Euclidean hypersurfaces isometric to spheres

1.   Introduction

Given an orientable immersed hypersurface $M^{n}$ in the Euclidean space $E^{n+1}$ with unit normal $\xi$ and shape operator $T$, we denote by $\psi :\, M^{n}\rightarrow E^{n+1}$ the immersion, by $g$ the induced metric, and denote the hypersurface by $\left(M^{n}, \psi, g, \xi, T\right)$. The eigenvalues $\mu _{1}, .., \mu _{n}$ of the shape operator $T$ are called principal curvatures of the hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ and play a very important role in the geometry^[1,2] as well as the topology of $\left(M^{n}, \psi, g, \xi, T\right)$ (cf. ^[3,4,5]). It is fascinating to see that constraints on principal curvatures also influence the topology of exterior $E^{n+1}\backslash M^{n}$. An interesting result in ^[6] proves that if a compact and connected hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$, $n\geq 2$, $A^{n}$ is the unbounded component of $E^{n+1}\backslash M^{n}$ and the principal curvatures satisfy $\mu _{1}+..+\mu _{n} < 0$, then $A^{n}$ simply connected. There are many important aspects of studying the geometry of hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ in $E^{n+1}$ and one of them is to study the geometry of $\left(M^{n}, \psi, g, \xi, T\right)$ under the condition $\Delta ^{2}\psi = 0$, and the study of such submanifolds was initiated by Chen ^[7,8,9], calling them biharrmonic hypersurfaces. Moreover, Chen conjectured that a biharmonic hypersurface of $E^{n+1}$ is minimal (cf. ^[10,11,12]). An interesting result in Euclidean submanifolds is that of Jacobowicz (cf. ^[6]), which states an $n$-dimensional Riemannian manifold $(M^{n}, g)$ with sectional curvatures less than a constant $\lambda ^{-2}$ admits an isometric immersion into the Euclidean space $E^{2n-1}$ can never be contained in a ball of radius $\lambda$ in $E^{2n-1}$ and this result is the generalization of the nonembeddability result due to Tompkins (cf. ^[3]). Taking clue from ^[6,13,14], in ^[2], the author generalized the result to a compact hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ in $E^{n+1}$, where it is proved that if the scalar curvature of $\left(M^{n}, g\right)$ is less than a constant $n(n-1)\lambda ^{-2}$, then no immersion $\psi :M^{n}\rightarrow E^{n+1}$ is contained in a ball of radius $\lambda$ in $E^{n+1}$. It is still open to show that if the Ricci curvatures of a compact Riemannian manifold $\left(M^{n}, g\right)$ are less than a constant $(n-1)\lambda ^{-2}$, then no immersion $\psi :\, M^{n}\rightarrow E^{n+1}$ is contained in a ball of radius $\lambda$ in $E^{n+1}$. The geometrical and topological properties^[15,16,17] of hypersurfaces are the branches of differential geometry ^[18,19], and a small portion of it is described above ^[20,21,22]. The submanifolds theory^[23,24] and soliton theory ^[25,26,27], etc., continue to inspire new insights and discoveries to help solve these problems and make it an active area of research in many branches of mathematics and physics.

Inspired by the previous results, in the present paper, we first intend to study the geometrical and topological properties of an orientable hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ in the Euclidean space $E^{n+1}$, and we express the vector $\psi$ as

where $\boldsymbol{\omega }$ is tangential projection of $\psi$ to $M^{n}$. We call $\boldsymbol{\omega }$ the basic vector field and the function $\sigma$ the support function of the hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$.

Secondly, the paper studied the static perfect fluid equation on a Riemannian manifold $(M^{n}, g)$, which is given by

where $Ric$ is the Ricci tensor, $Hes(\sigma)$ is the Hessian of $\sigma$, $\tau$ is the scalar curvature and $\Delta$ is the Laplace operator on $\left(M^{n}, g\right)$. It is known that the static perfect fluid equation has immense importance in mathematical physics, in particular in fluid dynamics and also in differential geometry (cf ^[28] and references therein).

Furthermore, we investigate the impacts on the geometry of the complete and simply connected hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ in the Euclidean space $E^{n+1}$ with support function $\sigma$ satisfying static perfect fluid equation and find conditions under which this hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ is isometric to the Euclidean sphere $S^{n}(c)$ of constant curvature $c$ (see Theorem 3.1).

We also show a compact and simply connected hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ in $E^{n+1}$ that has support function $\sigma$, mean curvature $H = \frac{1}{n}TraceT$, and the shape operator $T$ satisfies

where $\nabla \sigma$ is the gradient of the support function $\sigma$, if and only if $H$ is a constant and hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ $\ $is isometric to $S^{n}\left(H^{2}\right)$ (see Theorem 4.1).

For a vector field $\zeta$ on a Riemannian manifold $\left(M^{n}, g\right)$ to be incompressible, it is required that $\text{div}\zeta = 0$. This notion is borrowed from fluid mechanics (cf. ^[8,30]). In this paper, further we study compact hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ in the Euclidean space $E^{n+1}$ of positive Ricci curvature with basic vector field $\boldsymbol{\omega }$ incompressible and show that for such hypersurfaces the mean curvature $H$ is a constant and $\left(M^{n}, \psi, g, \xi, T\right)$ is isometric to $S^{n}\left(H^{2}\right)$ and also the converse holds (see Theorem 5.1).

2.   Preliminaries

Let $\left(M^{n}, \psi, g, \xi, T\right)$ be an orientable hypersurface in the Euclidean space $E^{n+1}$ with immersion $\psi :M^{n}\rightarrow E^{n+1}$. We denote by $\Omega \left(M^{n}\right)$ the space of smooth vector fields on $M^{n}$ and by $\nabla _{E}$ the covariant derivative in the direction of $E\in \Omega \left(M^{n}\right)$ with respect to the Riemannian connection of the induced metric $g$. Then, on differentiating Eq (1.1) with respect to $E\in \Omega \left(M^{n}\right)$ and using the Gauss-Weingarten formulae and equating akin parts, we obtain the following for the basic vector field $\boldsymbol{\omega }$ ^[10,29].

The curvature tensor of $\left(M^{n}, \psi, g, \xi, T\right)$ has the following expression:

and using the following formulas of the Ricci tensor and the mean curvature $H$ of $\left(M^{n}, \psi, g, \xi, T\right)$

where $\left\{ u_{\alpha }\right\} _{1}^{n}$ is a local orthonormal frame on $\left(M^{n}, \psi, g, \xi, T\right)$, we obtain through Eq (2.1)

The scalar curvature $\tau$ of $\left(M^{n}, \psi, g, \xi, T\right)$ is obtained by taking trace in the above equation, and we have the formula

where

The shape operator $T$ of the hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ of the Euclidean space $E^{n+1}$ satisfies the following Codazzi equation:

and here $\left(\nabla T\right) \left(E_{1}, E_{2}\right)$ means

On differentiating the expression for $H$ in Eq (2.3) and utilizing Eq (2.6), we confirm

and it accounts for the expression for the gradient $\nabla H$ of $H$ given by

For a compact hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ of the Euclidean space $E^{n+1}$ with support function $\sigma$ and mean curvature $H$, Minkowski's integral formula states

Recall that for a vector field $\boldsymbol{\omega }$ on a Riemannian manifold $(M^{n}, g)$, we say that an operator $S$ defined on $(M^{n}, g)$ is invariant under $\boldsymbol{\omega }$ if the following holds:

where $\left\{ \varphi _{i}\right\}$ is the local flow of $\boldsymbol{\omega }$. It follows that if $S$ is invariant under $\boldsymbol{\omega }$, then we have

where $£_{\boldsymbol{\omega }}$ is the Lie derivative with respect to $\boldsymbol{\omega }$.

Next, we discuss the model example of the sphere $S^{n}(c)$ of constant curvature $c$ as an embedded hypersurface $\psi :S^{n}(c)\rightarrow E^{n+1}$ given by

The unit normal $\xi$ to $S^{n}(c)$ is expressed by $\xi = \sqrt{c}\psi$ and the shape operator $T = -\sqrt{c}I$. Moreover, the support function $\sigma$ of $S^{n}(c)$ is given by $\sigma = \frac{1}{\sqrt{c}}$ and the basic vector field $\boldsymbol{\omega }$ of $S^{n}(c)$ is given by $\boldsymbol{\omega } = 0$. The mean curvature $H$ of $S^{n}(c)$ is given by $H = -\sqrt{c}$.

3.   Hypersurfaces with support function solution of static perfect fluid equation

Let $\left(M^{n}, \psi, g, \xi, T\right)$ be an orientable hypersurface in the Euclidean space $E^{n+1}$ with immersion $\psi :M^{n}\rightarrow E^{n+1}$ support function $\sigma$, basic vector field $\boldsymbol{\omega }$, mean curvature $H$, and scalar curvature $\tau$. We assume that the pressure on the support function $\sigma$ to satisfy the static perfect fluid equation, namely

where $Hes(\sigma)$ is the Hessian, defined by

and we also have Hessian operator $H^{\sigma }$ and the Ricci operator $Q$ defined by

The Laplace operator $\Delta$ on $\left(M^{n}, \psi, g, \xi, T\right)$ is defined by $\Delta \sigma = \text{div}\left(\nabla \sigma \right)$, and it is also the trace of $H^{\sigma }$. Moreover, we also assume that the hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ of the Euclidean space $E^{n+1}$ has a basic vector field $\boldsymbol{\omega }$ under which the operator $T$ is invariant, that is,

which amounts to the following

We intend to show that a complete and simply connected hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ of $E^{n+1}$ having positive Ricci curvature, subjected to conditions (3.1) and (3.2), gets ready to acquire the shape of a sphere, as seen in the following:

Theorem 3.1. A complete and simply connected hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ of the Euclidean space $E^{n+1}$, $n > 1$, with positive Ricci curvature satisfies the shape operator $T$ is invariant under the basic vector field $\boldsymbol{\omega }$, and the support function $\sigma$ satisfies the static perfect fluid equation if and only if the mean curvature $H$ is a constant and $\left(M^{n}, \psi, g, \xi, T\right)$ is isometric to the sphere $S^{n}\left(H^{2}\right)$.

Proof. Let $\left(M^{n}, \psi, g, \xi, T\right)$ be a complete and simply connected hypersurface of positive Ricci curvature in the Euclidean space $E^{n+1}$, $n > 1$, with support function $\sigma$ and the basic vector field $\boldsymbol{\omega }$ satisfying Eqs (3.1) and (3.2), respectively. On employing Eq (2.1) in Eq (3.2), we extract the following

Next, we wish to use Eq (2.1) in order to compute the Hessian operator $H^{\sigma }$ as follows:

which, in view of Coddazzi Eq (2.6) and Eqs (2.1) and (3.3), yields

Equation (2.4) has the form $QE = nHTE-T^{2}E$ and employing it in Eq (3.4), it turns out that

Defining the second fundamental form $\Xi$ of the hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$, by

Thus, Eq (3.5) now takes the shape

Next, taking trace in the Eq (3.4), we conclude

and using Eq (2.5) in above the equation, we have

Inserting this value in Eq (3.1), we confirm

Combining Eqs (3.6) and (3.7), we have

Assume that $\left(1+n\sigma H\right) = 0$ holds. Then, we conclude $\sigma \nabla H = -H\nabla \sigma$, which, on employing Eq (2.1) implies

Now, we use Eqs (2.6) and (3.3) to conclude $\left(\nabla T\right)\left(E, \boldsymbol{\omega }\right) = 0$, and taking the inner product with $E$ while using the symmetry of $T$ gives

On taking the sum in the above equation over a local orthonormal frame $\left\{u_{\alpha }\right\} _{1}^{n}$ and employing the (2.7), we conclude

The Eq (3.9), on taking the inner product with $\boldsymbol{\omega }$ and using Eq (3.10), we get

and utilizing it with Eq (2.4) in computing $Ric\left(\boldsymbol{\omega }, \boldsymbol{\omega }\right)$, we get

As $Ric > 0$, the above equation implies the basic vector field $\boldsymbol{\omega } = 0$, that is, $\text{div}\left({\boldsymbol{\omega }}\right) = n(1+\sigma H)$ (by virtue of Eq (2.1)) and we get $\sigma H = -1$, and combining it with our assumption $\left(1+n\sigma H\right) = 0$, we get $(n-1) = 0$, a contradiction as $n > 1$. Hence, we have $\left(1+n\sigma H\right) \neq 0$ and as $M^{n}$ is simply connected. It is connected and as such, Eq (3.8) now yields

that is equivalent to

which implies

Taking $E = u_{\alpha}$ in the above equation for an orthonormal frame $\left\{u_{\alpha }\right\} _{1}^{n}$, and summing while using Eq (2.7), we conclude $n\nabla H = \nabla H$, which, in view of $n > 1$, implies $H$ is a constant. Now, the Eq (2.2), gives

The above equation confirms that the hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ has constant curvature $H^{2}$. Note that the constant $H^{2} > 0$ as $Ric = (n-1)H^{2}g > 0$. Hence, the complete and simply connected $\left(M^{n}, \psi, g, \xi, T\right)$ is isometric to $S^{n}\left(H^{2}\right)$.

The converse is trivial, as for the support function $\sigma$ of $S^{n}\left(H^{2}\right)$ is a constant $\sigma = -\frac{1}{H}$ that satisfies Eq (3.1) and the basic vector field ${\boldsymbol{\omega }} = 0$ satisfies (3.2) and also that $S^{n}\left(H^{2}\right)$ has positive Ricci curvature. □

4.   Hypersurface with gradient of support function an eigenvector of shape operator

Let $\left(M^{n}, \psi, g, \xi, T\right)$ be a complete and simply connected hypersurface in $E^{n+1}$ with basic vector field $\boldsymbol{\omega }$, support function $\sigma$, and mean curvature $H$. At times, simple restrictions lead to very fundamental results. In this section, we are going to witness a similar situation. We are going to show that a simple condition like $T\left(\nabla \sigma \right) = H\left(\nabla \sigma \right)$, that is, the gradient $\nabla \sigma$ of $\ \sigma$ is an eigenvector of $T$ with eigenvalue function $H$, and an appropriate lower bound on the integral of $\left\Vert \nabla \sigma \right\Vert ^{2}$ leads to a characterization of the sphere. Indeed, we prove the following:

Theorem 4.1. A compact and simply connected hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ of the Euclidean space $E^{n+1}$, $n > 1$, with support function $\sigma$, basic vector field $\boldsymbol{\omega }$ and mean curvature $H$ satisfy

if and only if, $H$ is a constant and $\left(M^{n}, \psi, g, \xi, T\right)$ is isometric to the sphere $S^{n}\left(H^{2}\right)$.

Proof. Let $\left(M^{n}, \psi, g, \xi, T\right)$ be a compact and simply connected hypersurface $E^{n+1}$, $n > 1$, with support function $\sigma$, the basic vector field $\boldsymbol{\omega }$ and mean curvature $H$, satisfying

and

Inserting the Eq (2.1) namely $\nabla \sigma = -T \boldsymbol{\omega }$ in Eq (4.1) to reach $T^{2} \boldsymbol{\omega } = HT\boldsymbol{\omega }$, which by the inner product with $\boldsymbol{\omega }$, gives

Employing Eq (2.4) in the above equation, we arrive at

and combining it with Eq (2.1), confirms

Note that Eq (4.1) also implies $T\left(\nabla \sigma \right) = \nabla \left(H\sigma \right) -\sigma \nabla H$, which by the inner product with $\boldsymbol{\omega }$, implies

and employing Eq (2.1), we conclude

We insert the above equation in Eq (4.3), yielding

Next, we use a local frame $\left\{ u_{\alpha }\right\} _{1}^{n}$ and Eq (2.1), in computing

and

where the Lie derivative $£ _{\boldsymbol{\omega }}g$ is given by

Now, recall the following integral formula from ^[31], for the compact hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$

and inserting Eqs (4.4), (4.6) and (4.7) in the above equation have

Rearranging the above equation, we have

which, on employing Eq (4.5), gives

Now, employing the inequality (4.2) in the above equation results in

Further, note that it is due to Schwartz's inequality $\left\Vert T\right\Vert ^{2}\geq nH^{2}$, the integrand in above inequality is non-negative, therefore, we conclude

Note that the implication $\sigma = 0$ of the above equation is forbidden due to Minkowski's formula (2.8). Hence, $\sigma \neq 0$ on the connected $M^{n}$ forces Eq (4.8) to yield

which being an equality in Schwartz's inequality, holds if and only if $T = HI$, and it leads to $H$ a constant (see argument after Eq (3.11)). Hence, as seen in the proof of Theorem 1, the curvature tensor hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ is given by

Now, by a global argument that on a compact hypersurface in a Euclidean space, there exists a point where all sectional curvatures are positive, we see that the constant $H^{2} > 0$. Hence, the simply connected hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ being compact is also complete, and thus, the complete and simply connected hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ has constant positive curvature $H^{2}$. Hence, $\left(M^{n}, \psi, g, \xi, T\right)$ is isometric to $S^{n}\left(H^{2}\right)$. The converse is trivial. □

5.   Hypersurfaces with incompressible basic vector field

In this section, we study the geometry of a compact and simply connected hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ in a Euclidean space $E^{n+1}$ with basic vector field $\boldsymbol{\omega }$, support function $\sigma$ and mean curvature $H$ with basic vector field $\boldsymbol{\omega }$ incompressible, that is, satisfying $\text{div} \left(\boldsymbol{\omega }\right) = 0$. The notion that $\boldsymbol{\omega }$ is incompressible for the compact hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ is so strong that it alone suffices in forcing the hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ of positive Ricci curvature to acquire the shape of a sphere, as seen in the following:

Theorem 5.1. A compact and simply connected hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ of the Euclidean space $E^{n+1}$, $n > 1$, of positive Ricci curvature with support function $\sigma$, mean curvature $H$, has the basic vector field $\boldsymbol{\omega }$ incompressible if and only if $H$ is a constant and $\left(M^{n}, \psi, g, \xi, T\right)$ is isometric to the sphere $S^{n}\left(H^{2}\right)$.

Proof. Suppose $\left(M^{n}, \psi, g, \xi, T\right)$ is a compact and simply connected hypersurface of the Euclidean space $E^{n+1}$, $n > 1$, with the basic vector field $\boldsymbol{\omega }$ incompressible. Then, as $\text{div} \left(\boldsymbol{\omega }\right) = 0$, by Eq (4.5), $\sigma H = -1$ and, therefore, both functions $\sigma$ and $H$ are nowhere zero on $M^{n}$ and we have

and joining it with Eq (2.7), we conclude

Now, using $\nabla \sigma = -T\boldsymbol{\omega }$ from Eq (2.1), which on differentiation gives the following expression for $H^{\sigma }$

and taking trace, while using the symmetry of $T$, we conclude

Now using the above equation with Eq (4.5) in the form $\sigma H = -1$ and Eq (5.2), we confirm

Multiplying the above equation by $\sigma$ and then integrating by parts would lead us to

which could be rearranged as

Now, using $\sigma H = -1$ and (5.1), in the above equation, it yields

We observe that

and that as $\boldsymbol{\omega }$ is incompressible implies $\text{div} \left(\sigma \boldsymbol{\omega }\right) = \boldsymbol{\omega } \left(\sigma \right)$. Thus, we have

Also, we see by Eq (2.1), $\boldsymbol{\omega }\left(\sigma \right) = g\left(\boldsymbol{\omega }, \nabla \sigma \right) = -g\left(T\boldsymbol{\omega }, \boldsymbol{\omega }\right)$ and the above equation becomes

Next, we plug the above equation with Eq (2.4) and get

Plugging the above equation with Eq (5.3), we achieve

and by Eq (2.1), $\nabla \sigma = -T\boldsymbol{\omega }$, the above equation reduces to

Owing to Schwartz's inequality, the left-hand side in the above equation is non-negative, and since the Ricci curvature is positive, the left-hand side in the above equation is strictly negative. The only possible conclusion is

The second equation in Eq (5.4) together with $\nabla \sigma = -T \boldsymbol{\omega }$ implies that $\sigma$ is a constant. This constant $\sigma$ has to be non-zero, for otherwise both $\boldsymbol{\omega } = 0$ and $\sigma = 0$ would imply $\psi = \boldsymbol{\omega }+\sigma \xi = 0$ a contradiction. Hence, $\sigma$ is a non-zero constant and also simultaneously $H$ is a non-zero constant (owing to $\sigma H = -1$) and Eq (5.4) reduces to

Then, using the following Eq (4.8) as in the proof of Theorem 4.1, we conclude $\left(M^{n}, \psi, g, \xi, T\right)$ is isometric to $S^{n}\left(H^{2}\right)$. The converse follows trivially as the sphere $S^{n}\left(H^{2}\right)$ has positive Ricci curvature, and as a hypersurface the sphere has a basic vector field $\boldsymbol{\omega } = 0$, which is automatically incompressible.

□

6.   Conclusions

There is a lot to comment on each result in this paper and future scopes of their respective generalizations. However, we shall concentrate on the Theorem 3.1, where it is proved that a complete and simply connected hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ of the Euclidean space $E^{n+1}$, $n > 1$, with positive Ricci curvature, satisfies the shape operator $T$ is invariant under the basic vector field $\boldsymbol{\omega }$, and the support function $\sigma$ satisfies the static perfect fluid equation if and only if the mean curvature $H$ is a constant and $\left(M^{n}, \psi, g, \xi, T\right)$ is isometric to the sphere $S^{n}\left(H^{2}\right)$. There are natural questions tagged to this result, namely:

(a) Can we relax the condition that the hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ has positive Ricci curvature?

(b) Can we replace the condition in Theorem 3.1 that the operator $T$ is invariant under $\boldsymbol{\omega }$ by the condition $T\left(\boldsymbol{\omega }\right) = \frac{\tau }{n}\boldsymbol{\omega }$?

(c) Note that apart from the support function $\sigma$ of the hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$, there is yet another function $\delta: \, M^{n}\rightarrow R$ defined by

and this function $\delta$ satisfies $\nabla \delta = \boldsymbol{\omega }$. A natural question would be to find additional conditions under which the complete and simply connected hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ with function $\delta$ satisfying static perfect fluid equation

is isometric to the sphere $S^{n}\left(H^{2}\right)$?

These questions would be our focus for future studies on the hypersurface $\left(M^{n}, \psi, g, \xi, T\right)$ of the Euclidean space $E^{n+1}$.

Author contributions

Yanlin Li: Conceptualization, investigation, methodology, writing-review and editing; Nasser Bin Turki: Conceptualization, investigation, methodology, writing-review and editing; Sharief Deshmukh: Conceptualization, investigation, methodology, writing-review and editing; Olga Belova: Conceptualization, investigation, methodology, writing-review and editing. All authors of this article have been contributed equally. All authors have read and agreed to the published version of the manuscript.

Acknowledgments

This project was supported by the Researchers Supporting Project number (RSP2024R413), King Saud University, Riyadh, Saudi Arabia.

Conflict of interest

The authors declare no conflict of interest.