ∇De = ≤ i j ≤ . Let U = e + e + e , then 1, 2 ei j 0, 1 , 2 1 2 3

∇D e = ( − )e − e ∇D e = e ∇D e = e ∇D e = ( − )e − e e e1 1 λ1 λ2 1 λ2 2, e e1 2 λ1 1, e e2 1 λ1 2, e e2 2 λ1 λ2 2 λ2 1, (96) 1 1 Be(e , e ) = Be(e , e ) = −λ e , Be(e , e ) = e , Be(e , e ) = − e . 1 1 2 2 2 3 1 2 2 3 2 1 2 3 D 2 2 2 Re (e1, e2)e1 = (λ1 − λ2)e1 − (λ1 − λ2) e2, D 2 2 2 Re (e1, e2)e2 = (λ1 − λ2) e1 + (λ1 − λ2)e2,

D D so (D, g , ∇e ) is ﬂat when λ1 = λ2. Similarly, we have

∇D e = ( f + f )e ∇D e = f e + f e ∇D e = f e + f e ∇D e = ( f + f )e b e1 1 1 2 1, b e1 2 2 1 1 2, b e2 1 1 1 2 2, b e2 2 1 2 2, (97) D D 2 Rb (e1, e2)e1 = Rb (e1, e2)e2 = f2 (e1 − e2).

1 1 1,⊥ Example 4. Let M = R × H3 and D = span{e1, e2} and TH3 = D ⊕ D , where H3 is the ∞ Heisenberg group. Let f (t) 6= 0 ∈ C (R) for any t ∈ R. Let π1 : R × H3 → R; (t, x) → t and π2 : R × H3 → H3; (t, x) → x. Let

M ∗ 2 2 ∗ D1 ∗ D1,⊥ g f = π1 dt ⊕ f π2 g ⊕ π2 g ; (98) ∗ ∗ 1 D ∗ 2 2 ∗ D1 D = π1 (TR) ⊕ π2 D ; g = π1 dt ⊕ f π2 g .

f M The Levi–Civita connection ∇ of (M, g f ) is given by

0 0 0 f f f f f f f ∇ ∂t = 0, ∇ e1 = e1, ∇e ∂t = e1, ∇ e2 = e2, (99) ∂t ∂t f 1 f ∂t f 0 f f f f f f 0 ∇e ∂t = e2, ∇ e3 = ∇e ∂t = 0, ∇e e1 = ∇e e2 = − f f ∂t, 2 f ∂t 3 1 2 f 1 f 1 f e f 1 ∇ e = e , ∇ e = − e , ∇ e = − 2 , ∇ e = − e , e1 2 2 3 e2 1 2 3 e1 3 2 f 2 e3 1 2 f 2 2 f e f 1 f ∇ e = 1 , ∇ e = e , ∇ e = 0. e2 3 2 f 2 e3 2 2 f 2 1 e3 3

Let D = span{∂t, e1, e2}; by (99), we have

0 0 0 D D f D f D f ∇ ∂t = 0, ∇ e = e , ∇ ∂t = e , ∇ e = e , (100) ∂t ∂t 1 f 1 e1 f 1 ∂t 2 f 2 0 D f D D 0 ∇ ∂t = e , ∇ e = ∇ e = − f f ∂t, e2 f 2 e1 1 e2 2 ∇D e = ∇D e = e1 2 0, e2 1 0. Symmetry 2021, 13, 79 17 of 19

The results of the Ricci tensor on D are as follows:

2 f 00 RicD(∂ , ∂ ) = , RicD(e , e ) = RicD(e , e ) = f f 00 + ( f 0)2, (101) t t f 1 1 2 2 D D D D Ric (∂t, e1) = Ric (∂t, e2) = Ric (e1, ∂t) = Ric (e2, ∂t) = 0; D D Ric (e1, e2) = Ric (e2, e1) = 0.

D Theorem 12. (D, g ) is Einstein with the Einstein constant c0 if and only if (1) c0 = 0, f (t) = c1, q q c0 c0 2 t − 2 t (2) c0 > 0, f (t) = c1e or f (t) = c2e , where c1, c2 are constant.

D Proof. By (101), (D, g ) is Einstein with the Einstein constant c0 if and only if c f 00 − 0 f = 0, (102) 2 00 0 2 2 f f + ( f ) = c0 f . (103)

If c0 = 0, by (102), then f = c2x + c1. Using (103), then c2 = 0, and so we get case (1). q q c0 c0 2 t − 2 t 0 2 c0 2 If c0 > 0, by (102), then f = c1e + c2e . Using (4.28), then ( f ) = 2 f , so c1 = 0 or c2 = 0, and we get case (2). q q −c0 −c0 0 2 c0 2 If c0 < 0, by (102), then f = c1cos( 2 t) + c2sin( 2 t). Using ( f ) = 2 f , we get c1 = c2 = 0. However, f 6= 0; thus, in this case there is no solution.

D Theorem 13. (D, g ) is a distribution with a constant scalar curvature λ0 if and only if 2 (1) λ0 = 0, f (t) = (c2t + c1) 3 , q 3λ q 3λ 0 t − 0 t 2 (2) λ0 > 0, f (t) = (c1e 8 + c2e 8 ) 3 , q 3λ q 3λ 2 0 0 3 (3) λ0 < 0, f (t) = (c1cos( − 8 t) + c2sin( − 8 t)) , where c1, c2 are constant.

Proof. By (101), we have

f 00 ( f 0)2 sD = 4 + 2 = λ . (104) f f 2 0

2 3 00 3 Let f (t) = w(t) and by (104), we get w (t) − 8 λ0w(t) = 0. By the elementary methods for ordinary differential equations, we prove the above theorem.

Let U = ∂t, By (100), we get

0 0 0 D D f D f D f ∇e ∂t = (λ − λ )∂t, ∇e e = e , ∇e ∂t = ( + λ )e , ∇e e = e , (105) ∂t 1 2 ∂t 1 f 1 e1 f 1 1 ∂t 2 f 2 0 D f D D 0 2 ∇e ∂t = ( + λ )e , ∇e e = ∇e e = (− f f − λ f )∂t, e2 f 1 2 e1 1 e2 2 2 ∇D e = ∇D e = e e1 2 e e2 1 0.

D D Theorem 14. (D, g , ∇e ) is a distribution with constant scalar curvature λ0 for U = ∂t if and only if λ +λ λ +λ − 1 2 t − 1 2 t − (1) λ = ξ f (t) = (c e 2 + c te 2 ) ξ 0 , 1 √ 2 , √ −(λ +λ )+ η −(λ +λ )− η 1 2 t 1 2 t −ξ (2) λ0 > ξ, f (t) = (c1e 2 + c2e 2 ) , Symmetry 2021, 13, 79 18 of 19

√ √ λ1+λ2 −η λ1+λ2 −η − 2 t − 2 t −ξ (3) λ0 < ξ, f (t) = (c1e cos( 2 t) + c2e sin( 2 t)) , 2 2 2 2 2 3 where c1, c2 are constant and ξ = − 3 (4λ1 − 2λ2 − 7λ1λ2), η = 4λ1 − 2λ2 − 7λ1λ2 + 2 λ0.

Proof. By (105), we have

f 00 (λ + λ ) f 0 ( f 0)2 sD = 4 + 4 1 2 + 2 + (6λ λ − 2λ2 − 2λ2). (106) e f f f 2 1 2 1 2

2 3 00 0 3 2 2 Let f (t) = w(t) and by (106), we get w (t) + (λ1 + λ2)w (t) + 8 (6λ1λ2 − 2λ1 − 2λ2 − λ0)w(t) = 0. By the elementary methods for ordinary differential equations, we prove the above theorem.

By (100), we have

0 0 D D f D f ∇ ∂t = ( f + f )∂t, ∇ e = ( + f )e , ∇ ∂t = ( + f )e , (107) ∂t 1 2 ∂t 1 f 1 1 e1 f 2 1 0 0 D f D f D D 0 ∇ e = ( + f )e , ∇ ∂t = ( + f )e , ∇ e = ∇ e = − f f ∂t, ∂t 2 f 1 2 e2 f 2 2 e1 1 e2 2 D D ∇ e = ∇ e = e1 2 e2 1 0.

Then, we get

f 00 ( f 0)2 f f 0 sD = 4 + 2 + 2 − 2 f 2. (108) f f 2 f 2

By Theorem 14, we have

D D Theorem 15. (D, g , ∇ ) is a distribution with a constant scalar curvature λ0 for U = ∂t if and only if 8 2 8 2 − f2t − f2t 3 f2 (1) λ0 = − 3 f2 , f (t) = (c1e + c2te ) , q q 2 3λ0 2 3λ0 8 2 8 2 − f2+ 4 f + t − f2− 4 f + t f (2) λ0 > − f , f (t) = (c1e 2 2 + c2e 2 2 ) 3 2 , 3 2 q q 3λ 3λ 8 2 8 2 − f2t 2 0 − f2t 2 0 3 f2 (3) λ0 < − 3 f2 , f (t) = (c1e cos( −4 f2 − 2 t) + c2e sin( −4 f2 − 2 t)) , where c1, c2 are constant.

5. Conclusions and Future Research For a Riemannian manifold with a semi-symmetric non-metric connection, the induced connection on a submanifold is also a semi-symmetric non-metric connection. The Gauss, Codazzi, and Ricci equations for distributions are a generalization of the case of submanifolds. Therefore, in this paper, we give the definition of the first generalized semi-symmetric non-metric connection and the second generalized semi-symmetric non-metric connection. The distribution can be viewed as a submanifold, so the corresponding metric of the Riemannian manifold distribution and orthogonal distribution are obtained. Then, by the D definition of an non-integrable distribution, we define the curvature tensor ReD(or R ) on D with respect to ∇e D(or ∇). By computation, we obtain the Gauss, Codazzi, and Ricci equations for non-integrable distributions in a Riemannian manifold with the first generalized semi-symmetric non-metric connection and the second generalized semi-symmetric non-metric connection, respectively. For a two-plane section Π ⊂ D, we define the sectional D D curvature KeD(Π)(or K (Π) of D with the induced connection ∇e D(or ∇ ) and the scalar D D D D curvature τe (or τ )with respect to D and ∇e (or ∇ ). Then, we obtain the Chen inequalities in both cases and give the equality case. We also give the results of the integrable distribution. Moreover, some properties of a totally geodesic and umbilical distribution are discussed in this paper. In following research, we will focus on the Lorentzian metric of distributions. Symmetry 2021, 13, 79 19 of 19

Author Contributions: Writing—original draft, T.W.; Writing—review and editing, Y.W. All authors have read and agreed to the published version of the manuscript. Funding: This research was funded by National Natural Science Foundation of China: No.11771070. Institutional Review Board Statement: Not applicable. Informed Consent Statement: Not applicable. Data Availability Statement: Data sharing not applicable. Acknowledgments: The author was supported in part by NSFC No.11771070. The author thanks the referee for his (or her) careful reading and helpful comments. Conﬂicts of Interest: The authors declare no conﬂict of interest.

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