Weingarten map of the hypersurface in 4-dimensional Euclidean space and its applications

TL;DR

This paper develops a practical method for computing the Weingarten map of hypersurfaces in 4-dimensional Euclidean space, enabling deeper analysis of their geometric properties similar to surface theory in 3D.

Contribution

It introduces an efficient computational approach for the Weingarten map of hypersurfaces in E^4, extending surface theory concepts to higher dimensions.

Findings

Method for calculating Weingarten map in E^4

Introduction of Gaussian and mean curvatures for hypersurfaces

Analysis of fundamental forms and Dupin indicatrix in E^4

Abstract

In this paper, by taking into account the beginning of the hypersurface theory in Euclidean space $E^4$, a practical method for the matrix of the Weingarten map (or the shape operator) of an oriented hypersurface $M^3$ in $E^4$ is obtained. By taking this efficient method, it is possible to study of the hypersurface theory in $E^4$ which is analog the surface theory in $E^3$. Furthermore, the Gaussian curvature, mean curvature, fundamental forms and Dupin indicatrix of $M^3$ is introduced.

Equations

$$⟨x,y⟩=x_1{y}_1+x_2{y}_2+x_3{y}_3+x_4{y}_4,$$

$$x\otimes y\otimes z=\left|{\begin{array}[]{*{20}{c}}{{e_{1}}}&{{e_{2}}}&{{e_{3}}}&{{e_{4}}}\\ {{x_{1}}}&{{x_{2}}}&{{x_{3}}}&{{x_{4}}}\\ {{y_{1}}}&{{y_{2}}}&{{y_{3}}}&{{y_{4}}}\\ {{z_{1}}}&{{z_{2}}}&{{z_{3}}}&{{z_{4}}}\end{array}}\right|.$$

$${\left\|{x\otimes y\otimes z}\right\|^{2}}=\left|{\begin{array}[]{*{20}{c}}{\left\langle{x,x}\right\rangle}&{\left\langle{x,y}\right\rangle}&{\left\langle{x,z}\right\rangle}\\ {\left\langle{y,x}\right\rangle}&{\left\langle{y,y}\right\rangle}&{\left\langle{y,z}\right\rangle}\\ {\left\langle{z,x}\right\rangle}&{\left\langle{z,y}\right\rangle}&{\left\langle{z,z}\right\rangle}\end{array}}\right|$$

$$M^3=\left\{X\in E^4|f:U\subset E^4\stackrel{diff.}{\to }\mathbb{R},f(X)=const.\overrightarrow{\nabla f}|{}_P\ne 0,P\in M^3\right\}$$

$$S:\chi \left(M^3\right)\to \chi \left(M^3\right),S\left(X\right)=D_{X}N,$$

$$I^q\left(X,Y\right)=⟨S^{q-1}\left(X\right),Y⟩,\forall X,Y\in \chi \left(M^3\right).$$

$$\begin{array}[]{l}\phi:U\subset{\mathbb{R}^{3}}\to{E^{4}}\\ \,\,\,\,\,\,\,\,\left({u,v,w}\right)\to\phi\left({u,v,w}\right)=\left({{\varphi_{1}}\left({u,v,w}\right),{\varphi_{2}}\left({u,v,w}\right),{\varphi_{3}}\left({u,v,w}\right),{\varphi_{4}}\left({u,v,w}\right)}\right)\end{array}$$

$$N=\frac{{\varphi }_u\otimes {\varphi }_v\otimes {\varphi }_w}{\Vert {\varphi }_u\otimes {\varphi }_v\otimes {\varphi }_w\Vert }$$

$$⟨N,{\varphi }_u⟩=⟨N,{\varphi }_v⟩=⟨N,{\varphi }_w⟩=0.$$

$$\begin{array}[]{l}S\left({{\phi_{u}}}\right)={D_{{\phi_{u}}}}N=\frac{{\partial N}}{{\partial u}}\\ \\ S\left({{\phi_{v}}}\right)={D_{{\phi_{v}}}}N=\frac{{\partial N}}{{\partial v}}\\ \\ S\left({{\phi_{w}}}\right)={D_{{\phi_{w}}}}N=\frac{{\partial N}}{{\partial w}}.\end{array}$$

$$\begin{array}[]{l}S\left({{\phi_{u}}}\right)=a_{11}\,{\phi_{u}}+a_{21}\,{\phi_{v}}+a_{31}\,{\phi_{w}}\\ S\left({{\phi_{v}}}\right)=a_{12}\,{\phi_{u}}+a_{22}\,{\phi_{v}}+a_{32}\,{\phi_{w}}\\ S\left({{\phi_{w}}}\right)=a_{13}\,{\phi_{u}}+a_{23}\,{\phi_{v}}+a_{33}\,{\phi_{w}}\end{array}$$

$$S=\left({\begin{array}[]{*{20}{c}}a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{array}}\right),$$

$$\begin{array}[]{l}\left\{\begin{array}[]{l}\left\langle{S\left({{\phi_{u}}}\right),{\phi_{u}}}\right\rangle=a_{11}\phi_{11}+a_{21}\phi_{12}+a_{31}\phi_{13}\\ \left\langle{S\left({{\phi_{u}}}\right),{\phi_{v}}}\right\rangle=a_{11}\phi_{12}+a_{21}\phi_{22}+a_{31}\phi_{23}\\ \left\langle{S\left({{\phi_{u}}}\right),{\phi_{w}}}\right\rangle=a_{11}\phi_{13}+a_{21}\phi_{23}+a_{31}\phi_{33},\end{array}\right.\\ \\ \left\{\begin{array}[]{l}\left\langle{S\left({{\phi_{v}}}\right),{\phi_{u}}}\right\rangle=a_{12}\phi_{11}+a_{22}\phi_{12}+a_{32}\phi_{13}\\ \left\langle{S\left({{\phi_{v}}}\right),{\phi_{v}}}\right\rangle=a_{12}\phi_{12}+a_{22}\phi_{22}+a_{32}\phi_{23}\\ \left\langle{S\left({{\phi_{v}}}\right),{\phi_{w}}}\right\rangle=a_{12}\phi_{13}+a_{22}\phi_{23}+a_{32}\phi_{33},\end{array}\right.\\ \\ \left\{\begin{array}[]{l}\left\langle{S\left({{\phi_{w}}}\right),{\phi_{u}}}\right\rangle=a_{13}\phi_{11}+a_{23}\phi_{12}+a_{33}\phi_{13}\\ \left\langle{S\left({{\phi_{w}}}\right),{\phi_{v}}}\right\rangle=a_{13}\phi_{12}+a_{23}\phi_{22}+a_{33}\phi_{23}\\ \left\langle{S\left({{\phi_{w}}}\right),{\phi_{w}}}\right\rangle=a_{13}\phi_{13}+a_{23}\phi_{23}+a_{33}\phi_{33},\end{array}\right.\end{array}$$

$$\begin{array}[]{l}\left\langle{{\phi_{u}},{\phi_{u}}}\right\rangle=\phi_{11},\,\,\left\langle{{\phi_{u}},{\phi_{v}}}\right\rangle=\phi_{12},\,\,\left\langle{{\phi_{u}},{\phi_{w}}}\right\rangle=\phi_{13},\\ \left\langle{{\phi_{v}},{\phi_{v}}}\right\rangle=\phi_{22},\,\,\left\langle{{\phi_{v}},{\phi_{w}}}\right\rangle=\phi_{23},\,\,\left\langle{{\phi_{w}},{\phi_{w}}}\right\rangle=\phi_{33}.\end{array}$$

$${\left\|{{\phi_{u}}\otimes{\phi_{v}}\otimes{\phi_{w}}}\right\|^{2}}=\left|{\begin{array}[]{*{20}{c}}\phi_{11}&\phi_{12}&\phi_{13}\\ \phi_{12}&\phi_{22}&\phi_{23}\\ \phi_{13}&\phi_{23}&\phi_{33}\end{array}}\right|\neq 0.$$

$$\left|{\begin{array}[]{*{20}{c}}\phi_{11}&\phi_{12}&\phi_{13}\\ \phi_{12}&\phi_{22}&\phi_{23}\\ \phi_{13}&\phi_{23}&\phi_{33}\end{array}}\right|=\Delta.$$

$$\begin{array}[]{l}S\left(U\right)={c_{1}}\,U+{c_{2}}\,V+{c_{3}}\,W\\ S\left(V\right)={c_{2}}\,U+{c_{4}}\,V+{c_{5}}\,W\\ S\left(W\right)={c_{3}}U+{c_{5}}\,V+{c_{6}}\,W,\end{array}$$

$$S=\left({\begin{array}[]{*{20}{c}}{{c_{1}}}&{{c_{2}}}&{{c_{3}}}\\ {{c_{2}}}&{{c_{4}}}&{{c_{5}}}\\ {{c_{3}}}&{{c_{5}}}&{{c_{6}}}\end{array}}\right).$$

$$\begin{array}[]{l}{c_{1}}=\left\langle{S\left(U\right),U}\right\rangle=\frac{1}{{{{\left\|{{\phi_{u}}}\right\|}^{2}}}}\left\langle{\frac{{\partial N}}{{\partial u}},{\phi_{u}}}\right\rangle,\\ \\ {c_{2}}=\left\langle{S\left(U\right),V}\right\rangle=\frac{1}{{\left\|{{\phi_{u}}}\right\|}}\frac{1}{{\left\|{{\phi_{v}}}\right\|}}\left\langle{\frac{{\partial N}}{{\partial u}},{\phi_{v}}}\right\rangle,\\ \\ {c_{3}}=\left\langle{S\left(U\right),W}\right\rangle=\frac{1}{{\left\|{{\phi_{u}}}\right\|}}\frac{1}{{\left\|{{\phi_{w}}}\right\|}}\left\langle{\frac{{\partial N}}{{\partial u}},{\phi_{w}}}\right\rangle,\\ \\ {c_{4}}=\left\langle{S\left(V\right),V}\right\rangle=\frac{1}{{{{\left\|{{\phi_{v}}}\right\|}^{2}}}}\left\langle{\frac{{\partial N}}{{\partial v}},{\phi_{v}}}\right\rangle,\\ \\ {c_{5}}=\left\langle{S\left(V\right),W}\right\rangle=\frac{1}{{\left\|{{\phi_{v}}}\right\|}}\frac{1}{{\left\|{{\phi_{w}}}\right\|}}\left\langle{\frac{{\partial N}}{{\partial v}},{\phi_{w}}}\right\rangle,\\ \\ {c_{6}}=\left\langle{S\left(W\right),W}\right\rangle=\frac{1}{{{{\left\|{{\phi_{w}}}\right\|}^{2}}}}\left\langle{\frac{{\partial N}}{{\partial w}},{\phi_{w}}}\right\rangle.\end{array}$$

$$\begin{array}[]{l}\left\langle{\frac{{\partial N}}{{\partial u}},{\phi_{u}}}\right\rangle+\left\langle{N,{\phi_{uu}}}\right\rangle=0,\\ \\ \left\langle{\frac{{\partial N}}{{\partial u}},{\phi_{v}}}\right\rangle+\left\langle{N,{\phi_{uv}}}\right\rangle=0,\\ \\ \left\langle{\frac{{\partial N}}{{\partial u}},{\phi_{w}}}\right\rangle+\left\langle{N,{\phi_{uw}}}\right\rangle=0,\\ \\ \left\langle{\frac{{\partial N}}{{\partial v}},{\phi_{v}}}\right\rangle+\left\langle{N,{\phi_{vv}}}\right\rangle=0,\\ \\ \left\langle{\frac{{\partial N}}{{\partial v}},{\phi_{w}}}\right\rangle+\left\langle{N,{\phi_{vw}}}\right\rangle=0,\\ \\ \left\langle{\frac{{\partial N}}{{\partial w}},{\phi_{w}}}\right\rangle+\left\langle{N,{\phi_{ww}}}\right\rangle=0.\end{array}$$

$${\left\|{{\phi_{u}}\otimes{\phi_{v}}\otimes{\phi_{w}}}\right\|^{2}}=\left|{\begin{array}[]{*{20}{c}}\phi_{22}&0&0\\ 0&\phi_{11}&0\\ 0&0&\phi_{33}\end{array}}\right|={\left\|{{\phi_{u}}}\right\|^{2}}\,{\left\|{{\phi_{v}}}\right\|^{2}}\,{\left\|{{\phi_{w}}}\right\|^{2}}.$$

$$\begin{array}[]{l}{c_{1}}=-\frac{1}{{{{\left\|{{\phi_{u}}}\right\|}^{3}}}}\frac{1}{{\left\|{{\phi_{v}}}\right\|}}\frac{1}{{\left\|{{\phi_{w}}}\right\|}}\det\left({{\phi_{uu}},{\phi_{u}},{\phi_{v}},{\phi_{w}}}\right),\\ \\ {c_{2}}=-\frac{1}{{{{\left\|{{\phi_{u}}}\right\|}^{2}}}}\frac{1}{{{{\left\|{{\phi_{v}}}\right\|}^{2}}}}\frac{1}{{\left\|{{\phi_{w}}}\right\|}}\det\left({{\phi_{uv}},{\phi_{u}},{\phi_{v}},{\phi_{w}}}\right),\\ \\ {c_{3}}=-\frac{1}{{{{\left\|{{\phi_{u}}}\right\|}^{2}}}}\frac{1}{{\left\|{{\phi_{v}}}\right\|}}\frac{1}{{{{\left\|{{\phi_{w}}}\right\|}^{2}}}}\det\left({{\phi_{uw}},{\phi_{u}},{\phi_{v}},{\phi_{w}}}\right),\\ \\ {c_{4}}=-\frac{1}{{\left\|{{\phi_{u}}}\right\|}}\frac{1}{{{{\left\|{{\phi_{v}}}\right\|}^{3}}}}\frac{1}{{\left\|{{\phi_{w}}}\right\|}}\det\left({{\phi_{vv}},{\phi_{u}},{\phi_{v}},{\phi_{w}}}\right),\\ \\ {c_{5}}=-\frac{1}{{\left\|{{\phi_{u}}}\right\|}}\frac{1}{{{{\left\|{{\phi_{v}}}\right\|}^{2}}}}\frac{1}{{{{\left\|{{\phi_{w}}}\right\|}^{2}}}}\det\left({{\phi_{vw}},{\phi_{u}},{\phi_{v}},{\phi_{w}}}\right),\\ \\ {c_{6}}=-\frac{1}{{\left\|{{\phi_{u}}}\right\|}}\frac{1}{{\left\|{{\phi_{v}}}\right\|}}\frac{1}{{{{\left\|{{\phi_{w}}}\right\|}^{3}}}}\det\left({{\phi_{ww}},{\phi_{u}},{\phi_{v}},{\phi_{w}}}\right).\end{array}$$

$$S=\left({\begin{array}[]{*{20}{c}}{\frac{\varphi_{11}}{\phi_{11}}}&{\frac{\varphi_{12}}{{\sqrt{\phi_{11}\phi_{22}}}}}&{\frac{\varphi_{13}}{{\sqrt{\phi_{11}\phi_{33}}}}}\\ \\ {\frac{\varphi_{12}}{{\sqrt{\phi_{11}\phi_{22}}}}}&{\frac{\varphi_{22}}{\phi_{22}}}&{\frac{\varphi_{23}}{{\sqrt{\phi}_{22}\phi_{33}}}}\\ \\ {\frac{\varphi_{13}}{{\sqrt{\phi_{11}\phi_{33}}}}}&{\frac{\varphi_{23}}{{\sqrt{\phi}_{22}\phi_{33}}}}&{\frac{\varphi_{33}}{\phi_{33}}}\end{array}}\right).$$

$$\begin{array}[]{l}\varphi_{11}=-\left\langle{{\phi_{uu}},N}\right\rangle,\,\,\varphi_{12}=-\left\langle{{\phi_{uv}},N}\right\rangle,\,\,\varphi_{13}=-\left\langle{{\phi_{uw}},N}\right\rangle,\\ \varphi_{22}=-\left\langle{{\phi_{vv}},N}\right\rangle,\,\,\varphi_{23}=-\left\langle{{\phi_{vw}},N}\right\rangle,\,\,\varphi_{33}=-\left\langle{{\phi_{ww}},N}\right\rangle.\end{array}$$

$$K=\frac{{\phi }_{11}{\phi }_{22}{\phi }_{33}+2{\phi }_{12}{\phi }_{13}{\phi }_{23}-{\phi }_{12}^2{\phi }_{33}-{\phi }_{13}^2{\phi }_{22}-{\phi }_{23}^2{\phi }_{11}}{{\varphi }_{11}{\varphi }_{22}{\varphi }_{33}}$$

$$H=\frac{1}{3}\left(\frac{{\phi }_{11}}{{\varphi }_{11}}+\frac{{\phi }_{22}}{{\varphi }_{22}}+\frac{{\phi }_{33}}{{\varphi }_{33}}\right),$$

$$\varphi \left(u,v,w\right)=\left(u,\frac{1}{u},v,w\right).$$

$$\begin{array}[]{l}S\left({\frac{{{\phi_{u}}}}{{\left\|{{\phi_{u}}}\right\|}}}\right)=\frac{{2{u^{3}}}}{{{{\left({1+{u^{4}}}\right)}^{3/2}}}}\frac{{{\phi_{u}}}}{{\left\|{{\phi_{u}}}\right\|}},\\ S\left({\frac{{{\phi_{v}}}}{{\left\|{{\phi_{v}}}\right\|}}}\right)=0,\\ S\left({\frac{{{\phi_{w}}}}{{\left\|{{\phi_{w}}}\right\|}}}\right)=0.\end{array}$$

$$S=\left({\begin{array}[]{*{20}{c}}{\frac{{2{u^{3}}}}{{{{\left({1+{u^{4}}}\right)}^{3/2}}}}}&0&0\\ 0&0&0\\ 0&0&0\end{array}}\right).$$

$$\varphi \left(u,v,w\right)=\left(\sin u\cos v\sin w,\sin u\sin v\sin w,\cos u\sin w,\cos w\right).$$

Full text

††2010 Mathematics Subject Classification: 53A05, 53A07, 14Q10††Keywords:Weingarten map, shape operator, vector product, ternary product, Dupin indicatrix

1 Introduction

Let $x=\sum\limits_{i=1}^{4}{{x_{i}}{e_{i}}},\,\,y=\sum\limits_{i=1}^{4}{{y_{i}}{e_{i}}},\,\,z=\sum\limits_{i=1}^{4}{{z_{i}}{e_{i}}}$ be three vectors in $\mathbb{R}^{4}$, equipped with the standard inner product given by

[TABLE]

where $\{{e_{1}},{e_{2}},{e_{3}},{e_{4}}\}$ is the standard basis of $\mathbb{R}^{4}.$ The norm of a vector $x\in\mathbb{R}^{4}$ is given by $\left\|x\right\|=\sqrt{\left\langle{x,x}\right\rangle}.$ The vector product (or the ternary product or cross product) of the vectors $x,y,z\in\mathbb{R}^{4}$ is defined by

[TABLE]

Some properties of the vector product are given as follows: (for the vector product in $\mathbb{R}^{4}$, see [1, 2, 5]

• i.

\left\{{\begin{array}[]{*{20}{l}}{{e_{1}}\otimes{e_{2}}\otimes{e_{3}}=-{e_{4}}}\\ {{e_{2}}\otimes{e_{3}}\otimes{e_{4}}={\mkern 1.0mu}{\mkern 1.0mu}{\mkern 1.0mu}{\mkern 1.0mu}{\mkern 1.0mu}{e_{1}}}\\ {{e_{3}}\otimes{e_{4}}\otimes{e_{1}}=-{e_{2}}}\\ {{e_{4}}\otimes{e_{1}}\otimes{e_{2}}={\mkern 1.0mu}{\mkern 1.0mu}{\mkern 1.0mu}{\mkern 1.0mu}{\mkern 1.0mu}{e_{3}}}\\ {{e_{3}}\otimes{e_{2}}\otimes{e_{1}}={\mkern 1.0mu}{\mkern 1.0mu}{\mkern 1.0mu}{\mkern 1.0mu}{\mkern 1.0mu}{e_{4}}}\end{array}}\right.

• ii.

[TABLE]

• iii.

$\left\langle{x\otimes y\otimes z,t}\right\rangle=\det\left({x,y,z,t}\right)$.

Let   $\,M^{3}\,$   be an oriented  3- dimensional hypersurface in   4-dimensional Euclidean space $E^{4}$. Let examine the implicit and parametric equations of $M^{3}$. Firstly; the implicit equation of $M^{3}$ can be defined by

[TABLE]

where ${\overrightarrow{\nabla f}\left|{{}_{P}}\right.}$ is the gradient vector of $M^{3}.$ The unit normal vector field of $M^{3}$ is defined by $N=\frac{{\overrightarrow{\nabla f}}}{{\left\|{\overrightarrow{\nabla f}}\right\|}}$.

The Weingarten map (or the shape operator) of $M^{3}$ is defined by

[TABLE]

where $D$ is the connection of $E^{4}$ and $\chi\left(M^{3}\right)$ is the space of vector fields of $M^{3}$. Then the Gauss curvature $K$ and mean curvature $H$ of $M^{3}$ are given by $K=detS$ and $H=\frac{1}{3}TrS$, respectively. Also, the $q-th$ fundamental forms of $M^{3}$ are given by [3],

[TABLE]

Secondly, to examine parametric form of the hypersurface $M^{3}$ given by the implicit equation in the eq (3), let consider

[TABLE]

where $\left({u,v,w}\right)\in R\subset{\mathbb{R}^{3}}$ and ${\varphi_{i}},1\leq i\leq 4$ are the real functions defined on $R$.

${M^{3}}=\phi\left(R\right)\subset{E^{4}}$ is a hypersurface if only if the frame field $\left\{{{\phi_{u}},{\phi_{v}},{\phi_{w}}}\right\}$ of $M^{3}$ is linearly independent system. It can be also seen by taking the Jacobian matrix {\left[\phi\right]_{*}}=\left[{\begin{array}[]{*{20}{c}}{{\phi_{u}}}&{{\phi_{v}}}&{{\phi_{w}}}\end{array}}\right] of the differential map of $\phi$. It is clear that if rank ${\left[\phi\right]_{*}}=3$, then the vector system $\left\{{{\phi_{u}},{\phi_{v}},{\phi_{w}}}\right\}$ is linearly independent. Furthermore, ${{\phi_{u}},{\phi_{v}},{\phi_{w}}}$ are the tangent vectors of the parameter curves $\alpha\left(u\right)=\phi\left({u,{v_{0}},{w_{0}}}\right)$, $\beta\left(v\right)=\phi\left({{u_{0}},v,{w_{0}}}\right)$ and $\gamma\left(w\right)=\phi\left({{u_{0}},{v_{0}},w}\right)$, respectively. Then the unit normal vector field of $M^{3}$ is defined by

[TABLE]

and it has the following properties:

[TABLE]

By using the Weingarten operator the below equalities can be written

[TABLE]

2 The matrix of the Weingarten map of hypersurface $M^{3}$ in $E^{4}$

In this original section, a practical method for the matrix of the Weingarten map of hypersurface $M^{3}$ in $E^{4}$ is introduced.

Let $M^{3}$ be an oriented hypersurface with the parametric equation $\phi\left({u,v,w}\right)$. Then $\{{\phi_{u}},{\phi_{v}},{\phi_{w}}\}$ is linearly independent and we also can write

[TABLE]

and the Weingarten matrix is given by

[TABLE]

where $a_{ij}\in\mathbb{R},1\leq i,j\leq 3.$ Using the equation (7), we have the following systems of linear equations:

[TABLE]

where

[TABLE]

Since the system $\{{\phi_{u}},{\phi_{v}},{\phi_{w}}\}$ is linearly independent, using the equations (2) and (9), we have

[TABLE]

Also, 3-linear equation systems given by the equation 8 have the determinant

[TABLE]

Because of the property ${\left\|{{\phi_{u}}\otimes{\phi_{v}}\otimes{\phi_{w}}}\right\|^{2}}=\Delta\neq 0$, these 3-linear equations systems can be solved by Cramer method. Then using the equations (6), (8) and (9) the matrix $S$ of the Weingarten map in $M^{3}$ can be found. Although $S$ is a symmetric linear operator, the matrix presentation $(a_{ij})$ of $S$ with respect to $\{{\phi_{u}},{\phi_{v}},{\phi_{w}}\}$ is not necessary to be symmetric because the system $\{{\phi_{u}},{\phi_{v}},{\phi_{w}}\}$ is not orthonormal.

2.1 Special Case

If we take the orthogonal frame field $\{{\phi_{u}},{\phi_{v}},{\phi_{w}}\}$ of the hypersurface $M^{3},$ then we have $\phi_{12}=\phi_{13}=\phi_{23}=0$ from the equation (9). Then, the system $\left\{{U=\frac{{{\phi_{u}}}}{{\left\|{{\phi_{u}}}\right\|}},\,\,\,V=\frac{{{\phi_{v}}}}{{\left\|{{\phi_{v}}}\right\|}},\,\,W=\frac{{{\phi_{w}}}}{{\left\|{{\phi_{w}}}\right\|}}}\right\}$ is an orthonormal frame field. Furthermore, we can write the following equations

[TABLE]

then, the matrix of the Weingarten map can be calculated as follows:

[TABLE]

By using the equations (4), (6) and (10), the coefficients $c_{i}\in\mathbb{R},$ $1\leq i\leq 6$ can be calculated as follows:

[TABLE]

By using the equation (5), we can also write six equations as below:

[TABLE]

Also, by using the equations (2) and (9), we find

[TABLE]

Hence we find the coefficients $c_{1},c_{2},c_{3},c_{4},c_{5},c_{6}$ of the Weingarten matrix in the equation (10) as follows:

[TABLE]

So, by taking into account the equations (4), (13) and (14) we have the symmetric Weingarten matrix

[TABLE]

where

[TABLE]

Finally the following theorem can be given for hypersurface $M^{3}$ in $E^{4}$:

Theorem 1

Let $M^{3}$ be an oriented hypersurface in $E^{4}.$ Then the Gaussian curvature and the mean curvature of $M^{3}$ can be given by:

[TABLE]

and

[TABLE]

respectively.

Proof. By using the equation (15) and the definitions of the Gaussian curvature $K$ and the mean curvature $H$, the theorem can be easily proved.

Example 2

Let $M^{3}$ be an oriented hypersurface with the implicit equation $xy=1$ in $E^{4}$. The parametric equation of $M^{3}$ can be given by

[TABLE]

*Then, we obtain ${\phi_{u}}\otimes{\phi_{v}}\otimes{\phi_{w}}=\left({-\frac{1}{{{u^{2}}}},-1,0,0}\right)$ and the unit normal field

$N=\frac{1}{{\sqrt{1+{u^{4}}}}}\left({-1,-{u^{2}},0,0}\right)$. By using the orthonormal basis $\left\{{\frac{{{\phi_{u}}}}{{\left\|{{\phi_{u}}}\right\|}},\,\,\,\frac{{{\phi_{v}}}}{{\left\|{{\phi_{v}}}\right\|}},\,\,\frac{{{\phi_{w}}}}{{\left\|{{\phi_{w}}}\right\|}}}\right\},$

we have*

[TABLE]

So, we find the Weingarten matrix $S$ as:

[TABLE]

Example 3

Let $S^{3}$ be a hypersphere with the implicit equation ${x^{2}}+{y^{2}}+{z^{2}}+{t^{2}}=1$ in $E^{4}$. The parametric equation of $S^{3}$ can be given by

[TABLE]

Then, $\{{\phi_{u}},{\phi_{v}},{\phi_{w}}\}$ is an orthogonal system. Also we have the orthonormal basis $\{U,V,W\}$ of $S^{3}$ such that

[TABLE]

Furthermore, the unit normal vector field $N$ can be found:

[TABLE]

Then using the equation (15), we obtain $S={I_{3}}.$

Theorem 4

Let $M^{3}$ be an oriented hypersurface in $E^{4}$ and let $\left\{{{X_{P}},{Y_{P}},{Z_{P}}}\right\}$ be a linearly independent vector system of the tangent space ${T_{{M^{3}}}}\left(P\right)$. Then, we have

[TABLE]

where $K$ and $H$ are the Gaussian curvature and the mean curvature of $M^{3}$, respectively.

Proof. By using (i), (ii) parts of the equation (2) and considering the definitions of the Gaussian curvature $K$ and the mean curvature $H$ the theorem can be easily proved.    In [4], it is proved that these equations are also provided for closed hypersurfaces.

Theorem 5

Let $M^{3}$ be an oriented hypersurface in $E^{4}$ and let $I^{q}$, $K$, $H$ be the $q$-th fundamental forms, the Gaussian curvature and the mean curvature, respectively. Then we have

[TABLE]

where $h$ is the harmonic mean of the non-zero principal curvatures of $M^{3}.$

Proof. Let ${k_{1}},{k_{2}},{k_{3}}$ be the characteristic values of the Weingarten map $S$ (or the principal curvatures of $M^{3}$ ). Then we obtain the characteristic polynomial ${P_{S}}\left(\lambda\right)$ of the Weingarten map $S$ of $M^{3}$ as

[TABLE]

By using the Cayley-Hamilton theorem, we obtain

[TABLE]

By using the definitions of the $q-th$ fundamental forms, the Gaussian curvature, the mean curvature and the harmonic mean

[TABLE]

of the principal curvature ${k_{1}},{k_{2}},{k_{3}}$,

we obtain the equation (16).

3 Dupin indicatrix of the hypersurface in $E^{4}$

Let $X,Y,Z$ be three principal vectors according to the principal curvatures ${k_{1}},{k_{2}},{k_{3}}$ of $M^{3}$. If we consider the orthonormal basis $\{X,Y,Z\}$ of $M^{3}$ then for any tangent vector $W_{P}\in T_{M^{3}}(P),$ we can write ${W_{P}}=x\,{X_{P}}+y\,{Y_{P}}+z\,{Z_{P}},\,\,$ where $x,y,z\in\mathbb{R},$ and

[TABLE]

Here, the Dupin indicatrix $\mathbb{D}$ of $M^{3}$ can be defined by

[TABLE]

In another words, the Dupin indicatrix corresponds to a

hypercylinder which has the equation

[TABLE]

Now, we will examine the Dupin indicatrix according to the Gaussian curvature $K:$

1)   Let $K\left(P\right)>0.$

• •

If ${k_{1}},\,{k_{2}},\,{k_{3}}>0$ then for equation of the Dupin indicatrix, we can write ${k_{1}}{x^{2}}+{k_{2}}{y^{2}}+{k_{3}}{z^{2}}=\pm 1.$ Hence, the Dupin indicatrix is the ellipsoidal class and this equation is called ellipsoidal cylinder in $E^{4}.$ In this condition,    $P\in M^{3}$ is called an ellipsoidal point.

• •

If ${k_{1}}>0,\,\,{k_{2}},\,{k_{3}}<0$ or ${k_{2}}>0,\,\,{k_{1}},\,{k_{3}}<0$ or ${k_{3}}>0$ ${k_{1}},\,{k_{2}}<0$ then for equation of the Dupin indicatrix, we can write ${k_{1}}{x^{2}}-{k_{2}}{y^{2}}-{k_{3}}{z^{2}}=\pm 1.$ Hence, the Dupin indicatrix is the hyperboloidical class and this equation is called hyperboloidical cylinder one or two sheets in $E^{4}$. In this condition, $P\in M^{3}$ is called a hyperboloidical point.

2)   Let $K\left(P\right)<0.$

• •

If only one of ${k_{i}}$’s,  ${i=1,2,3}$ is negative, then for the equation of the Dupin indicatrix, we can write

[TABLE]

The above equations are called one or two sheeted hyperboloidical cylinder in $E^{4}.$ Then $P\in M^{3}$ is called a hyperboloidical point.

• •

If ${k_{1}},{k_{2}},{k_{3}}<0$ then the Dupin indicatrix is the ellipsoidal class and this equation is called ellipsoidal cylinder in $E^{4}.$ So $P\in M^{3}$ is called a ellipsoidal point.

3. Let $K\left(P\right)=0$.

• •

If ${k_{1}}=0$ or ${k_{2}}=0\,\ or\,\ {k_{3}}=0$, then for the equation of the Dupin indicatrix for each case, we get

• i

If ${k_{1}}=0,\,{k_{2}},\,{k_{3}}$ are the same or different signs then ${k_{2}}{y^{2}}+{k_{3}}{z^{2}}=\pm 1$.

• ii

If ${k_{2}}=0,\,{k_{1}},{k_{3}}$ are the same or different signs then ${k_{1}}{x^{2}}+{k_{3}}{z^{2}}=\pm 1$.

• iii

If ${k_{3}}=0,\,{k_{1}},\,{k_{2}}$ are the same or different signs then ${k_{1}}{x^{2}}+{k_{2}}{y^{2}}=\pm 1.$

These equations are called elliptic cylinder or hyperbolic cylinder in $E^{4}.$ In this condition, $P\in M^{3}$ is called an elliptic cylinder or hyperbolic cylinder point.

• •

If ${k_{1}}={k_{2}}={k_{3}}=0$ then the point $P\in M^{3}$ is a flat point.

• •

If any two of ${k_{i}}$’s,  ${i=1,2,3}$ are zero and other positive or negative then ${k_{3}}{z^{2}}=\pm 1$ or ${k_{2}}{y^{2}}=\pm 1$ or ${k_{1}}{x^{2}}=\pm 1.$