Some notes on the tangent bundle with a Ricci quarter-symmetric metric connection

Yanlin Li; Aydin Gezer; Erkan Karakaş; Yanlin Li; Aydin Gezer; Erkan Karakaş

doi:10.3934/math.2023886

AIMS Mathematics

2023, Volume 8, Issue 8: 17335-17353. doi: 10.3934/math.2023886

Previous Article Next Article

Research article

Some notes on the tangent bundle with a Ricci quarter-symmetric metric connection

1.
School of Mathematics, Key Laboratory of Cryptography of Zhejiang Province, Hangzhou Normal University, Hangzhou 311121, China
2.
Faculty of Science, Department of Mathematics, Ataturk University, Erzurum 25240, Turkey

Received: 27 February 2023 Revised: 11 May 2023 Accepted: 15 May 2023 Published: 19 May 2023
MSC : 53B20, 53C07, 53C35

Let $(M, g)$ be an $n$ -dimensional (pseudo-)Riemannian manifold and $TM$ be its tangent bundle $TM$ equipped with the complete lift metric $^{C}g$ . First, we define a Ricci quarter-symmetric metric connection $\overline{\nabla }$ on the tangent bundle $TM$ equipped with the complete lift metric $^{C}g$ . Second, we compute all forms of the curvature tensors of $\overline{\nabla }$ and study their properties. We also define the mean connection of $\overline{\nabla }$ . Ricci and gradient Ricci solitons are important topics studied extensively lately. Necessary and sufficient conditions for the tangent bundle $TM$ to become a Ricci soliton and a gradient Ricci soliton concerning $\overline{\nabla }$ are presented. Finally, we search conditions for the tangent bundle $TM$ to be locally conformally flat with respect to $\overline{\nabla }$ .

Keywords:

complete lift metric,
Ricci quarter-symmetric metric connection,
Ricci soliton,
tangent bundle,
vector field

Citation: Yanlin Li, Aydin Gezer, Erkan Karakaş. Some notes on the tangent bundle with a Ricci quarter-symmetric metric connection[J]. AIMS Mathematics, 2023, 8(8): 17335-17353. doi: 10.3934/math.2023886

Related Papers:

[1]	Muhammad Ali Khan, Saleem Abdullah, Alaa O. Almagrabi . Analysis of deep learning technique using a complex spherical fuzzy rough decision support model. AIMS Mathematics, 2023, 8(10): 23372-23402. doi: 10.3934/math.20231188
[2]	Tehreem, Harish Garg, Kinza Ayaz, Walid Emam . Multi attribute decision-making algorithms using Hamacher Choquet-integral operators with complex intuitionistic fuzzy information. AIMS Mathematics, 2024, 9(12): 35860-35884. doi: 10.3934/math.20241700
[3]	Tahir Mahmood, Azam, Ubaid ur Rehman, Jabbar Ahmmad . Prioritization and selection of operating system by employing geometric aggregation operators based on Aczel-Alsina t-norm and t-conorm in the environment of bipolar complex fuzzy set. AIMS Mathematics, 2023, 8(10): 25220-25248. doi: 10.3934/math.20231286
[4]	Shichao Li, Zeeshan Ali, Peide Liu . Prioritized Hamy mean operators based on Dombi t-norm and t-conorm for the complex interval-valued Atanassov-Intuitionistic fuzzy sets and their applications in strategic decision-making problems. AIMS Mathematics, 2025, 10(3): 6589-6635. doi: 10.3934/math.2025302
[5]	Muhammad Qiyas, Neelam Khan, Muhammad Naeem, Saleem Abdullah . Intuitionistic fuzzy credibility Dombi aggregation operators and their application of railway train selection in Pakistan. AIMS Mathematics, 2023, 8(3): 6520-6542. doi: 10.3934/math.2023329
[6]	Chunxiao Lu, Zeeshan Ali, Peide Liu . Selection of artificial neutral networks based on cubic intuitionistic fuzzy Aczel-Alsina aggregation operators. AIMS Mathematics, 2024, 9(10): 27797-27833. doi: 10.3934/math.20241350
[7]	Rana Muhammad Zulqarnain, Xiao Long Xin, Muhammad Saeed . Extension of TOPSIS method under intuitionistic fuzzy hypersoft environment based on correlation coefficient and aggregation operators to solve decision making problem. AIMS Mathematics, 2021, 6(3): 2732-2755. doi: 10.3934/math.2021167
[8]	Talha Midrar, Saifullah Khan, Saleem Abdullah, Thongchai Botmart . Entropy based extended TOPOSIS method for MCDM problem with fuzzy credibility numbers. AIMS Mathematics, 2022, 7(9): 17286-17312. doi: 10.3934/math.2022952
[9]	Wajid Azeem, Waqas Mahmood, Tahir Mahmood, Zeeshan Ali, Muhammad Naeem . Analysis of Einstein aggregation operators based on complex intuitionistic fuzzy sets and their applications in multi-attribute decision-making. AIMS Mathematics, 2023, 8(3): 6036-6063. doi: 10.3934/math.2023305
[10]	Jian Qi . Artificial intelligence-based intelligent computing using circular q-rung orthopair fuzzy information aggregation. AIMS Mathematics, 2025, 10(2): 3062-3094. doi: 10.3934/math.2025143

Abstract

1. Introduction

Fibonacci polynomials and Lucas polynomials are important in various fields such as number theory, probability theory, numerical analysis, and physics. In addition, many well-known polynomials, such as Pell polynomials, Pell Lucas polynomials, Tribonacci polynomials, etc., are generalizations of Fibonacci polynomials and Lucas polynomials. In this paper, we extend the linear recursive polynomials to nonlinearity, that is, we discuss some basic properties of the bi-periodic Fibonacci and Lucas polynomials.

The bi-periodic Fibonacci $\left \{ f_{n} \left (t\right) \right \}$ and Lucas $\left \{ l_{n} \left (t \right) \right \}$ polynomials are defined recursively by

$\begin{equation*} f_{0}\left (t \right ) = 0, \quad f_{1}\left (t\right ) = 1, \quad {f_{n}\left ( t\right ) = } \begin{cases} ayf_{n-1}\left ( t \right ) +f_{n-2}\left ( t \right ) \quad & n \equiv 0 \pmod{2}, \\ byf_{n-1}\left (t \right ) +f_{n-2}\left (t \right ) \quad & n \equiv 1 \pmod{2}, \end{cases} \quad n\ge 2, \end{equation*}$

and

$\begin{equation*} l_{0}\left ( t \right ) = 2, \quad l_{1}\left (t \right ) = at, \quad {l_{n}\left ( t \right ) = } \begin{cases} byl_{n-1}\left ( t \right ) +l_{n-2}\left ( t \right ) \quad & n \equiv 0 \pmod{2}, \\ ayl_{n-1}\left ( t \right ) +l_{n-2}\left ( t \right ) \quad & n \equiv 1 \pmod{2}, \end{cases} \quad n\ge 2, \end{equation*}$

where $a$ and $b$ are nonzero real numbers. For $t = 1$ , the bi-periodic Fibonacci and Lucas polynomials are, respectively, well-known bi-periodic Fibonacci $\left \{f_{n} \right \}$ and Lucas $\left \{l_{n} \right \}$ sequences. We let

$\begin{equation*} {\varsigma \left ( n \right ) = } \begin{cases} 0 \quad & n \equiv 0 \pmod{2}, \\ 1 \quad & n \equiv 1 \pmod{2}, \end{cases} \quad n\ge 2. \end{equation*}$

In ^[1], the scholars give the Binet formulas of the bi-periodic Fibonacci and Lucas polynomials as follows:

$\begin{equation} f_{n}\left (t \right ) = \frac{a^{\varsigma \left (n+1 \right) }}{ \left( ab \right )^{\left \lfloor \frac{n}{2} \right \rfloor } }\left (\frac{\sigma^{n} \left ( t\right ) -\tau ^{n}\left ( t \right ) }{\sigma\left (t \right ) -\tau \left ( t \right ) }\right), \end{equation}$

(1.1)

and

$\begin{equation} l_{n}\left (t \right ) = \frac{a^{\varsigma \left (n \right) }}{ \left( ab \right )^{\left \lfloor \frac{n+1}{2} \right \rfloor } }\left (\sigma ^{n}\left (t\right ) +\tau ^{n}\left ( t\right ) \right), \end{equation}$

(1.2)

where $n\ge 0$ , $\sigma\left (t \right)$ , and $\tau \left (t \right)$ are zeros of $\lambda ^{2}-abt\lambda -ab$ . This is $\sigma \left (t \right) = \frac{abt +\sqrt{a^{2} b^{2} t^{2}+4ab} }{2}$ and $\tau \left (t \right) = \frac{abt -\sqrt{a^{2} b^{2} t^{2}+4ab} }{2}$ . We note the following algebraic properties of $\sigma\left (t \right)$ and $\tau \left (t \right)$ :

$\begin{equation*} \sigma \left ( t \right ) +\tau \left (t \right ) = abt, \quad \sigma \left (t \right ) -\tau \left ( t \right ) = \sqrt{a^{2} b^{2} t^{2} +4ab}, \quad \sigma\left ( t \right ) \tau \left ( t\right ) = -ab. \end{equation*}$

Many scholars studied the properties of bi-periodic Fibonacci and Lucas polynomials; see ^[2,3,4,5,6]. In addition, many scholars studied the power sums problem of second-order linear recurrences and its divisible properties; see ^[7,8,9,10].

Taking $a = b = 1$ and $t = 1$ , we obtain the Fibonacci $\left \{ F_{n} \right \}$ or Lucas $\left \{ L_{n} \right \}$ sequence. Melham ^[11] proposed the following conjectures:

Conjecture 1. Let $m\ge 1$ be an integer, then the sum

$\begin{equation*} L_{1} L_{3} L_{5}\cdots L_{2m+1}\sum\limits_{k = 1}^{n} F_{2k}^{2m+1} \end{equation*}$

can be represented as $\left (F_{2n+1}-1 \right)^{2}R_{2m-1}\left (F_{2n+1} \right)$ , including $R_{2m-1}\left (t \right)$ as a polynomial with integer coefficients of degree $2m-1$ .

Conjecture 2. Let $m\ge 1$ be an integer, then the sum

$\begin{equation*} L_{1} L_{3} L_{5}\cdots L_{2m+1}\sum\limits_{k = 1}^{n} L_{2k}^{2m+1} \end{equation*}$

can be represented as $\left (L_{2n+1}-1 \right)Q_{2m}\left (L_{2n+1} \right)$ , where $Q_{2m}\left (t \right)$ is a polynomial with integer coefficients of degree $2m$ .

In ^[12], the authors completely solved the Conjecture $2$ and discussed the Conjecture $1$ . Using the definition and properties of bi-periodic Fibonacci and Lucas polynomials, the power sums problem and their divisible properties are studied in this paper. The results are as follows:

Theorem 1. We get the identities

$\begin{align} & \sum\limits_{k = 1}^{n}f_{2k}^{2m+1} \left (t\right ) = \frac{a^{2m+1} }{b\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits _{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m+1}{m-j}\left ( \frac{f_{\left ( 2n+1 \right )\left ( 2j+1\right ) }\left ( t \right ) - f_{2j+1}\left ( t \right ) }{l_{2j+1} \left ( t \right ) } \right ), \end{align}$

(1.3)

$\begin{align} & \sum\limits_{k = 1}^{n}f_{2k+1}^{2m+1} \left ( t \right ) = \frac{\left ( ab \right ) ^{m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits_{j = 0}^{m} \binom{2m+1}{m-j} \left ( \frac{f_{\left ( 2n+2\right )\left (2j+1 \right ) }\left (t\right ) -f_{2\left ( 2j+1 \right ) }\left (t\right ) }{l_{2j+1}\left ( t \right ) } \right ), \end{align}$

(1.4)

$\begin{align} & \sum\limits_{k = 1}^{n}l_{2k}^{2m+1} \left (t \right ) = \sum\limits_{j = 0}^{m} \binom{2m+1}{m-j} \left ( \frac{l_{\left ( 2n+1 \right )\left ( 2j+1 \right ) }\left (t \right ) -l_{2j+1}\left ( t\right ) }{l_{2j+1}\left ( t\right ) } \right ), \end{align}$

(1.5)

$\begin{align} &\sum\limits_{k = 1}^{n}l_{2k+1}^{2m+1} \left (t\right ) = \frac{ a ^{m+1} }{b^{m+1} } \sum\limits_{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m+1}{m-j} \left ( \frac{l_{\left ( 2n+2\right )\left ( 2j+1\right ) }\left ( t \right ) -l_{2\left ( 2j+1 \right ) }\left ( t \right ) }{l_{2j+1}\left ( t \right ) } \right ), \end{align}$

(1.6)

where $n$ and $m$ are positive integers.

Theorem 2. We get the identities

$\begin{align} & \sum\limits_{k = 1}^{n}f_{2k}^{2m} \left ( t \right ) = \frac{a^{2m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits_{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m}{m-j} \frac{f_{2j\left ( 2n+1 \right ) }\left (t \right ) }{f_{2j}\left (t\right ) } -\frac{a^{2m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} }\binom{2m}{m}\left ( -1 \right )^{m}\left ( n+ \frac{1}{2 } \right ) , \end{align}$

(1.7)

$\begin{align} & \sum\limits_{k = 1}^{n}f_{2k+1}^{2m} \left ( t \right ) = \frac{\left ( ab \right ) ^{m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits_{j = 0}^{m} \binom{2m}{m-j} \left ( \frac{f_{2j\left ( 2n+2\right ) }\left ( t \right ) -f_{4j }\left ( t \right ) }{f_{2j}\left ( t \right ) } \right )-\frac{\left ( ab \right ) ^{m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} }\binom{2m}{m}n , \end{align}$

(1.8)

$\begin{align} & \sum\limits_{k = 1}^{n}l_{2k}^{2m} \left ( t \right ) = \sum\limits_{j = 0}^{m} \binom{2m}{m-j} \frac{f_{2j\left ( 2n+1 \right ) }\left ( t \right ) }{l_{2j+1}\left (t \right ) }-2^{2m-1}-\binom{2m}{m}\left ( n+\frac{1}{2} \right ) , \end{align}$

(1.9)

$\begin{align} &\sum\limits_{k = 1}^{n}l_{2k+1}^{2m} \left ( t \right ) = \frac{ a ^{m} }{b^{m} } \sum\limits_{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m}{m-j} \left ( \frac{f_{2j\left ( 2n+2\right ) }\left ( t \right ) -f_{4j }\left (t \right ) }{f_{2j}\left ( t \right ) } \right ) -\frac{a^{m} }{b^{m} } \binom{2m}{m} \left ( -1 \right )^{m}n , \end{align}$

(1.10)

where $n$ and $m$ are positive integers.

As for application of Theorem 1, we get the following:

Corollary 1. We get the congruences:

$\begin{equation} bl_{1} \left ( t \right ) l_{3} \left ( t \right ) \cdots l_{2m+1}\left ( t \right )\sum\limits _{k = 1}^{n}f_{2k}^{2m+1}\left ( t \right ) \equiv 0\pmod{f_{2n+1}\left ( t \right ) -1}, \end{equation}$

(1.11)

and

$\begin{equation} al_{1} \left ( t\right ) l_{3} \left (t \right ) \cdots l_{2m+1}\left ( t \right )\sum\limits _{k = 1}^{n}l_{2k}^{2m+1}\left ( t \right ) \equiv 0\pmod{l_{2n+1}\left (t \right ) -at}, \end{equation}$

(1.12)

where $n$ and $m$ are positive integers.

Taking $t = 1$ in Corollary 1, we have the following conclusions for bi-periodic Fibonacci $\left \{ f_{n} \right \}$ and Lucas $\left \{ l_{n} \right \}$ sequences.

Corollary 2. We get the congruences:

$\begin{equation} bl_{1} l_{3} \cdots l_{2m+1} \sum\limits _{k = 1}^{n}f_{2k}^{2m+1} \equiv 0\pmod{f_{2n+1} -1}, \end{equation}$

(1.13)

and

$\begin{equation} al_{1} l_{3} \cdots l_{2m+1} \sum\limits _{k = 1}^{n}l_{2k}^{2m+1} \equiv 0\pmod{l_{2n+1} -a }, \end{equation}$

(1.14)

where $n$ and $m$ are nonzero real numbers.

Taking $a = b = 1$ and $t = 1$ in Corollary 1, we have the following conclusions for bi-periodic Fibonacci $\left \{ F_{n} \right \}$ and Lucas $\left \{ L_{n} \right \}$ sequences.

Corollary 3. We get the congruences:

$\begin{equation} L_{1} L_{3} \cdots L_{2m+1} \sum\limits _{k = 1}^{n}F_{2k}^{2m+1} \equiv 0\pmod{F_{2n+1} -1}, \end{equation}$

(1.15)

and

$\begin{equation} L_{1} L_{3} \cdots L_{2m+1} \sum\limits _{k = 1}^{n}L_{2k}^{2m+1} \equiv 0\pmod{L_{2n+1} -1 }, \end{equation}$

(1.16)

where $n$ and $m$ are nonzero real numbers.

2. Proofs of theorems

To begin, we will give several lemmas that are necessary in proving theorems.

Lemma 1. We get the congruence

$\begin{equation*} f_{\left ( 2n+1 \right )\left ( 2j+1\right ) } \left ( t \right ) -f_{2j+1 } \left ( t \right ) \equiv 0 \pmod{f_{2n+1 } \left ( t \right )-1}, \end{equation*}$

where $n$ and $m$ are nonzero real numbers.

Proof. We prove it by complete induction for $j\ge 0$ . This clearly holds when $j = 0$ . If $j = 1$ , we note that $abf_{3\left (2n+1 \right) } \left (t \right) = \left (a^{2} b^{2} t^{2} +4ab \right) f_{2n+1}^{3}\left (t \right)-3abf_{2n+1} \left (t\right)$ and we obtain

$\begin{equation*} \begin{split} & f_{3\left ( 2n+1 \right ) }\left ( t \right )-f_{3 }\left ( t \right ) = \left ( abt^{2}+4 \right ) f_{2n+1}^{3}\left ( t\right )-3f_{2n+1} \left (t \right )-\left ( abt^{2}+4 \right )f_{1}^{3} \left (t \right )+3f_{1} \left (t\right ) \\ & = \left ( abt^{2}+4 \right )\left ( f_{2n+1}\left ( t \right )-f_{1}\left ( t \right )\right )\left ( f_{2n+1}^{2}\left (t \right )+ f_{2n+1}\left ( t \right )f_{1}\left ( t \right )+ f_{1}^{2}\left ( t \right ) \right )-3\left ( f_{2n+1} \left ( t\right ) -f_{1}\left ( t \right ) \right ) \\ & = \left ( abt^{2}+4 \right )\left ( f_{2n+1}\left ( t\right )-1\right )\left ( f_{2n+1}^{2}\left (t \right )+ f_{2n+1}\left ( t \right )f_{1}\left ( t \right )+ f_{1}^{2}\left ( t \right ) \right )-3\left ( f_{2n+1} \left ( t \right ) -1 \right ) \\ & \equiv 0 \pmod{f_{2n+1}\left (t \right ) -1 } . \end{split} \end{equation*}$

This is obviously true when $j = 1$ . Assuming that Lemma 1 holds if $j = 1, 2, \dots, k$ , that is,

$\begin{equation*} f_{\left ( 2n+1 \right )\left ( 2j+1\right ) } \left ( t \right ) -f_{2j+1 } \left ( t \right ) \equiv 0 \pmod{f_{2n+1 } \left ( t \right )-1}. \end{equation*}$

If $j = k+1 \ge 2$ , we have

$\begin{equation*} l_{2\left ( 2n+1 \right ) } \left ( t \right ) f_{\left ( 2n+1 \right )\left ( 2j+1 \right ) }\left ( t \right ) = f_{\left ( 2n+1 \right )\left ( 2j+3 \right ) }\left ( t \right ) +abf_{\left ( 2n+1 \right )\left ( 2j-1 \right ) }\left ( t \right ), \end{equation*}$

and

$\begin{equation*} \begin{split} abl_{2\left ( 2n+1 \right ) } \left ( t \right ) = \left ( a^{2} b^{2} t^{2} +4ab \right )f_{2n+1}^{2}\left ( t \right )-2ab \equiv \left ( a^{2} b^{2} t^{2} +4ab \right )f_{1}^{2}\left ( t \right )-2ab \pmod{f_{2n+1}\left ( t \right ) -1 }. \end{split} \end{equation*}$

We have

$\begin{equation*} \begin{split} &f_{\left ( 2n+1 \right )\left ( 2k+3 \right ) } \left (t\right ) -f_{ 2k+3 } \left ( t\right )\\ & = l_{2\left ( 2n+1 \right ) }\left ( t \right ) f_{\left ( 2n+1 \right )\left ( 2k+1 \right ) }\left ( t \right )-abf_{\left (2n+1\right )\left ( 2k-1 \right ) } \left ( t \right ) -l_{2} \left ( t \right )f_{2k+1}\left ( t \right )+abf_{2k-1} \left ( t \right ) \\ & \equiv \left ( \left ( abt^{2}+4 \right )f_{ 1}^{2} \left (t \right ) -2 \right )f_{\left ( 2n+1 \right )\left ( 2k+1 \right ) }\left (t \right )-abf_{\left ( 2n+1 \right )\left ( 2k-1 \right ) } \left (t \right ) \\ &\quad -\left ( \left ( abt^{2}+4 \right )f_{ 1}^{2} \left ( t \right ) -2 \right ) f_{2k+1} \left ( t \right )+abf_{2k-1}\left ( t \right ) \\ & \equiv \left ( \left ( abt^{2}+4 \right )f_{ 1}^{2} \left ( t \right ) -2 \right )\left ( f_{\left ( 2n+1 \right ) \left ( 2k+1 \right ) }\left ( t \right )-f_{2k+1}\left ( t\right ) \right ) -ab\left ( f_{\left ( 2n+1 \right )\left ( 2k-1 \right ) } \left (t \right ) -f_{2k-1}\left ( t \right ) \right ) \\ & \equiv 0 \pmod{f_{2n+1}\left (t \right ) -1 } . \end{split} \end{equation*}$

This completely proves Lemma 1. □

Lemma 2. We get the congruence

$\begin{equation*} al_{\left ( 2n+1 \right )\left ( 2j+1\right ) } \left ( t \right ) -al_{2j+1 } \left ( t \right ) \equiv 0 \pmod{l_{2n+1 } \left ( t \right )-at}, \end{equation*}$

where $n$ and $m$ are nonzero real numbers.

Proof. We prove it by complete induction for $j\ge 0$ . This clearly holds when $j = 0$ . If $j = 1$ , we note that $al_{3\left (2n+1 \right) } \left (t \right) = b l_{2n+1}^{3}\left (t \right)+3al_{2n+1} \left (t \right)$ and we obtain

$\begin{equation*} \begin{split} & a l_{3\left ( 2n+1 \right ) }\left (t \right )-al_{3 }\left ( t \right ) = b l_{2n+1}^{3}\left ( t \right )+3al_{2n+1} \left ( t \right )-bl_{1}^{3} \left (t \right )-3al_{1} \left ( t \right ) \\ & = \left ( l_{2n+1}\left ( t \right )-l_{1}\left ( t \right )\right )\left ( bl_{2n+1}^{2}\left ( t \right )+ bl_{2n+1}\left ( t \right )l_{1}\left ( t \right )+ bl_{1}^{2}\left (t \right ) \right )-3a\left ( l_{2n+1} \left ( t \right ) -l_{1}\left ( t\right ) \right ) \\ & = \left ( l_{2n+1}\left ( t \right )-at\right )\left ( bl_{2n+1}^{2}\left ( t \right )+ bayl_{2n+1}\left ( t \right )+ ba^{2}t^{2} \right )-3a\left ( l_{2n+1} \left ( t \right ) -at \right )\\ & \equiv 0 \pmod{l_{2n+1}\left ( t \right ) -at } . \end{split} \end{equation*}$

This is obviously true when $j = 1$ . Assuming that Lemma 2 holds if $j = 1, 2, \dots, k$ , that is,

$\begin{equation*} al_{\left ( 2n+1 \right )\left ( 2j+1\right ) } \left (t \right ) -al_{2j+1 } \left ( t \right ) \equiv 0 \pmod{l_{2n+1 } \left ( t \right )-at}. \end{equation*}$

If $j = k+1 \ge 2$ , we have

$\begin{equation*} l_{2\left ( 2n+1 \right ) } \left (t \right ) l_{\left ( 2n+1 \right )\left ( 2j+1 \right ) }\left ( t\right ) = l_{\left ( 2n+1 \right )\left ( 2j+3 \right ) }\left ( t\right ) +l_{\left ( 2n+1 \right )\left ( 2j-1 \right ) }\left ( t \right ), \end{equation*}$

and

$\begin{equation*} \begin{split} & al_{2\left ( 2n+1 \right ) } \left ( t \right ) = bl_{2n+1}^{2}\left ( t \right )+2a \equiv bl_{1}^{2}\left ( t \right )+2a \pmod{l_{2n+1}\left ( t \right ) -at }. \end{split} \end{equation*}$

We have

$\begin{equation*} \begin{split} &al_{\left ( 2n+1 \right )\left ( 2k+3 \right ) } \left ( t\right ) -al_{\left ( 2k+3 \right ) } \left ( t \right )\\ & = a\left ( l_{2\left ( 2n+1 \right ) }\left ( t \right ) l_{\left ( 2n+1 \right )\left ( 2k+1 \right ) }\left (t \right )-l_{\left (2n+1\right )\left ( 2k-1 \right ) } \left ( t \right ) \right )-a\left ( l_{2} \left (t \right )l_{2k+1}\left ( t \right )-l_{2k-1} \left (t \right ) \right ) \\ & \equiv \left ( bl_{ 1}^{2} \left ( t \right ) +2a \right )l_{\left ( 2n+1 \right )\left ( 2k+1 \right ) }\left ( t \right )-al_{\left ( 2n+1 \right )\left ( 2k-1 \right ) } \left ( t \right ) -\left ( bl_{ 1}^{2} \left ( t \right ) +2a \right ) l_{2k+1} \left ( t \right )+al_{2k-1}\left ( t \right ) \\ & \equiv \left ( abt^{2} +2 \right )\left ( al_{\left ( 2n+1 \right ) \left ( 2k+1 \right ) }\left ( t\right )-al_{2k+1}\left ( t \right ) \right ) - \left ( al_{\left ( 2n+1 \right )\left ( 2k-1 \right ) } \left ( t\right ) -al_{2k-1}\left (t \right ) \right ) \\ & \equiv 0 \pmod{l_{2n+1}\left ( t \right ) -at } . \end{split} \end{equation*}$

This completely proves Lemma 2. □

Proof of Theorem 1. We only prove (1.3), and the proofs for other identities are similar.

$\begin{equation*} \begin{split} \sum\limits _{k = 1}^{n}f_{2k}^{2m+1}\left ( t \right )& = \sum\limits_{k = 1}^{n}\left (\frac{a^{\varsigma \left ( 2k+1 \right ) } }{\left (ab \right )^{ \left \lfloor \frac{2k}{2} \right \rfloor} }\cdot \left ( \frac{\sigma ^{2k}\left ( t \right ) - \tau ^{2k} \left (t \right ) }{\sigma\left ( t \right ) - \tau \left (t \right ) } \right ) \right ) ^{2m+1} \\ & = \frac{a^{2m+1} }{\left ( \sigma\left (t \right ) -\tau \left ( t \right ) \right )^{2m+1} }\sum\limits _{k = 1}^{n}\frac{\left ( \sigma ^{2k}\left (t \right ) - \tau ^{2k} \left ( t \right ) \right )^{2m+1} }{\left ( ab \right )^{\left ( 2m+1 \right )k } } \\ & = \frac{a^{2m+1} }{\left ( \sigma\left (t\right ) -\tau \left ( t \right ) \right )^{2m+1} }\sum\limits _{k = 1}^{n}\sum\limits _{j = 0}^{2m+1}\left ( -1 \right )^{j} \binom{2m+1}{j} \frac{ \sigma^{2k\left ( 2m+1-j \right ) } \left ( t \right ) \tau ^{2kj} \left ( t \right ) }{\left ( ab \right )^{\left ( 2m+1 \right )k } } \\ & = \frac{a^{2m+1} }{\left ( \sigma\left ( t \right ) -\tau \left (t \right ) \right )^{2m+1} }\sum\limits _{j = 0}^{2m+1}\left ( -1 \right )^{j} \binom{2m+1}{j} \left ( \frac{1-\frac{\sigma ^{2n\left ( 2m+1-2j \right ) }\left (t \right ) }{\left ( ab \right )^{\left ( 2m+1-2j \right )n } } }{\frac{\left ( ab \right )^{2m+1-2j} }{\sigma ^{2\left ( 2m+1-2j \right ) }\left (t \right ) } -1} \right )\\ & = \frac{a^{2m+1} }{\left (\sigma\left (t \right ) -\tau \left (t\right ) \right )^{2m+1} }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m+1}{j} \left ( \frac{1-\frac{\sigma ^{2n\left ( 2m+1-2j \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( 2m+1-2j \right )n } } }{\frac{\left ( ab \right )^{2m+1-2j} }{\sigma ^{2\left ( 2m+1-2j \right ) }\left (t \right ) } -1} - \frac{1-\frac{\sigma ^{2n\left ( 2j-1-2m \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( 2j-1-2m\right )n } } }{\frac{\left ( ab \right )^{2j-1-2m} }{\sigma ^{2\left ( 2j-1-2m \right ) }\left ( t \right ) } -1}\right )\\ & = \frac{a^{2m+1} }{\left ( \sigma\left ( t \right ) -\tau \left ( t \right ) \right )^{2m+1} }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m+1}{j} \left ( \frac{\frac{\sigma^{2 \left ( 2m+1-2j \right ) }\left ( t\right ) }{\left ( ab \right )^{ 2m+1-2j } }- \frac{\sigma ^{\left ( 2n+2 \right ) \left ( 2m+1-2j \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( n+1 \right ) \left ( 2m+1-2j \right ) } } +1- \frac{\sigma ^{2n \left ( 2j-1-2m \right ) }\left (t \right ) }{\left ( ab \right )^{ \left ( 2j-1-2m \right ) n }} }{1- \frac{\sigma^{2 \left ( 2m+1-2j \right ) }\left ( t \right ) }{\left ( ab \right )^{ \left ( 2m+1-2j \right ) }}} \right )\\ & = \frac{a^{2m+1} }{\left ( \sigma\left ( t \right ) -\tau \left (t \right ) \right )^{2m+1} }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m+1}{j}\\ & \quad \times \left ( \frac{\sigma^{2m+1-2j}\left ( t \right ) - \tau ^{2m+1-2j}\left ( t \right ) -\frac{\sigma^{\left ( 2n+1 \right ) \left ( 2m+1-2j \right ) }\left ( t \right )}{\left ( ab \right )^{\left ( 2m+1-2j \right )n } } +\frac{\tau ^{\left ( 2n+1 \right ) \left ( 2m+1-2j \right ) }\left (t \right )}{\left ( ab \right )^{ \left ( 2m+1-2j \right )n } }}{-\sigma ^{2m+1-2j}\left ( t \right ) -\tau ^{2m+1-2j}\left ( t \right ) } \right ) \\ & = \frac{a^{2m+1} }{b\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits _{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m+1}{m-j}\left ( \frac{f_{\left ( 2n+1 \right )\left ( 2j+1\right ) }\left (t \right ) - f_{2j+1}\left (t \right ) }{l_{2j+1} \left ( t \right ) } \right ) . \end{split} \end{equation*}$

□

Proof of Theorem 2. We only prove (1.7), and the proofs for other identities are similar.

$\begin{equation*} \begin{split} \sum\limits _{k = 1}^{n}f_{2k}^{2m }\left ( t \right )& = \sum\limits_{k = 1}^{n}\left (\frac{a^{\varsigma \left ( 2k +1\right ) } }{\left (ab \right )^{ \left \lfloor \frac{2k}{2} \right \rfloor} }\cdot \left ( \frac{\sigma ^{2k}\left ( t\right ) - \tau ^{2k} \left ( t \right ) }{\sigma\left ( t\right ) - \tau \left (t \right ) } \right ) \right ) ^{2m} \\ & = \frac{a^{2m} }{\left ( \sigma\left (t \right ) -\tau \left (t \right ) \right )^{2m} }\sum\limits _{k = 1}^{n}\frac{\left ( \sigma^{2k}\left ( t\right ) - \tau ^{2k} \left ( t \right ) \right )^{2m} }{\left ( ab \right )^{ 2mk } } \\ & = \frac{a^{2m} }{\left ( \sigma\left ( t \right ) -\tau \left (t \right ) \right )^{2m} }\sum\limits _{k = 1}^{n}\sum\limits _{j = 0}^{2m}\left ( -1 \right )^{j} \binom{2m}{j} \frac{ \sigma^{2k\left ( 2m-j \right ) } \left (t \right ) \tau ^{2kj} \left ( t \right ) }{\left ( ab \right )^{2mk } }\\ & = \frac{a^{2m} }{\left (\sigma\left (t \right ) -\tau \left ( t \right ) \right )^{2m} }\sum\limits _{j = 0}^{2m}\left ( -1 \right )^{j} \binom{2m }{j} \left ( \frac{1-\frac{\sigma ^{2n\left ( 2m-2j \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( 2m-2j \right )n } } }{\frac{\left ( ab \right )^{2m-2j} }{\sigma ^{2\left ( 2m-2j \right ) }\left ( t \right ) } -1} \right ) \end{split} \end{equation*}$

$\begin{equation*} \begin{split} & = \frac{a^{2m} }{\left (\sigma\left (t \right ) -\tau \left ( t \right ) \right )^{2m} }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m}{j} \left ( \frac{1-\frac{\sigma ^{2n\left ( 2m-2j \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( 2m-2j \right )n } } }{\frac{\left ( ab \right )^{2m-2j} }{\sigma ^{2\left ( 2m-2j \right ) }\left ( t \right ) } -1} +\frac{1-\frac{\sigma ^{2n\left ( 2j-2m \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( 2j-2m\right )n } } }{\frac{\left ( ab \right )^{2j-2m} }{\sigma ^{2\left ( 2j-2m \right ) }\left (t\right ) } -1}\right )\\ & \quad +\frac{a^{2m} }{\left ( \sigma \left ( t \right ) -\tau \left (t \right ) \right )^{2m} } \left ( -1 \right ) ^{m+1}\binom{2m}{m}n\\ & = \frac{a^{2m} }{\left (\sigma\left ( t \right ) -\tau \left (t \right ) \right )^{2m} }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m}{j} \left ( \frac{\frac{\sigma ^{2 \left ( 2m -2j \right ) }\left ( t \right ) }{\left ( ab \right )^{ 2m -2j } }- \frac{\sigma ^{\left ( 2n+2 \right ) \left ( 2m -2j \right ) }\left ( t \right ) }{\left ( ab \right )^{\left ( n+1 \right ) \left ( 2m -2j \right ) } } -1+\frac{\sigma^{2n \left ( 2j -2m \right ) }\left ( t \right ) }{\left ( ab \right )^{ \left ( 2j -2m \right ) n }} }{1- \frac{\sigma^{2 \left ( 2m -2j \right ) }\left ( t \right ) }{\left ( ab \right )^{ 2m-2j }}} \right )\\ & \quad +\frac{a^{2m} }{\left (\sigma \left ( t \right ) -\tau \left ( t \right ) \right )^{2m} } \left ( -1 \right ) ^{m+1}\binom{2m}{m}n \\ & = \frac{a^{2m } }{\left ( \sigma\left ( t \right ) -\tau \left ( t \right ) \right )^{2m } }\sum\limits _{j = 0}^{m}\left ( -1 \right )^{j} \binom{2m }{j} \left ( \frac{\sigma^{2m -2j}\left ( t \right ) - \tau ^{2m -2j}\left (t\right ) -\frac{\sigma^{\left ( 2n+1 \right ) \left ( 2m -2j \right ) }\left (t \right )}{\left ( ab \right )^{n\left ( 2m -2j \right ) } } +\frac{\tau ^{\left ( 2n+1 \right ) \left ( 2m-2j \right ) }\left (t \right )}{\left ( ab \right )^{n\left ( 2m-2j \right ) } }}{ \tau ^{2m -2j}\left ( t \right )-\sigma ^{2m -2j}\left ( t \right ) } \right ) \\ & \quad +\frac{a^{2m} }{\left ( \sigma \left (t \right ) -\tau \left ( t \right ) \right )^{2m} } \left ( -1 \right ) ^{m+1}\binom{2m}{m}n\\ & = \frac{a^{2m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \sum\limits _{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m}{m-j}\left ( \frac{f_{2j\left ( 2n+1 \right ) }\left ( t \right ) - f_{ 2j}\left (t \right ) }{f_{ 2j} \left (t \right ) } \right ) +\frac{a^{2m} }{\left ( a^{2} b^{2} t^{2} +4ab \right )^{m} } \left ( -1 \right ) ^{m+1}\binom{2m}{m}n . \end{split} \end{equation*}$

□

Proof of Corollary 1. First, from the definition of $f_{n}\left (t \right)$ and binomial expansion, we easily prove $\left (f_{2n+1}\left (t \right)-1, a^{2} b^{2} t^{2} +4ab \right) = 1$ . Therefore, $\left (f_{2n+1}\left (t \right)-1, \left (a^{2} b^{2} t^{2} +4ab \right)^{m} \right) = 1$ . Now, we prove (1.11) by Lemma 1 and (1.3):

$\begin{equation*} \begin{split} &b l_{1} \left ( t\right ) l_{3} \left ( t \right ) \cdots l_{2m+1}\left ( t \right )\sum\limits _{k = 1}^{n}f_{2k}^{2m+1}\left ( t\right ) \\ & = l_{1} \left ( t\right ) l_{3} \left (t \right ) \cdots l_{2m+1}\left ( t \right ) \left ( \frac{a^{2m+1} }{\left ( \sigma\left (t \right ) -\tau \left (t \right ) \right )^{2m} } \sum\limits_{j = 0}^{m}\left ( -1 \right )^{m-j} \binom{2m+1}{m-j} \left ( \frac{f_{\left ( 2n+1 \right )\left ( 2j+1 \right ) }\left ( t \right ) -f_{2j+1}\left ( t \right ) }{l_{2j+1}\left (t\right ) } \right ) \right )\\ & \equiv 0\pmod{f_{2n+1}\left ( t \right ) -1}. \end{split} \end{equation*}$

Now, we use Lemma 2 and (1.5) to prove (1.12):

$\begin{equation*} \begin{split} &al_{1} \left ( t\right ) l_{3} \left ( t \right ) \cdots l_{2m+1}\left ( t\right )\sum\limits _{k = 1}^{n}l_{2k}^{2m+1}\left ( t \right ) \\ & = l_{1} \left ( t \right ) l_{3} \left (t \right ) \cdots l_{2m+1}\left ( t \right ) \left ( \sum\limits_{j = 0}^{m} \binom{2m+1}{m-j} \left ( \frac{al_{\left ( 2n+1 \right )\left ( 2j+1 \right ) }\left ( t\right ) -al_{2j+1}\left (t \right ) }{l_{2j+1}\left ( t \right ) } \right ) \right )\\ & \equiv 0\pmod{l_{2n+1}\left ( t \right ) -at}. \end{split} \end{equation*}$

□

3. Conclusions

In this paper, we discuss the power sums of bi-periodic Fibonacci and Lucas polynomials by Binet formulas. As corollaries of the theorems, we extend the divisible properties of the sum of power of linear Fibonacci and Lucas sequences to nonlinear Fibonacci and Lucas polynomials. An open problem is whether we extend the Melham conjecture to nonlinear Fibonacci and Lucas polynomials.

Use of AI tools declaration

The authors declare that they did not use Artificial Intelligence (AI) tools in the creation of this paper.

Acknowledgments

The authors would like to thank the editor and referees for their helpful suggestions and comments, which greatly improved the presentation of this work. All authors contributed equally to the work, and they have read and approved this final manuscript. This work is supported by Natural Science Foundation of China (12126357).

Conflict of interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

References

[1]	A. Friedmann, J. A. Schouten, Über die geometrie der halbsymmetrischen Übertragungen, Math. Z., 21 (1924), 211–223. https://doi.org/10.1007/BF01187468 doi: 10.1007/BF01187468
[2]	H. A. Hayden, Sub-spaces of a space with torsion, Proc. London Math. Soc., S2-34 (1932), 27–50. https://doi.org/10.1112/plms/s2-34.1.27 doi: 10.1112/plms/s2-34.1.27
[3]	S. K. Hui, On $\phi$ -pseudo symmetric Kenmotsu manifolds with respect to quarter-symmetric metric connection, Appl. Sci., 15 (2013), 71–84.
[4]	S. K. Hui, R. Prasad, D. Chakraborty, Ricci solitons on Kenmotsu Manifolds with respect to quarter symmetric non-metric $\phi$ -connection, Ganita, 67 (2017), 195–204.
[5]	S. K. Hui, R. S. Lemence, On generalized $\phi$ -recurrent Kenmotsu manifolds with respect to quarter-symmetric metric connection, Kyungpook Math. J., 58 (2018), 347–359. http://dx.doi.org/10.5666/KMJ.2018.58.2.347 doi: 10.5666/KMJ.2018.58.2.347
[6]	S. K. Hui, L. I. Piscoran, T. Pal, Invariant submanifolds of (LCS)n-manifolds with respect to quarter symmetric metric connection, Acta Math. Univ. Comenianae, 87 (2018), 205–221.
[7]	S. K. Hui, A. Abolarinwa, M. A. Khan, F. Mofarreh, A. Saha, S. Bhattacharyya, Li-Yau-type gradient estimate along geometric flow, Mathematics, 11 (2023), 1364. https://doi.org/10.3390/math11061364 doi: 10.3390/math11061364
[8]	S. K. Hui, J. Roy, Invariant and anti-invariant submanifolds of special quasi-Sasakian manifolds, J. Geom., 109 (2018), 37. https://doi.org/10.1007/s00022-018-0442-2 doi: 10.1007/s00022-018-0442-2
[9]	S. K. Hui, S. Uddin, A. H. Alkhaldi, P. Mandal, Invariant submanifolds of generalized Sasakian-space-forms, Int. J. Geom. Methods Mod. Phys., 15 (2018), 1850149. https://doi.org/10.1142/S0219887818501499 doi: 10.1142/S0219887818501499
[10]	S. K. Hui, D. Chakraborty, Ricci almost solitons on Concircular Ricci pseudosymmetric $\beta$ -Kenmotsu manifolds, Hacettepe J. Math. Stat., 47 (2018), 579–587.
[11]	S. Golab, On semi-symmetric and quarter-symmetric linear connections, Tensor (N.S.), 29 (1975), 249–254.
[12]	D. Kamilya, U. C. De, Some properties of a Ricci quarter-symmetric metric connection in a Riemannian manifold, Indian J. Pure Appl. Math., 26 (1995), 29–34.
[13]	Y. Li, S. Liu, Z. Wang, Tangent developables and Darboux developables of framed curves, Topol. Appl., 301 (2021), 107526. http://dx.doi.org/10.1016/j.topol.2020.107526 doi: 10.1016/j.topol.2020.107526
[14]	Y. Li, M. Erdogdu, A. Yavuz, Differential geometric approach of Betchov-Da Rios soliton equation, Hacet. J. Math. Stat., 52 (2023), 114–125. https://doi.org/10.15672/hujms.1052831 doi: 10.15672/hujms.1052831
[15]	Y. Li, K. Eren, K. Ayvacı, S. Ersoy, The developable surfaces with pointwise 1-type Gauss map of Frenet type framed base curves in Euclidean 3-space, AIMS Math., 8 (2023), 2226–2239. http://dx.doi.org/10.3934/math.2023115 doi: 10.3934/math.2023115
[16]	Y. Li, Z. Chen, S. Nazra, R. A. Abdel-Baky, Singularities for timelike developable surfaces in Minkowski 3-space, Symmetry, 15 (2023), 277. http://dx.doi.org/10.3390/sym15020277 doi: 10.3390/sym15020277
[17]	Y. Li, M. T. Aldossary, R. A. Abdel-Baky, Spacelike circular surfaces in Minkowski 3-space, Symmetry, 15 (2023), 173. http://dx.doi.org/10.3390/sym15010173 doi: 10.3390/sym15010173
[18]	Y. Li, A. Abdel-Salam, M. Saad, Primitivoids of curves in Minkowski plane, AIMS Math., 8 (2023), 2386–2406. http://dx.doi.org/10.3934/math.2023123 doi: 10.3934/math.2023123
[19]	Y. Li, O. Tuncer, On (contra)pedals and (anti)orthotomics of frontals in de Sitter 2-space, Math. Meth. Appl. Sci., 46 (2023), 11157–11171. http://dx.doi.org/10.1002/mma.9173 doi: 10.1002/mma.9173
[20]	Y. Li, A. Abolarinwa, A. Alkhaldi, A. Ali, Some inequalities of Hardy type related to Witten-Laplace operator on smooth metric measure spaces, Mathematics, 10 (2022), 4580. http://dx.doi.org/10.3390/math10234580 doi: 10.3390/math10234580
[21]	Y. Li, A. Alkhaldi, A. Ali, R. A. Abdel-Baky, M. Saad, Investigation of ruled surfaces and their singularities according to Blaschke frame in Euclidean 3-space, AIMS Math., 8 (2023), 13875–13888. http://dx.doi.org/10.3934/math.2023709 doi: 10.3934/math.2023709
[22]	Y. Li, D. Ganguly, Kenmotsu metric as conformal $\eta$ -Ricci soliton, Mediterr. J. Math., 20 (2023), 193. https://doi.org/10.1007/s00009-023-02396-0 doi: 10.1007/s00009-023-02396-0
[23]	Y. Li, S. K. Srivastava, F. Mofarreh, A. Kumar, A. Ali, Ricci soliton of $\mathscr{CR}$ -warped product manifolds and their classifications, Symmetry, 15 (2023), 976. https://doi.org/10.3390/sym15050976 doi: 10.3390/sym15050976
[24]	Y. Li, P. Laurian-Ioan, L. Alqahtani, A. Alkhaldi, A. Ali, Zermelo's navigation problem for some special surfaces of rotation, AIMS Math., 8 (2023), 16278–16290. https://doi.org/10.3934/math.2023833 doi: 10.3934/math.2023833
[25]	Y. Li, Z. Wang, Lightlike tangent developables in de Sitter 3-space, J. Geom. Phys., 164 (2021), 104188. https://doi.org/10.1016/j.geomphys.2021.104188 doi: 10.1016/j.geomphys.2021.104188
[26]	Y. Li, Z. Wang, T. H. Zhao, Geometric algebra of singular ruled surfaces, Adv. Appl. Clifford Algebras, 31 (2021), 19. https://doi.org/10.1007/s00006-020-01097-1 doi: 10.1007/s00006-020-01097-1
[27]	Y. Li, A. Çalışkan, Quaternionic shape operator and rotation matrix on ruled surfaces, Axioms, 12 (2023), 486. https://doi.org/10.3390/axioms12050486 doi: 10.3390/axioms12050486
[28]	Y. Li, S. G. Mazlum, S. Şenyurt, The Darboux trihedrons of timelike surfaces in the Lorentzian 3-space, Int. J. Geom. Methods Mod. Phys., 20 (2023), 2350030-82. https://doi.org/10.1142/S0219887823500305 doi: 10.1142/S0219887823500305
[29]	S. G. Mazlum, S. Şenyurt, L. Grilli, The invariants of dual parallel equidistant ruled surfaces, Symmetry, 15 (2023), 206. https://doi.org/10.3390/sym15010206 doi: 10.3390/sym15010206
[30]	S. G. Mazlum, Geometric properties of timelike surfaces in Lorentz-Minkowski 3-space, Filomat, 37 (2023), 5735–5749. https://doi.org/10.2298/FIL2317735G doi: 10.2298/FIL2317735G
[31]	S. G. Mazlum, S. Şenyurt, L. Grilli, The dual expression of parallel equidistant ruled surfaces in Euclidean 3-space, Symmetry, 14 (2022), 1062. https://doi.org/10.3390/sym14051062 doi: 10.3390/sym14051062
[32]	S. Şenyurt, S. G. Mazlum, Spacelike surface geometry, Int. J. Geom. Methods Mod. Phys., 14 (2017), 1750118. https://doi.org/10.1142/S0219887817501183
[33]	Z. I. Szabó, Structure theorems on Riemannian spaces satisfying $R(X, Y)\cdot R = 0$ . I, J. Differ. Geom., 17 (1982), 531–582.
[34]	K. Yano, On semi symmetric metric connection, Rev. Roum. Math. Pures Appl., 15 (1970), 1579–1586.
[35]	K. Yano, J. Imai, Quarter-symmetric connection and their curvature tensors, Tensor (N.S.), 38 (1982), 13–18.
[36]	K. Yano, S. Ishihara, Tangent and cotangent bundles, New York: Marcel Dekker, Inc., 1973.

Reader Comments

Your name:*

Email:*
© 2023 the Author(s), licensee AIMS Press. This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0)

通讯作者: 陈斌, bchen63@163.com

1.
沈阳化工大学材料科学与工程学院沈阳 110142

AIMS Mathematics

1.8 3.4

Metrics

Article views(1792) PDF downloads(174) Cited by(23)

Preview PDF

Download XML

Export Citation

Article outline

Show full outline

AIMS Mathematics

Some notes on the tangent bundle with a Ricci quarter-symmetric metric connection

Related Papers:

Abstract

1. Introduction

2. Proofs of theorems

3. Conclusions

Use of AI tools declaration

Acknowledgments

Conflict of interest

References

Reader Comments

通讯作者: 陈斌, bchen63@163.com

Metrics

Other Articles By Authors

Catalog

AIMS Mathematics

Some notes on the tangent bundle with a Ricci quarter-symmetric metric connection

Related Papers:

Abstract

1. Introduction

2. Proofs of theorems

3. Conclusions

Use of AI tools declaration

Acknowledgments

Conflict of interest

References

Reader Comments

通讯作者: 陈斌, bchen63@163.com

Metrics

Other Articles By Authors

Related pages

Tools

Export File

Citation

Format

Content

Catalog