Hypersurfaces immersed in a golden Riemannian manifold

Abstract

The purpose of this paper is to study properties of hypersurfaces of a Riemannian manifold equipped with a golden structure. We discuss the necessary and sufficient conditions for a hypersurface to be a golden Riemannian manifold in terms of invariance. The totally geodesics and totally umbilical hypersurfaces in golden Riemannian manifolds are analyzed. Also, we construct an example on hypersurface of the Euclidean space endowed with a golden Riemannian structure.

1 Introduction

The theory of submanifolds was initiated by Fermat in his study of geometry of plane curves. Nowadays, the theory of submanifolds is an active and vast research field playing an important role in the development of modern differential geometry. Moreover, the theory of hypersurfaces of an almost contact manifold have been studied by geometers in [1, 3, 6, 19, 20].

The notion of golden structure was introduced by Crasmareanu and Hretcanu in [7] and they studied its properties in [8, 17]. Then several geometers studied golden Riemannian manifolds (see [9,10,11,12, 17, 21]). Crasmareanu and Hretcanu [18] also defined and studied metallic structure as generalization of golden stucture. Then, the metallic structures were studied by many geometers (see [5, 15, 16]).

Gezer et al. [13] investigated the integrability conditions of the golden Riemannian manifold. Hretcanu studied submanifolds of Riemannian manifold with golden structure [14]. Bahadir and Uddin [4] studied slant submanifolds of golden Riemannian manifolds. Ahmad and Qayyoom [2] studied submanifolds in a Riemannian manifold with golden structure.

Motivated by above studies, in this paper we study some properties of hypersurfaces of a golden Riemannian manifold. The paper is organized as follows.

In Sect. 2, we define golden Riemannian manifolds.

In Sect. 3, we establish several properties of induced structure $\sum = (J, g,\xi ,u,a)$ on the hypersurface immersed in golden Riemannian manifold. In last section, we construct an example of golden Riemannian structure on Euclidean space and obtain its hypersurfaces.

2 Definitions and preliminaries

Definition 2.1

[7] Let $({\overline{M}},g)$ be a Riemannian manifold. A golden structure on $({\overline{M}}, g)$ is a non-null tensor $\overline{{J}}$ of type (1, 1) which satisfies the equation

$$\begin{aligned} \overline{{J}}^{2} = \overline{{J}}+I, \end{aligned}$$

(2.1)

where I is the identity transformation. We say that the metric g is $\overline{{J}}$-compatible if

$$\begin{aligned} g(\overline{{J}}X,Y)=g(X,\overline{{J}}Y) \end{aligned}$$

(2.2)

for all X,Y vector fields on ${\overline{M}}$. If we substitute $\overline{{J}}X$ into X in (2.2), then we have

$$\begin{aligned} g(\overline{{J}}X, \overline{{J}}Y)= g(\overline{{J}}X,Y)+g(X,Y). \end{aligned}$$

(2.3)

The Riemannian metric (2.2) is called $\overline{{J}}$-compatible and $({\overline{M}},\overline{{J}},g)$ is called a Golden Riemannian structure.

Definition 2.2

If $({\overline{M}}, g, {\overline{J}})$ is a golden Riemannian manifold and ${\overline{J}}$ is parallel with respect to Levi-Civita connection ${\overline{\nabla }}$ on M $(i.e. {\overline{\nabla }}_{}{\overline{J}}) = 0,$ then we call $({\overline{M}}, g, {\overline{J}})$ is a locally golden Riemannian manifold.

Proposition 2.3

[7] The eigenvalues of a golden structure ${\overline{J}}$ are the golden ratio $\phi$ and $1-\phi$.

Let M be an n-dimensional hypersurface isometrically immersed in an $(n+1)$-dimensional golden Riemannian manifold $({\overline{M}},g,\overline{{J}})$ with $n \in {\mathbb {N}}.$ We denote by $T_xM$ the tangent space of M at a point $x{\in }M$ and $T_x^{\bot }M$ the normal space of x in M. Let $i_{*}$ be the differential of immersion $i_{*}:M\rightarrow {{\overline{M}}}.$ The induced Riemannian metric g on M is given by $g(X,Y)=g(i_{{*}}X,i_{{*}}Y)$ for every $X,Y{\in }{\chi }(M)$.

We consider a local orthonormal basis N of the normal space $T_x^{\bot }M.$ For every $X{\in }T_xM$, the vector fields $\overline{{J}}(i_{{*}}X)$ and ${\overline{J}}(N)$ can be decomposed in tangential and normal components as follows:

$$\begin{aligned} {\overline{J}}(i_{{*}}X)= & {} i_{ {*}} ({J}X)+u(X)N, \end{aligned}$$

(2.4)

$$\begin{aligned} {\overline{J}}(N)= & {} i_{{*}} (\xi )+ aN, \end{aligned}$$

(2.5)

where J is a (1, 1) tensor field on M,  $\xi$ is tangent vector field on M,  u is $1-form$ on M and a is $1\times 1$ matrix of smooth real function on M.

We denote the covariant derivative in ${\overline{M}}$ by ${\overline{\nabla }}$ and covariant derivative in M by $\nabla$. We denote by A the Weingarten operator on TM with respect to the local unit normal vector field N of M in ${\overline{M}}.$ The Gauss and Weingarten formulae are given by

$$\begin{aligned} {\overline{\nabla }}_{X}Y= & {} \nabla _{X}Y+h(X,Y){N}, \end{aligned}$$

(2.6)

$$\begin{aligned} {\overline{\nabla }}_{X}N= & {} -AX + \nabla _{X}^{\bot }{N}, \end{aligned}$$

(2.7)

where $h(X,Y)=g(AX,Y)$ is the second fundamental form in $T^{\bot }M$ and $X, Y \in \chi (M).$

Proposition 2.4

[17] If M is a hypersurface isometrically immersed in a golden Riemannian manifold $({\overline{M}}, g, {\overline{J}}),$ from (2.4) and (2.5), we obtain the structure $\sum = (J, g, u, \xi , a)$ induced on M by the golden structure ${\overline{J}},$ satisfies the following inequalities:

$$\begin{aligned} {J}^{2}(X)= & {} {J}(X)+X-u(X)\xi , \end{aligned}$$

(2.8)

$$\begin{aligned} u({J}X)= & {} (1-a)u(X), \end{aligned}$$

(2.9)

$$\begin{aligned} u(\xi )= & {} 1+a-a^2, \end{aligned}$$

(2.10)

$$\begin{aligned} {J}(\xi )= & {} (1-a)\xi , \end{aligned}$$

(2.11)

$$\begin{aligned} u(X)= & {} g(X,\xi ), \end{aligned}$$

(2.12)

$$\begin{aligned} g({J}X,Y)= & {} g(X,{J}Y), \end{aligned}$$

(2.13)

$$\begin{aligned} g({J}X,{J}Y)= & {} g({J}X,Y)+g(X,Y)-u(X)u(Y). \end{aligned}$$

(2.14)

for every $X,Y\in \chi (M)$.

Proposition 2.5

[17] If M is a hypersurface in a golden Riemannian manifold $({\overline{M}},g,{\overline{J}})$ and ${\overline{J}}$ is parallel with respect to the Levi Civita connection ${\overline{\nabla }}$ on ${\overline{M}}$ $( {{\overline{\nabla }}}_{\;} {\overline{J}}=0),$ then the structure $\sum =(J,g,u,\xi ,a)$ induced on M by the structure ${\overline{J}}$ has the following properties:

$$\begin{aligned} (\nabla _X{J})(Y)= & {} g(AX,Y)\xi +u(Y)AX, \end{aligned}$$

(2.15)

$$\begin{aligned} (\nabla _{X}u)(Y)= & {} -g(AX,{{J}}Y)+ag(AX,Y), \end{aligned}$$

(2.16)

$$\begin{aligned} \nabla _X\xi= & {} -{{J}}(AX)+aAX, \end{aligned}$$

(2.17)

$$\begin{aligned} X(a)=-2u(AX)= & {} -2g(AX,\xi )=-2g(X,A\xi ) \end{aligned}$$

(2.18)

for every $X,Y\in \chi (M).$

Theorem 2.6

[17] If the structure $\sum =(J,g,\xi ,u,a)$ induced by structure ${\overline{J}}$ on an umbilical hypersurface M in a golden Riemannian manifold $({\overline{M}},g,\overline{{J}})$ with ${\overline{\nabla }}_{} \overline{{J}}=0$ we have

$$\begin{aligned} (\nabla _X{J})(Y)= & {} \lambda [g(X,Y)\xi +g(Y,\xi )X], \end{aligned}$$

(2.19)

$$\begin{aligned} (\nabla _X u)(Y)= & {} \lambda [ag(X,Y)-g(X,{J}Y)], \end{aligned}$$

(2.20)

$$\begin{aligned} \nabla _X\xi= & {} \lambda (aX-{J}(X)), \end{aligned}$$

(2.21)

$$\begin{aligned} \nabla _{\xi }\xi= & {} \lambda (2a-1)\xi , \end{aligned}$$

(2.22)

$$\begin{aligned} X(a)= & {} -2\lambda g(X,\xi ) \end{aligned}$$

(2.23)

for any X, Y $\in$ $\chi (M)$.

Corollary 2.7

Let M be a totally umbilical submanifold in a golden Riemannian manifold $({\overline{M}},g,\overline{{J}})$ with structure $\sum = ({J},g,\xi ,u,a)$ induced by $\overline{{J}}$ on M and ${\overline{\nabla }}_{} \overline{{J}}=0,$ follows that

$$\begin{aligned} (\nabla _X{J})(\xi )= & {} \lambda (1+a-a^2)X, \end{aligned}$$

(2.24)

$$\begin{aligned} (\nabla _\xi {{J}})(Y)= & {} 2\lambda g(Y,\xi )\xi ), \end{aligned}$$

(2.25)

$$\begin{aligned} (\nabla _Xu)\xi= & {} 2a\lambda g(X,\xi )-\lambda g(X,\xi ), \end{aligned}$$

(2.26)

for any $X, Y \in \chi (M)$.

Proof

For totally umbilical manifold, $A=\lambda I$, then equation (2.15) reduces to

$$\begin{aligned} (\nabla _X{J})(Y)= \lambda [g({X},Y)+ u(Y)X]. \end{aligned}$$

Taking $Y=\xi ,$ we get

$$\begin{aligned} (\nabla _X{J})\xi = \lambda [g({X},\xi )\xi +u(\xi ){X}]. \end{aligned}$$

Using equality (2.10), we obtain

$$\begin{aligned} (\nabla _X{J})\xi =\lambda (1+a-a^2)X, \end{aligned}$$

which is (2.24). Using equality (2.19) and if $X=\xi ,$ we obtain

$$\begin{aligned} (\nabla _\xi {J})(Y)=\lambda [g(Y,\xi )\xi + g(\xi ,Y)\xi ] \end{aligned}$$

$$\begin{aligned} (\nabla _{\xi }{J})Y=2\lambda g(Y,\xi )\xi , \end{aligned}$$

which gives (2.25). Using (2.20) and taking $Y=\xi ,$ we have

$$\begin{aligned} (\nabla _Xu)\xi =\lambda [ag(X,\xi )-g(X,{J}(\xi ))]. \end{aligned}$$

Using equality (2.11), we obtain

$$\begin{aligned} (\nabla _Xu)\xi =2\lambda ag(X,\xi )-\lambda g(X,\xi ). \end{aligned}$$

which proves (2.26). $\square$

3 Properties of induced structures on hypersurfaces in golden Riemannian manifolds

Proposition 3.1

Let M be a hypersurface on a golden Riemannian manifold $({\overline{M}},g,\overline{{J}})$ with structure $\sum = ({J},g,\xi ,u,a)$ induced on M by the structure ${\overline{J}}$ with $\xi \ne 0$. A necessary and sufficient condition for M to be totally geodesic in M is that $\nabla _{X}{J}=0$ for any $X\in \chi (M)$.

Proof

If M is totally geodesic, then $A=0$. Using equality (2.15), we obtain

$$\begin{aligned} \nabla _{X}{J}=0. \end{aligned}$$

Conversely, we suppose that $\nabla _{X}{J}=0$ and from (2.15), we have

$$\begin{aligned} g(AX,Y)\xi +g(Y,\xi )AX=0. \end{aligned}$$

We may have one of the following conditions:

(i) If AX and $\xi$ are linearly dependent vector fields, then there exist a real number $\alpha$ such that $AX=\alpha \xi$ and from this we obtain

$$\begin{aligned} g(\alpha \xi ,Y)\xi +g(Y,\xi )\alpha \xi =0. \end{aligned}$$

That is,

$$\begin{aligned} g(Y,\xi )=0 \end{aligned}$$

for any $Y\in \chi (M)$. Then for $Y=\xi ,$ we obtain $g(\xi ,\xi )=0$, which is equivalent with $\xi =0$. But this is impossible.

(ii) If AX and $\xi$ are linearly independent vector fields, then

$$\begin{aligned} g(AX,Y)=0 \end{aligned}$$

for any $X,Y\in \chi (M).$ Thus, $A=0$ and from this we have that M is a totally geodesic hypersurface in ${\overline{M}}.$ $\square$

Proposition 3.2

If M is a hypersurface in a golden Riemannian manifold $({\overline{M}},g,\overline{{J}})$ with structure $\sum = ({J},g,\xi ,u,a)$ induced on M by $\overline{{J}}$, then following equalities are equivalent:

$$\begin{aligned} \nabla _{X}u=0 \Leftrightarrow \nabla _X\xi =0 \end{aligned}$$

for each $X\in \chi (M).$

Proof

$\nabla _Xu=0$, then by using equality (2.16) we obtain

$$\begin{aligned} g(AX,{J}Y)=ag((AX),Y). \end{aligned}$$

Using (2.13) in above equation, we get

$$\begin{aligned} g({J}(AX)-aAX,Y)=0 \end{aligned}$$

for any $X,Y\in \chi (M).$ By use of (2.17), we have

$$\begin{aligned} g(\nabla _X\xi ,Y)=0. \end{aligned}$$

As $y\ne 0,$ we get

$$\begin{aligned} \nabla _X\xi =0. \end{aligned}$$

Conversely, we suppose that$\nabla \xi =0$ and we have

$$\begin{aligned} g(\nabla _X\xi ,Y)=0. \end{aligned}$$

By using equality (2.17), we get

$$\begin{aligned} g({J}(AX)-aAX,Y)=0, \end{aligned}$$

$$\begin{aligned}&g({J}(AX),Y)-g(aAX,Y)=0, \\&\quad g(AX,{J}Y)-ag(AX,Y)=0. \end{aligned}$$

In view of (2.16), above equation reduces to

$$\begin{aligned} \nabla _Xu=0. \end{aligned}$$

$\square$

Proposition 3.3

If M is a totally umbilical hypersurface in a golden Riemannian manifold $({\overline{M}},g,\overline{{J}})$ with the structure $\sum = ({J},g,\xi ,u,a)$ induced on M by ${\overline{J}}.$ Then the $1-form$ u is closed.

Proof

As M is totally umbilical hypersurface, that is $A=\lambda I.$ Then

$$\begin{aligned} (\nabla _Xu)(Y)=-\lambda g(X,JY)+a\lambda g(X,Y). \end{aligned}$$

(3.1)

Since,

$$\begin{aligned} du(X,Y)=(\nabla _Xu)(Y)-(\nabla _Yu)(X). \end{aligned}$$

Using (2.13) and (3.1) in above equation, we get

$$\begin{aligned} du(X,Y)= & {} [-\lambda g(X,{J}Y)+a\lambda g(X,Y)]-[-\lambda g(Y,{J}X)+a\lambda g(Y,X)]\nonumber \\ du(X,Y)= & {} -\lambda g(X,{J}Y)+\lambda {g}(Y,{J}X)\nonumber \\ du(X,Y)= & {} \lambda [-g({J}X,Y)+g({J}X,Y)] \nonumber \\ du(X,Y)= & {} 0. \end{aligned}$$

(3.2)

Thus, $1-form$ u is closed. $\square$

Proposition 3.4

Let M be a hypersurface in a golden Riemannian manifold $({{\overline{M}}},g,\overline{{J}})$ with ${\overline{\nabla }}_{} \overline{{J}}=0$ and the structure $\sum = ({J},g ,\xi ,u,a)$ induced on M by $\overline{{J}}$. Then $\xi$ is a Killing vector field with respect to g on M if and only if we have

$$\begin{aligned} 2aA={J}A+A{J}, \end{aligned}$$

(3.3)

where A is the Weingarten operator on M.

Proof

We have that $\xi$ is a Killing vector field on M if and only if

$$\begin{aligned} (L_\xi g)(Y,Z)=0 \end{aligned}$$

for all $Y,Z\in \chi (M).$ Then

$$\begin{aligned}&L_\xi g(Y,Z)-g(L_\xi Y,Z)-g(Y,L_\xi Z)=0 \\&\quad \xi g(Y,Z)-g([\xi ,Y],Z)-g(Y,[\xi ,Z])=0 \\&\quad \xi g(Y,Z)-g(\nabla _\xi Y-\nabla _Y\xi ,Z)-g(Y,\nabla _\xi Z-\nabla _Z\xi )=0 \\&\quad \xi g(Y,Z)-g(\nabla _\xi Y,Z)-g(Y,\nabla _\xi Z)+g(\nabla _Y\xi ,Z)+g(Y,\nabla _Z\xi )=0 \\&\quad g(\nabla _Y\xi ,Z)+g(Y,\nabla _Z\xi )=0 \end{aligned}$$

for all $Y,Z \in \chi (M).$

Using equality (2.17) in above equation we obtain

$$\begin{aligned} g(-{J}AY+aAY,Z)+g(-{J}AZ+aAZ,Y)=0 \end{aligned}$$

or,

$$\begin{aligned} g(2aAY-{J}AY-A({J}Y),Z)=0, \end{aligned}$$

which is equivalent to

$$\begin{aligned} 2aA={J}A+A{J}. \end{aligned}$$

$\square$

Theorem 3.5

[16] Let M be a hypersurface in a locally golden Riemannian manifold $({\overline{M}}, g, {\overline{J}}).$ If the normal connection $\nabla ^{\bot }$ vanishes identically on the normal bundle $T^{\bot }(M)$ and M is a non-invariant hypersurface with respect to the golden structure J then the induced structure $\sum = (J, g, \xi , u, a)$ on M is normal if and only if the tensor field J commutes with every Weingarten operator A $(i.e. \; JA = AJ).$

Proposition 3.6

Let M be a hypersurface of a golden Riemannian manifold $({\overline{M}},g,\overline{{J}})$ with ${\overline{\nabla }}_{\;} \overline{{J}}=0$ and the normal induced structure $\sum = ({J},g,\xi ,u,a)$ on M by ${\overline{J}}$. Then $\xi$ is a Killing vector field with respect to g on M if only if

$$\begin{aligned} aA={J}A=A{J}, \end{aligned}$$

where A is the Weingarten operator on M.

Proof

A necessary and sufficient condition for the normality of structure $\sum = ({J},g,\xi ,u,a)$ is that the tensor field J commutes by the Weingarten operator A. That is, ${J}A = A{J}.$ Then the proof of the proposition is a consequence of (3.3). $\square$

Proposition 3.7

Let M be a hypersurface of a golden Riemannian manifold $({\overline{M}},g,\overline{{J}})$ with ${\overline{\nabla }}_{\;} \overline{{J}} = 0$ and structure $\sum = ({J},g,\xi ,u,a)$ induced on M by $\overline{{J}}$ and $\xi$ is a Killing vector field. If $a\ne \phi$, then rank $A=1$ and $\xi$ is an eigenvector of the Weingarten operator A with the eigenvalue $\frac{\xi (a)}{2(a^2-a-1)}$.

Proof

We have ${J}A=aA$, then ${J}^2(AX)=a^2(AX)$ for all $X\in \chi (M).$ Using equality (2.6), we get

$$\begin{aligned} -u(AX)\xi =(a^2-a-1)AX. \end{aligned}$$

Using equality (2.18) in above equation, we obtain

$$\begin{aligned} \frac{X(a)}{2}\xi =(a^2-a-1)AX \end{aligned}$$

or,

$$\begin{aligned} AX=\frac{X(a)}{2(a^2-a-1)}\xi \end{aligned}$$

for all $X\in \chi (M).$ If we put $X=\xi$ in above equation, we obtain

$$\begin{aligned} A(\xi ) = \frac{\xi (a)}{2(a^2-a-1)}\xi \end{aligned}$$

for all $X\in \chi (M).$

Thus, $\xi$ is an eigenvector of Weingarten operator A and its eigenvalue is $\frac{\xi (a)}{2(a^2-a-1)}$. $\square$

Theorem 3.8

Let M is a hypersurface of a golden Riemannian manifold $({\overline{M}},g, {\overline{J}})$. Then M is invariant if and only if the normal of M is an eigenvector of the matrix ${\overline{J}}.$

Proof

Suppose $a=\phi$. Using equality (2.10), we have

$$\begin{aligned} u(\xi )=1+\phi -\phi ^{2} \end{aligned}$$

or,

$$\begin{aligned} u(\xi )=g(X,\xi )=0, \end{aligned}$$

which is equivalent to $X\bot \xi .$

As X and $\xi$ are both tangential, then there is only one possibility that $\xi =0$ because $X\ne 0$. From (2.5), we have

$$\begin{aligned} \overline{{J}}N=i_*\xi +aN, \end{aligned}$$

which gives

$$\begin{aligned} \overline{{J}}N=aN \end{aligned}$$

for $\xi =0.$ Hence N is the eigenvector for ${\overline{J}}$. $\square$

Remark 3.9

Let M be a hypersurface in a golden Riemannian manifold $({\overline{M}},g,\overline{{J}})$ with the induced structure $\sum = ({J},g,\xi ,u,a)$ on M by ${\overline{J}}$. We suppose that $(e_1,\ldots ,e_n)$ is an orthonormal basis of the tangent space $T_xM$ for any $x\in M$. Then

$$\begin{aligned} div\xi = trace (e_i\rightarrow \nabla _{e_j}\xi ) \end{aligned}$$

using $\nabla _X\xi = -\lambda {J}X+a\lambda X,$ we obtain

$$\begin{aligned} \nabla _{e_i}\xi = \lambda (aI-{J})(e_i). \end{aligned}$$

So,

$$\begin{aligned} div\xi= & {} \lambda trace(aI-{J}),\\ div\xi= & {} \lambda (na-trace{J}). \end{aligned}$$

4 Example of induced structure on a hypersurface in a golden Riemannian manifold

Example 4.1

We suppose that the ambient space is $E^{2a+b}$ (a, b $\epsilon$ $N^{*}$) and for any point of $E^{2a+b},$ we have its coordinates

$$\begin{aligned} (x^1,\ldots ,x^a,y^1,\ldots ,y^a,z^1,\ldots ,z^b)= (x^i,y^i,z^j), \end{aligned}$$

where i $\epsilon$ ${1,\ldots ,a}$ and j $\epsilon$ ${1,\ldots ,b}$. The tangent space $T_x(E^{2a+b})$ is isomorphic with $E^{2a+b}$.

Let ${\overline{J}}:E^{2a+b}\rightarrow E^{2a+b}$ be the golden structure on $E^{2a+b}$ defined as.

$$\begin{aligned}&\overline{{J}}(x^1,\ldots ,x^a,y^1,\ldots ,y^a,z^1,\ldots ,z^b)=(\phi x^1,\ldots ,\phi x^a,\phi y^1,\ldots ,\phi y^a,\\&(1-\phi )z^1,\ldots ,(1-\phi )z^b). \end{aligned}$$

Then

$$\begin{aligned}&\overline{{J}}^2(x^1,\ldots ,x^a,y^1,\ldots ,y^a,z^1,\ldots ,z^b) \nonumber \\&\quad =(\phi ^2 x^1,\ldots ,\phi ^2 x^a,\phi ^2 y^1,\ldots ,\phi ^2 y^a, (1-\phi )^2z^1,\ldots ,(1-\phi )^2z^b). \end{aligned}$$

(4.1)

Since $\phi$ and $(1-\phi )$ are roots of $x^2=x+1$ then $\phi ^2=\phi +1$ and $(1-\phi )^2=(1-\phi )+1.$ Then, (4.1) gives

$$\begin{aligned}&\overline{{J}}^2(x^1,\ldots ,x^a,y^1,\ldots ,y^a,z^1,\ldots ,z^b)= \\&\quad ((\phi +1)x^1,\ldots ,(\phi +1)x^a,(1+\phi )y^1,\ldots ,\\&\quad (1+\phi )y^a,((1-\phi )+1)z^1,\ldots ,((1-\phi )+1)z^b) \\&\quad \overline{{J}}^2(x^1,\ldots ,x^a,y^1,\ldots ,y^a,z^1,\ldots ,z^b)= \\&\quad (\phi x^1,\ldots ,\phi x^a,\phi y^1,\ldots \phi y^a (1-\phi )z^1,\ldots ,(1-\phi )z^b)\\&\quad +(x^1,\ldots ,x^a,y^1,\ldots ,y^a,z^1,\ldots ,z^b) \\&\quad \overline{{J}}^2(x^i,y^i,z^j)={J}(x^i,y^i,z^j)+I(x^i,y^i,z^j) \end{aligned}$$

or,

$$\begin{aligned} \overline{{J}}^2= \overline{{J}}+I. \end{aligned}$$

It follows that $( E^{2a+b},g,\overline{{J}})$ is a golden Riemannian manifold.

In $E^{2a+b},$ we can get the hypersphere

$$\begin{aligned} S^{2a+b-1}(r)={(x^1,\ldots ,x^a,y^1,\ldots ,y^a,\ldots ,z^1,\ldots ,z^b),\sum _{i=1}^{a}{x^i}^2+\sum _{i=1}^{a}{y^i}^2+\sum _{j=1}^{b}{z^j}^2=R^2}, \end{aligned}$$

where R is its radius and $(x^1,\ldots ,x^a,y^1,\ldots ,y^a,z^1,\ldots ,z^b)$ are the coordinates of any point of $S^{2a+b-1}(R)$. We use following notations

$\sum _{i=1}^{a}(x^i)^2=r_1^2$, $\sum _{i=1}^{a}(y^i)^2=r_2^2,$ $\sum _{j=1}^{b}(z^j)^2=r_3^2$ and ${r_1^2+r_2^2=r^2}.$ Thus, we have ${r^2+r_3^2=R^3}.$ We remark that $N_1=\frac{1}{R}(x^i,y^i,z^j)$, i $\epsilon$ $(1,\ldots ,a)$ , j $\epsilon$ $(1,\ldots ,b)$ is a unit normal vector field on sphere $S^{2a+b-1}(R)$ and $J(N_1)=\frac{1}{R}(\phi y^i,\phi x^i,(1-\phi )z^j).$

For a tangent vector field X on $S^{2a+b-1}(R),$ we use the following notation

$$\begin{aligned} X=(X^1,\ldots ,X^a,Y^1,\ldots ,Y^a,Z^1,\ldots ,Z^b)=(X^i,Y^i,Z^j). \end{aligned}$$

Hence, we have

$$\begin{aligned} \sum _{i=1}^{a}x^iX^i+\sum _{i=1}^{a}y^iY^i+\sum _{j=1}^{b}z^jZ^j=0. \end{aligned}$$

If we decompose $\overline{{J}}(N)$ and $\overline{{J}}(X^i,Y^i,Z^j)$ respectively in tangential and normal components on $T_(x,y,z)S^{2a+b-1}(R),$ we obtain

$$\begin{aligned} \overline{{J}}(N)=\xi +AN , \overline{{J}}(X^i,Y^i,Z^j)={J}(X^i,Y^i,Z^j)+ u(X^i,Y^i,Z^j), \end{aligned}$$

where $(X^i,Y^i,Z^j)$ is a tangent vector field on $S^{2a+b-1}(R)$, u is $1-form$ on $S^{2a+b-1}(R)$ and A is smooth real function on $S^{2a+b-1}(R)$.

Using $A=\langle \overline{{J}}(N), N\rangle$, $\xi =\overline{{J}}(N)-AN$ , $u(X^i,Y^i,Z^j)=\langle (X^i,Y^i,Z^j),\xi \rangle$ and ${J}(X^i,Y^i,Z^j)=\overline{{J}}(X^i,Y^i,Z^J)-u(X^i,Y^i,Z^j)N,$ the elements of the induced structure $\sum =({J},g,\xi ,u,A)$ on $S^{2a+b-1}(R)$ by the golden Riemannian structure $(\overline{{J}},g)$ on $E^{2a+b}$ are given as follows:

• (i) $A=\frac{\phi r^2+(1-\phi )r_3^2}{R^2},$

• (ii) $\xi =\frac{1}{R}[\phi y^i-\frac{\phi r^2+(1-\phi )r_3^2}{R^2}x^i,\phi x^i-\frac{\phi r^2+(1-\phi )r_3^2}{R^2}y^i,\frac{(1-2\phi )r^2}{R^2}z^j],$

• (iii) $u(X)=\frac{1}{R}[\phi \sum _{i=1}^{a}(y^iX^i+x^iY^i)+(1-\phi )\sum _{j=1}^{b}z^jZ^j],$

• (iv) ${J}(X)=(\phi Y^i-\frac{1}{R}u(X)x^i,\phi X^i-\frac{1}{R}u(X)y^i,(1-\phi )Z^j-\frac{1}{R}u(X)z^j).$

Now, we have

$$\begin{aligned}&A=\langle \overline{{J}}(N),N\rangle \\&\quad \langle \overline{{J}}(N),N\rangle =\langle \xi +AN,N\rangle . \\&\quad \frac{1}{R^2}\langle \overline{{J}}(x^i,y^i,z^j),(x^i,y^i,z^j)\rangle =A\langle N,N\rangle \\&\quad \frac{1}{R^2}\langle ( \phi x^1,\ldots ,\phi x^a,\phi y^1,\ldots ,\phi \\&\quad y^a,(1-\phi )z^1,\ldots ,(1-\phi )z^b),(x^1,\ldots ,x^a,y^1,\ldots ,y^a,z^1,\ldots ,z^b)\rangle \\&\quad =\;\frac{A}{R^2}\langle (x^i,y^i,z^j),(x^i,y^i,z^j)\rangle \\&\quad \frac{1}{R^2}(\phi \sum _{i=1}^{a}(x^i)^2+\sum _{i=1}^{a}(y^i)^2+(1-\phi )\sum _{j=1}^{b}(z^j)^2)\\&\quad =\frac{A}{R^2}(\sum _{i=1}^{a}(x^i)^2+\sum _{i=1}^{a}(y^i)^2+\sum _{j=1}^{a}(z^j)^2) \\&\quad \phi r_1^2+\phi r_2^2+(1-\phi )r_3^2=A(r_1^2+r_2^2+r_3^2) \\&\quad \phi r^2+(1-\phi )r_3^2=AR^2 \end{aligned}$$

or,

$$\begin{aligned} A=\frac{\phi r^2+(1-\phi )r_3^2}{R^2}. \end{aligned}$$

Since,

$$\begin{aligned} \xi= & {} \overline{{J}}(N)-AN \\ \xi= & {} \overline{{J}}(\frac{1}{R}(x^i,y^i,z^j)-A(\frac{1}{R}(x^i,y^i,z^j)) \\ \xi= & {} \frac{1}{R}[(\phi x^1,\ldots ,\phi x^a,\phi y^1,\ldots ,\phi y^a,(1-\phi )z^1,\ldots ,(1-\phi )z^b)- \\&\frac{\phi r^2+(1-\phi )r_3^2}{R^2}(x^1,\ldots ,x^a,y^1,\ldots ,y^a,z^1,\ldots ,z^b)] \\ \xi= & {} \frac{1}{R}((\phi x^i,\phi y^i,(1-\phi )z^j)-\frac{\phi r^2+(1-\phi )r_3^2}{R^2}(x^i,y^i,z^j)) \\ \xi= & {} \frac{1}{R}(\frac{(2\phi -1)r_3^2}{R^2}x^i,\frac{(2\phi -1)r_3^2}{R^2}y^i,\frac{(1-2\phi )r^2}{R^2}z^j) \\ \xi= & {} \frac{(2\phi -1)}{R^3}(r_3^2x^i,r_3^2y^i,-r^2z^j). \end{aligned}$$

Also,

$$\begin{aligned} u(X)= & {} u(X^i,Y^i,Z^j)=\langle (X^i,Y^i,Z^j),\xi \rangle \\ u(X)= & {} \langle (X^i,Y^i,Z^j),{\overline{J}}(N)-AN\rangle \\ u(X)= & {} \langle (X^1,\ldots ,X^a,Y^1,\ldots ,Y^a,Z^1,\ldots ,Z^b),\frac{1}{R} \\&(\phi x^1,\ldots ,\phi x^a,\phi y^1,\ldots ,\phi y^a,(1-\phi )z^1,\ldots ,(1-\phi )z^b)\rangle \\ u(X)= & {} \frac{1}{R}(\sum _{i=1}^{a}\phi x^iX^i+\sum _{i=1}^{a}\phi y^iY^i+\sum _{j=1}^{b}(1-\phi )z^jZ^j) \\ u(X)= & {} \frac{1}{R}[\phi \sum _{i=1}^{a}(x^iX^i+y^iY^i)+(1-\phi )\sum _{j=1}^{b}z^jZ^j)]. \end{aligned}$$

Since,

$$\begin{aligned} {J}(X)= & {} {J}(X^i,Y^i,Z^j) \\ {J}(X)= & {} \overline{{J}}(X^i,Y^i,Z^j)-u(X^i,Y^i,Z^j).N \\ {J}(X)= & {} (\phi X^i,\phi Y^i,(1-\phi ) Z^j)-\frac{1}{R}u(X)(x^i,y^i,z^j) \\ {J}(X)= & {} (\phi X^i-\frac{1}{R}u(X)x^i,\phi Y^i-\frac{1}{R}u(X)y^i,(1-\phi )Z^j-\frac{1}{R}u(X)z^j), \end{aligned}$$

where $X=(X^i,Y^i,Z^j)$ is a tangent vector on sphere at any point $(x^i,y^i,z^j)$. Therefore, from the above relations we have $({J},<>,\xi ,u,a)$ induced structure by $\overline{{J}}$ from $E^{2a+b}$ on the sphere $S^{2a+b-1}(R)$ of codimension 1 in Euclidean space $E^{2a+b}(R)$.

In conclusion, $S^{2a+b-1}(r)$ is a totally umbilical hypersurface in $E^{2a+b}$.