\begin{document}

\maketitle

\begin{abstract}
The aim of this paper is to study the T-tensor of  $(\alpha,\beta)$-metrics. We characterize the case in which the $(\alpha,\beta)$-metric satisfies the T-condition; that is,  the T-tensor vanishes. Also, the case which the $(\alpha,\beta)$-metric admitting functions $\sigma_h(x)$ such that $\sigma_hT^h_{ijk}=0$, or simply $\sigma$T-condition,  is characterized. In both cases, by solving some ODEs, we find explicit formulas for the $(\alpha,\beta)$-metrics. Both of the two classes are  obtained by Z. Shen  in a completely different approach. The  $(\alpha,\beta)$-metrics  satisfying the T-condition are positively almost regular  Berwald  metrics and those   $(\alpha,\beta)$-metrics  satisfying the $\sigma$T-condition provide a class of almost regular  Landsberg non-Berwaldian spaces.  
\end{abstract}


\section{Introduction}


 The T-tensor plays an important role in Finsler geometry. For example, it was proved by  Szabo Zoltan \cite{Szabo_T-tensor} that a positive definite Finsler metric with vanishing T-tensor is Riemannian. Also,  Hashiguchi \cite{hashiguchi}, showed   that a  Landsberg space remains  Landsberg by every conformal change if and only if the T-tensor vanishes identically. For further results and studies on the T-tensor, we refer for example to \cite{Asanov_T-tensor,r2.12,r2.8,matsumoto}.

For a Finsler manifold $(M,F)$, the conformal transformation of $F$ is defined by 
$$ \overline{F}=e^{\sigma(x)}F,$$
where $\sigma(x)$ is a function on the manifold $M$.     Under  a conformal transformation,  Landsberg space remains  Landsberg if and only if    
 $$\sigma_r T^r_{jkh}=0, \quad \sigma_r:=\frac{\partial \sigma}{\partial x^r}.$$
  It is obvious to note that if the  condition $\sigma_r T^r_{jkh}=0$   holds for every $\sigma(x)$, then the T-tensor vanishes.  But using the result of Szabo Zoltan \cite{Szabo_T-tensor}, the space will be Riemannian.  So, what will be more beneficial is to consider the case when  Landsberg space remains  Landsberg under some conformal transformations. 
  In \cite{Elgendi_BL}, the author focused on the case which the condition $\sigma_r T^r_{jkh}=0$ is satisfied for some $\sigma(x)$ (some conformal transformations). 
  
  \bigskip
  
In this paper, we study the T-tensor of the $(\alpha,\beta)$-metrics. Any $(\alpha,\beta)$-metric $F$ takes the form $F=\alpha\phi(s)$, $s:=\frac{\beta}{\alpha}$ and $\beta$ is one form on the base manifold $M$.   We start by studying the Cartan tensor $C_{ijk}$ of $(\alpha,\beta)$-metrics. We show that the Cartan tensor $C_{ijk}$ vanishes identically; that is, the space is Riemannian, if and only if $\phi(s)= \sqrt{k_1+k_2s^2}$, where $k_1$ and $k_2$ are constants.

We calculate the T-tensor of the $(\alpha,\beta)$-metrics. We find the necessary and sufficient conditions for $(\alpha,\beta)$-metrics to satisfy the T-condition; that is, the T-tensor vanishes identically. By solving some ODEs, we  show that an  $(\alpha,\beta)$-metric satisfies the T-condition if and only if it is Riemannian or $\phi(s)$ has the following form
 $$\phi(s)=
      c_3s^{\frac{c b^2-1}{cb^2}}(cb^2-cs^2)^{\frac{1}{2cb^2}}. $$

We introduce the notion of $\sigma$T-condition; we say that a Finsler space satisfies the  $\sigma$T-condition if it is admitting functions $\sigma_h(x)$ such that $\sigma_hT^h_{ijk}=0$. We find the necessary and sufficient conditions for an $(\alpha,\beta)$-metric in order  to satisfy the $\sigma$T-condition. Moreover, we show that  the $(\alpha,\beta)$-metrics satisfy the  $\sigma$T-condition if and only if the T-tensor vanishes (this is the trivial case) or $\phi(s)$ is given by
$$\phi(s)=c_3 \,exp\left(\int \frac{c_1\sqrt{b^2-s^2}+c_2 s}{s(c_1\sqrt{b^2-s^2}+c_2 s)+1}ds\right). $$


It is worthy  to mention that the above  special $(\alpha,\beta)$-metrics are obtained by Z. Shen \cite{Shen_example}. Namely, the  formulas of $\phi(s)$ that characterized the T-condition produce  positively almost regular Berwald metrics. One can predict that the metric, in this case, is not regular because the T-tensor vanishes (by Szabo Zoltan's result). In his paper, Shen showed this almost regular property.  The non trivial formula that characterized the $\sigma$T-condition (with some restrictions) provides the class of (almost regular) Landsberg metrics  which are not Berwaldian.

   

  
 In \cite{Elgendi_BL}, the author claimed  that the long existing problem of regular Landsberg  non Berwaldian spaces  is (closely) related to   the question:\\
 
 \textit{Is there  any  Finsler space admitting a function $\sigma(x)$ such that $\sigma_rT^r_{ijk}=0$, $\sigma_r=\frac{\partial \sigma}{\partial x^r}$?}\\


  
   And in this paper we confirm this claim, in the almost regular case, since the class of $(\alpha,\beta)$-metrics that satisfies the $\sigma$T-condition is as same as the class of Landsberg non Berwaldian metrics which is obtained by Z. Shen \cite{Shen_example}.

\section{The Cartan tensor and T-tensor of $(\alpha,\beta)$-metrics}

Let  $M$ be  an n-dimensional smooth manifold.   Let $(x^i)$ be
the coordinates of any  point of the base manifold $M$  and $(y^i)$ a
supporting element at the same point.  We mean by $T_xM$ the tangent space at $x\in M$ and by $TM=\undersymm{\bigcup}{x\in M}\,\,T_{x}M$ the tangent bundle of $M$.

An $(\alpha,\beta)$-metric is a function   $F$ on  $TM$ defined by
$$F=\alpha \phi(s), \quad s:=\frac{\beta}{\alpha},$$
where $\phi=\phi(s)$ is a $C^\infty$ function on $(-b_0,b_0)$ and $\alpha=\sqrt{a_{ij}y^iy^j}$ is a Riemannian metric and $\beta=b_i(x)y^i$ is a one form on $M$. For a Riemannian metric $\alpha$ and a one form $\beta$ such that $\|\beta_x\|_\alpha<b_0$,
\begin{equation}\label{phi_conditions}
\phi(s)>0, \quad \phi(s)-s\phi'(s)+(t^2-s^2)\phi''(s)>0, \quad |s|<t<b_0,
\end{equation}
 the function $F$ is a (positive definite) Finsler metric and such $(\alpha,\beta)$-metric is said to be \textit{regular}. And if $\|\beta_x\|_\alpha\leq b_0$ for all $x\in M$, then the $(\alpha,\beta)$-metric is said to be \textit{almost regular}. If $\phi=\phi(s)$ is a $C^\infty$  on $(0,b_0)$, then $F$ is called \textit{postively almost regular}. For more details, we refer to \cite{Shen_example}.

For an $(\alpha,\beta)$-metric $F=\alpha\phi(s)$, the metric tensor $g_{ij}=\frac{1}{2}\frac{\partial^2}{\partial y^i\partial y^j} F^2$, it can be calculated as follows 
\begin{equation}\label{a_ij}
g_{ij}=\rho a_{ij}+\rho_0 b_ib_j+\rho_1(b_i\alpha_j+b_j\alpha_i)+\rho_2 \alpha_i\alpha_j,
\end{equation}
where $\alpha_i:=\frac{\partial \alpha}{\partial y^i}=\frac{a_{ij}y^j}{\alpha}$.
\begin{eqnarray*}
\rho&:=&\phi^2-s\phi\phi', \\
\rho_0&:=&\phi'^2+\phi \phi'',\\
\rho_1&:=&\phi\phi'-s(\phi'^2+\phi \phi''),\\
\rho_2&:=&s^2(\phi'^2+\phi \phi'')-s\phi\phi',
\end{eqnarray*}
where  $\phi':=\frac{d\phi}{ds}$, $\phi'':=\frac{d^2\phi}{ds^2}$ and so on.  


\begin{proposition} \label{inverse-metric}
The inverse $g^{ij}$ of the metric $g_{ij}$ is given by
$$g^{ij}=\frac{1}{\rho}a^{ij}+\mu_ob^ib^j+\mu_1(b^i\alpha^j+b^j\alpha^j)+\mu_2\alpha^i\alpha^j,$$
where
$\mu_o:=-\frac{\phi\phi''}{\rho(\rho+\phi\phi''m^2)}$,\quad$\mu_1:=-\frac{\rho_1}{\rho(\rho+\phi\phi''m^2)}$\, and \,$\mu_2:=\frac{\rho_1(s\rho+(\rho_1+s\phi\phi'')m^2)}{\rho^2(\rho+\phi\phi''m^2)}$.
\end{proposition}
\begin{proof}
To determine the inverse of\,
 ${g}_{ij} $, let
\begin{equation}\label{inverse}
{g}^{\,ir}=(1/\rho){a}^{ir}+\mu_0b^ib^r+\mu_1(b^i\alpha^r+b^r\alpha^i)+\mu_2\alpha^i\alpha^r,
\end{equation}
where $b^i=a^{ij}b_j\,,b^2=b^ib_i$. The functions $\mu_0$, $\mu_1$ and $\mu_2$
are to be determined. From  (\ref{inverse}), \eqref{a_ij} and using the fact that $s\rho_1+\rho_2=0$, we have
 \begin{eqnarray*}
 {g}^{\,ir}\,{g}_{jr}&=&
  \delta^r_j+b^rb_j(\rho_0/\rho+(\rho+\rho_0b^2+s\rho_1)\mu_0+(\rho_1+s\rho_0)\mu_1\\
   &&+b^r\alpha_j(\rho_1/\rho+(b^2 \rho_1+s\rho_2)\mu_0+\rho \mu_1)\\
   &&+\alpha^rb_j(\rho_1/\rho+(b^2 \rho_0+\rho+s\rho_2)\mu_1+(\rho_1+s\rho_0) \mu_2)\\
   &&+\alpha^r\alpha_j(\rho_2/\rho+(b^2 \rho_1+s\rho_2)\mu_1+\rho \mu_2)\\
\end{eqnarray*}
The coefficients of $b^ib_j$, $b^i\alpha_j$, $\alpha^ib_j$ and $\alpha^i\alpha_j$ must
vanish so that  we get the following system of equations:
\begin{equation}\label{shs1-5}
\left.
    \begin{array}{rcl}
(\rho+\rho_0b^2+s\rho_1)\mu_0+(\rho_1+s\rho_0)\mu_1&=&-\rho_0/\rho\\
(b^2 \rho_1+s\rho_2)\mu_0+\rho \mu_1&=&-\rho_1/\rho\\
(b^2 \rho_0+\rho+s\rho_2)\mu_1+(\rho_1+s\rho_0) \mu_2&=&-\rho_1/\rho\\
(b^2 \rho_1+s\rho_2)\mu_1+\rho \mu_2&=&-\rho_2/\rho.
    \end{array}
  \right.
  \end{equation}
   Solving the above  system for  $\mu_0$,
  $\mu_1$ and $\mu_2$, the result follows.
  \end{proof}
\begin{remark}\label{denomaintor}
It is to be noted that if  $\rho+\phi\phi''m^2=0$, the quantity appeared in the denominators of $\mu_0$, $\mu_1$ and $\mu_2$, then we get
$$\phi(s)-s\phi'(s)+(b^2-s^2)\phi''(s)=0.$$
The solution of the above ODE is given by
$$\phi(s)=c_1s+c_2\sqrt{b^2-s^2}, \quad c_1 \,\, \text{and} \,\, c_2  \,\,\text{are  constants} .$$
But, by making use of the condition \eqref{phi_conditions}, such a $\phi$ is excluded.
\end{remark}

\bigskip

Since the Cartan tensor $C_{ijk}$ is  defined by 
$$C_{ijk}=\frac{1}{2}\frac{\partial g_{ij}}{\partial y^k},$$
we have the following.
\begin{lemma}\label{C_ijk}
The Cartan tensor $C_{ijk}$ of an $(\alpha,\beta)$-metric is given by
$$C_{ijk}=\frac{\rho_1}{2\alpha}(h_{ij}m_k+h_{jk}m_i+h_{ik}m_j)+\frac{\rho_0'}{2\alpha}m_im_jm_k,$$
where $h_{ij}=a_{ij}-\alpha_i\alpha_j$ and $m_i:=b_i-s\alpha_i$.
\end{lemma}

\begin{remark} The covariant vector $m_i$ satisfies the  properties
$$m_i\neq 0, \quad y^im_i=0, \quad  m^2:=m^im_i=b^im_i\neq 0, \quad b^ih_{ij}=m_j, $$
{where},  $m^2=b^2-s^2$.
\end{remark}


\begin{theorem}For $n\geq 3$, the following assertions are equivalent:
\begin{description}
\item[(a)] $\rho_1=0$.

\item[(b)] $\rho_2=0$.

\item[(c)] $(\alpha,\beta)$-metric is Riemannian.

\item[(d)] $\phi=\sqrt{k_1s^2+k_2}$.
\end{description}
\end{theorem}
\begin{proof}~\\
\noindent\textbf{(a)$\Longleftrightarrow $(d):} Let $\rho_1=0$.  One can write $\rho_1=\phi\phi'-s\rho_0$ and also  $\rho_0=(\phi\phi')'$. So, we have 
\begin{eqnarray*}
\rho_1=0&\Longleftrightarrow&s\rho_0=\phi\phi'\\
&\Longleftrightarrow&s(\phi\phi')'=\phi\phi'\\
&\Longleftrightarrow&\phi\phi'=k_1s \quad  (\textsl{by integration})\\
&\Longleftrightarrow&\phi=\sqrt{k_1s^2+k_2},
\end{eqnarray*}
where  $k_1$ and $k_2$ are arbitrary constants.

\noindent\textbf{(a)$\Longleftrightarrow $(c):} Let $\rho_1=0$.  By making use of the above equivalences and Lemma \ref{C_ijk}, we get 
\begin{eqnarray*}
\rho_1=0&\Longleftrightarrow&\phi\phi'=ks \\
&\Longrightarrow&(\phi\phi')'=k\\
&\Longrightarrow&(\phi\phi')''=\rho_0'=0 \\
&\Longrightarrow& C_{ijk}=0,
\end{eqnarray*}
and hence the space $(M,F)$ is Riemannian.  Now, assume that $(M,F)$ is Riemannian, we have 
$$\rho_1(h_{ij}m_k+h_{jk}m_i+h_{ik}m_j)+\rho_0'm_im_jm_k=0.$$
Contracting the above equation by $b^ib^j$ and using the fact that $m^2\neq 0$, we obtain
$$3m^2\rho_1+\rho_0'm^4=0.$$
And the contraction by $g^{ij}$ gives
$$(n+1)\rho_1+\rho_0'm^2=0.$$
Hence, by making use of the fact that $n\geq 3$, we get that $\rho_1=0$ and $\rho_0'=0$.


\noindent\textbf{(a)$\Longleftrightarrow $(b):}  Since  $\rho_2=s\rho_1$, then the proof is obvious.
\end{proof}


For a Finsler manifold $(M,F)$, the T-tensor is defined by \cite{ttensor}
\begin{equation} \label{T-tensor}
T_{rijk}=FC_{rijk}-F(C_{sij}C^{s}_{rk}+C_{sjr}C^{s}_{ik}+C_{sir}C^{s}_{jk})
+C_{rij}\ell_k+C_{rik}\ell_j +C_{rjk}\ell_i+C_{ijk}\ell_r,
\end{equation}
where $\ell_j:=\dot{\partial}_j F$, where $\dot{\partial}_j$ is the differentiation with respect to $y^j$.

The T-tensor is totally symmetric in all of its indices.


\begin{theorem}\label{t}
The T-tensor of an $(\alpha,\beta)$-metric  takes the form:
\begin{eqnarray*}
  {T}_{hijk} &=& \Phi (h_{hi}h_{jk}+h_{hj}h_{ik}+h_{hk}h_{ij})\\
  &&+\Psi(h_{hk}m_im_j+h_{hj}m_im_k+h_{hi}m_jm_k+h_{ij}m_hm_k+h_{jk}m_im_h+h_{ik}m_jm_h)\\
   &&+\Omega m_hm_im_jm_k
\end{eqnarray*}
where
$$\Phi:=-\frac{\rho_1\phi}{2\alpha}
  (s+\alpha k_1m^2),\quad \Psi:=\frac{\rho_1\phi'}{\alpha}-\frac{\rho_1^2\phi}{\alpha\rho}-\frac{s\rho_0'\phi}{2\alpha}-\frac{\rho_1\phi m^2K_2}{2},$$
 $$\Omega:=\frac{\rho_0''\phi}{2\alpha}+\frac{2\rho_0'\phi'}{\alpha}-3\phi(k_2(\rho_1+\frac{\rho_0'm^2}{2})+\frac{\rho_1\rho_0'}{2\alpha\rho}),$$  
  $$K_1:=\frac{\rho_1(1+\rho\mu_0m^2)}{2\alpha\rho}=\frac{\rho_1}{2\alpha(\rho+m^2\phi\phi'')}, $$
  $$  K_2:=\frac{\rho_0'(1+\rho\mu_0m^2)}{2\alpha\rho}+\frac{\rho_1\mu_o}{\alpha}=\frac{\rho\rho_0'-2\rho_1\phi\phi''}{2\alpha\rho(\rho+m^2\phi\phi'')}.$$
\end{theorem}
\begin{proof}
By using Lemma \ref{C_ijk} and making use of the fact that $\dot{\partial}_is=\frac{m_i}{\alpha}$, we have
\begin{eqnarray}\label{lamda}
  \dot{\partial}_h{C}_{ijk}  &=& 
\nonumber     -\frac{\rho_1}{2\alpha^2}(h_{ik}n_{jh}+h_{jk}n_{ih}+h_{ij}n_{kh}
   +h_{jh}n_{ik}+h_{kh}n_{ij}+h_{ih}n_{jk}) \\
\nonumber    &&-\frac{s\rho_1}{2\alpha^2}(h_{ik}h_{jh}+h_{jk}h_{ih}+h_{kh}h_{ij})
   -\frac{s\rho_0'}{2\alpha^2}(h_{ij}m_hm_k+h_{ki}m_jm_h+h_{hk}m_im_j\\
\nonumber    &&+h_{jh}m_km_i+h_{ih}m_jm_k+h_{kj}m_im_h)+\frac{\rho_0''}{2\alpha^2}m_im_jm_hm_k\\
   &&-\frac{\rho_0'}{2\alpha^2}(n_{ij}m_hm_k+n_{kh}m_im_j)
\end{eqnarray}
where  $n_{ij}:=\alpha_im_j+\alpha_jm_i$. By making use of the fact that  $K_1$ and $K_2$ satisfy 
$$\frac{\rho_1}{2\alpha}\left(K_2m^2+\frac{\rho_1}{\alpha\rho}\right)=K_1\left(\frac{\rho_1}{\alpha}+\frac{\rho_0'm^2}{2\alpha}\right),$$ 
we have
\begin{eqnarray}\label{cc}
\nonumber {C}_{ijr} {C}^r_{hk}+{C}_{jkr} {C}^r_{hi}+{C}_{ikr} {C}^r_{hj} &=& \left(
   \frac{\rho_1K_2m^2}{2\alpha}+\frac{\rho_1^2}{\alpha^2\rho}\right)
   (h_{ij}m_hm_k+h_{ki}m_jm_h+h_{hk}m_im_j\\
   \nonumber&&{\hspace{-4cm}}
+h_{jh}m_km_i+h_{ih}m_jm_k+h_{kj}m_im_h)+3\left( \frac{\rho_1\rho_0'}{2\alpha^2\rho}+K_2\left( \frac{\rho_1}{\alpha}+ \frac{\rho_0'm^2}{2\alpha
   }\right)\right)m_im_jm_hm_k\\
 \nonumber  &&{\hspace{-4cm}}+ \frac{\rho_1K_1m^2}{2\alpha}(h_{ik}h_{jh}+h_{jk}h_{ih}+h_{kh}h_{ij}).
\end{eqnarray}
Since $ \ell_i:=\dot{\partial}_iF= \phi \alpha_i+\phi'm_i$, we get 
\begin{eqnarray*}
  C_{hij}\ell_k+C_{hik}\ell_j +C_{hjk}\ell_i+C_{ijk}\ell_h&=& \frac{\rho_0'\phi}{2\alpha}(m_im_jn_{kh}+m_km_hn_{ij})
 +\frac{2\rho_0'\phi'}{\alpha}m_im_jm_hm_k\\
   &&{\hspace{-4cm}}+\frac{\rho_1\phi'}{\alpha}(h_{ij}m_hm_k+h_{ki}m_jm_h+h_{hk}m_im_j+h_{jh}m_km_i+h_{ih}m_jm_k+h_{kj}m_im_h)\\
   &&{\hspace{-4cm}}  
   +\frac{\rho_1\phi}{2\alpha} (h_{ik}n_{jh}+h_{jk}n_{ih}+h_{ij}n_{kh}
   +h_{jh}n_{ik}+h_{kh}n_{ij}+h_{ih}n_{jk}).
\end{eqnarray*}
Now, taking the fact that $F=\alpha \phi$ into account, the T-tensor of the space $(M,F)$ is given by
\begin{eqnarray*}
  {T}_{hijk} &=&FC_{hijk}-F(C_{sij}C^{s}_{hk}+C_{hjr}C^{s}_{ik}+C_{sih}C^{s}_{jk})
+C_{hij}\ell_k+C_{hik}\ell_j +C_{hjk}\ell_i+C_{ijk}\ell_h\\
  &=& \Phi (h_{hi}h_{jk}+h_{hj}h_{ik}+h_{hk}h_{ij})\\
  &&+\Psi(h_{hk}m_im_j+h_{hj}m_im_k+h_{hi}m_jm_k+h_{ij}m_hm_k+h_{jk}m_im_h+h_{ik}m_jm_h)\\
   &&+\Omega\, m_hm_im_jm_k.
\end{eqnarray*}
\end{proof}
For an $(\alpha,\beta)$-metric, one can calculate $\Phi$, $\Psi$ and $\Omega$ to obtain the formula for its T-tensor. Or one can, easily,   use Maple program for these calculations, for example we have the following corollary. 
\begin{corollary} The T-tensor  of  Kropina metric, $(F=\frac{\alpha}{s}  ,\phi(s)=1/s)$, is given by
\begin{eqnarray*}\label{kropina2}
  {T}_{hijk} &=&
  \frac{2}{\alpha^2b^2s^2}(h_{hi}h_{jk}+h_{hj}h_{ik}+h_{hk}h_{ij})+
  \frac{2}{\alpha b^2 s^3}(h_{hi}m_{j}m_k+h_{hj}m_{i}m_k+h_{ij}m_{h}m_k\\
   &&+h_{jk}m_{i}m_h+h_{hk}m_{i}m_j+h_{ik}m_{j}m_h)+\frac{6 }{\alpha b^2 s^5}m_hm_im_jm_k.
\end{eqnarray*}
It is to be noted  that the above result is  also obtained by
Shibata \cite{r2.12} and \cite{beta2}.

The T-tensor  of Randers metric, ($F={\alpha}(1+s), \phi(s)=1+s)$, is given by
\begin{eqnarray*}
  {T}_{hijk} &=& -
  \frac{b^2+s^2+2s}{4\alpha}(h_{hi}h_{jk}+h_{hj}h_{ik}+h_{hk}h_{ij}),
\end{eqnarray*}
\end{corollary}
The above case has been studied by Matsumoto \cite{r2.8}.
\section{The T-condition and $\sigma$T-conditions}

It is well known that the Finsler spaces with vanishing T-tensor are called Finsler spaces satisfying the \textit{T-condition}. In a similar manner, we call the Finsler spaces admitting functions $\sigma_h(x)$ such that $\sigma_hT^h_{ijk}=0$ Finsler spaces satisfying   the \textit{$\sigma$T-condition}. 
In this section, we characterize the $(\alpha,\beta)$-metrics which satisfy the T-condition and the $\sigma$T-condition.  
\begin{theorem}\label{T-condition}
The $(\alpha,\beta)$-metrics with $n\geq 3$ satisfy the T-condition; that is, the T-tensor  vanishes identically, if and only if $\Phi=0$.  
\end{theorem}

\begin{proof}
 Let $T_{hijk}=0$, then we have
 \begin{eqnarray}\label{T=0}
\nonumber && \Phi (h_{hi}h_{jk}+h_{hj}h_{ik}+h_{hk}h_{ij})+\Psi(h_{hk}m_im_j+h_{hj}m_im_k+h_{hi}m_jm_k\\
  &&+h_{ij}m_hm_k+h_{jk}m_im_h+h_{ik}m_jm_h)
   +\Omega\, m_hm_im_jm_k=0.
\end{eqnarray}
By contracting the above equation by $b^h$, we get
  $$(\Phi+m^2\Psi) (h_{jk}m_i+h_{ik}m_j+h_{ij}m_k)+(3\Psi+m^2\Omega)m_im_jm_k=0,$$
  from which together with the fact that $n\geq 3$, we have 
\begin{eqnarray}\label{T=0_1}
  \Phi+m^2\Psi=0,\quad 3\Psi+m^2\Omega=0.
\end{eqnarray}
Again, contraction \eqref{T=0} by $a^{hi}$, we obtain
$$
  ((n+1)\Phi+m^2\Psi) h_{jk}+((n+3)\Psi+m^2\Omega)m_jm_k=0.
$$
Then, taking the fact that $n\geq 3$ into account, we get 
\begin{eqnarray}\label{T=0_2}
  (n+1)\Phi+m^2\Psi=0, \quad (n+3)\Psi+m^2\Omega=0.
\end{eqnarray}
 Now, solving the equations \eqref{T=0_1} and \eqref{T=0_2} for $\Phi$, $\Psi$ and $\Omega$, we have $\Phi=0$, $\Psi=0$ and $\Omega=0$.
 
Conversely, let $\Phi=0$, then we have either $\rho_1=0$ or $s+\alpha k_1m^2=0$. If $\rho_1=0$ (the space is Riemannian), then $\rho_0'=0$ and hence $\Psi=0$ and $\Omega=0$. And if  $s+\alpha k_1m^2=0$, one can conclude  that $\Psi=0$ and $\Omega=0$ (see Theorem \ref{Theorem_T-condition}). 
\end{proof}


\begin{proposition}\label{raised-tensor} The T-tensor $T^h_{ijk}:=g^{hr}T_{rijk}$ is given by
\begin{eqnarray*}
T^h_{ijk}&=&\frac{\Phi}{\rho}(h^h_{i}h_{jk}+h^h_{j}h_{ik}+h^h_{k}h_{ij})+\frac{\Psi}{\rho}(h^h_{k}m_im_j+h^h_{j}m_im_k+h^h_{i}m_jm_k+h_{ij}m^hm_k\\
&&+h_{jk}m_im^h+h_{ik}m_jm^h)+\frac{\Omega}{\rho} m^hm_im_jm_k+(\mu_0b^h+\mu_1\alpha^h)(\Phi(h_{ik}m_{j}+h_{ij}m_{k}+h_{jk}m_{i})\\
&&+\Psi(m^2(h_{ik}m_{j}+h_{ij}m_{k}+h_{jk}m_{i})+3m_im_jm_k)+\Omega m^2 m_im_jm_k)
\end{eqnarray*}
\end{proposition}
\begin{proof}
The proof is a straightforward calculations by using Proposition \ref{inverse-metric}.
\end{proof}

\begin{theorem}\label{sigmaT-condition}
The $(\alpha,\beta)$-metrics with $n\geq 3$ admits  function $\sigma_i(x)$ such that $\sigma_hT^h_{ijk}=0$ if and only if
\begin{description}
\item[(a)] $\Phi+m^2\Psi=0.$

\item[(b)] $m^2\Omega+3 \Psi=0.$

\item[(c)]$\sigma_j-\frac{\sigma_0}{ s\alpha}b_j=0$.
\end{description}
\end{theorem}

\begin{proof}  By using Proposition \ref{raised-tensor}, we have
\begin{eqnarray*}
\sigma_h T^h_{ijk}&=&\frac{\Phi}{\rho}\left(\left(\sigma_i-\frac{\sigma_0}{\alpha}\alpha_i\right)h_{jk}+\left(\sigma_j-\frac{\sigma_0}{\alpha}\alpha_j\right)h_{ik}+\left(\sigma_k-\frac{\sigma_0}{\alpha}\alpha_k\right)h_{ij}\right)\\
&&+\frac{\Psi}{\rho}\Big(\left(\sigma_k-\frac{\sigma_0}{\alpha}\alpha_k\right)m_im_j+\left(\sigma_j-\frac{\sigma_0}{\alpha}\alpha_j\right)m_im_k+\left(\sigma_i-\frac{\sigma_0}{\alpha}\alpha_i\right)m_jm_k\\
&&+\left(\sigma_\beta-s\frac{\sigma_0}{\alpha}\right)\left(h_{ij}m_k
+h_{jk}m_i+h_{ik}m_j\right)\Big)
+\frac{\Omega}{\rho} \left(\sigma_\beta-s\frac{\sigma_0}{\alpha}\right)m_im_jm_k\\
&&+\left(\mu_0\sigma_\beta+\mu_1\frac{\sigma_0}{\alpha}\right)\Big(\Phi(h_{ik}m_{j}+h_{ij}m_{k}+h_{jk}m_{i})\\
&&+\Psi(m^2(h_{ik}m_{j}+h_{ij}m_{k}+h_{jk}m_{i})+3m_im_jm_k)+\Omega m^2 m_im_jm_k\Big),
\end{eqnarray*}
where, $\sigma_0:=\sigma_iy^i$ and $\sigma_\beta:=\sigma_ib^i$.
Using the fact that $m_i=b_i-s\alpha_i$, we get
\begin{eqnarray*}
\sigma_h T^h_{ijk}&=&\frac{\Phi}{\rho}\left(\left(\sigma_i-\frac{\sigma_0}{s\alpha}b_i\right)h_{jk}+\left(\sigma_j-\frac{\sigma_0}{s\alpha}b_j\right)h_{ik}+\left(\sigma_k-\frac{\sigma_0}{s\alpha}b_k\right)h_{ij}\right)\\
&&
\frac{\sigma_0\Phi}{s\alpha\rho}(m_ih_{jk}+m_jh_{ik}+m_kh_{ij})\\
&&+
\frac{\Psi}{\rho}\left(\left(\sigma_k-\frac{\sigma_0}{s\alpha}b_k\right)m_im_j+\left(\sigma_i-\frac{\sigma_0}{s\alpha}b_i\right)m_km_j+\left(\sigma_j-\frac{\sigma_0}{s\alpha}b_j\right)m_im_k\right)\\
&&+\frac{\Psi}{\rho}\left(\sigma_\beta-s\frac{\sigma_0}{\alpha}\right)(h_{ij}m_k
+h_{jk}m_i+h_{ik}m_j)
\\
&&+3\frac{\sigma_0\Psi}{s\alpha\rho}m_km_im_j+\frac{\Omega}{\rho} \left(\sigma_\beta-s\frac{\sigma_0}{\alpha}\right)m_im_jm_k\\
&&+\left(\mu_0\sigma_\beta+\mu_1\frac{\sigma_0}{\alpha}\right)(\Phi(h_{ik}m_{j}+h_{ij}m_{k}+h_{jk}m_{i})\\
&&+\Psi(m^2(h_{ik}m_{j}+h_{ij}m_{k}+h_{jk}m_{i})+3m_im_jm_k)+\Omega m^2 m_im_jm_k).
\end{eqnarray*}
The above equation can be written in the following  form
\begin{eqnarray*}
\sigma_h T^h_{ijk}&=&\left(\frac{\sigma_0\Phi}{s\alpha\rho}+\frac{\Psi}{\rho}\left(\sigma_\beta-s\frac{\sigma_0}{\alpha}\right)+(\Phi+m^2\Psi)\left(\mu_0\sigma_\beta+\mu_1\frac{\sigma_0}{\alpha}\right)\right)(h_{ik}m_{j}+h_{ij}m_{k}+h_{jk}m_{i})\\
&&+ \left(\frac{\Omega}{\rho} \left(\sigma_\beta-s\frac{\sigma_0}{\alpha}\right)+ 3\frac{\sigma_0\Psi}{s\alpha\rho}+(\Omega m^2+3\Psi )\left(\mu_0\sigma_\beta+\mu_1\frac{\sigma_0}{\alpha}\right)\right)m_km_im_j \\
&&+\frac{\Phi}{\rho}\left( h_{jk}\left(\sigma_i-\frac{\sigma_0}{s\alpha}b_i\right)+h_{ij}\left(\sigma_k-\frac{\sigma_0}{s\alpha}b_k\right)+ h_{ik}\left(\sigma_j-\frac{\sigma_0}{ s\alpha}b_j\right)\right)\\
&&+\frac{\Psi}{\rho}\left( m_jm_k\left(\sigma_i-\frac{\sigma_0}{ s\alpha}b_i\right)+m_im_j\left(\sigma_k-\frac{\sigma_0}{s\alpha}b_k\right)+ m_im_k\left(\sigma_j-\frac{\sigma_0}{ s\alpha}b_j\right)\right).\\
\end{eqnarray*}


Now, putting $\sigma_h T^h_{ijk}=0$, one can conclude that

$$\frac{\sigma_0\Phi}{s\alpha\rho}+\frac{\Psi}{\rho}\left(\sigma_\beta-s\frac{\sigma_0}{\alpha}\right)+(\Phi+m^2\Psi)\left(\mu_0\sigma_\beta+\mu_1\frac{\sigma_0}{\alpha}\right)=0,$$
$$\frac{\Omega}{\rho} \left(\sigma_\beta-s\frac{\sigma_0}{\alpha}\right)+ 3\frac{\sigma_0\Psi}{s\alpha\rho}+(\Omega m^2+3\Psi )\left(\mu_0\sigma_\beta+\mu_1\frac{\sigma_0}{\alpha}\right)=0,$$
$$\sigma_k-\frac{\sigma_0}{s\alpha}b_k=0.$$

Therefore, $\sigma_\beta=\frac{\sigma_0b^2}{s\alpha}$ and  taking the fact that $\sigma\neq 0$ into account,  we get 

$$\left(\frac{1}{s\alpha\rho}+\mu_0\frac{b^2}{s\alpha}+\frac{\mu_1}{\alpha}\right)(\Phi+m^2\Psi)=0,$$
$$\left( \frac{1}{s\alpha\rho}+\mu_0\frac{b^2}{s\alpha}+\frac{\mu_1}{\alpha}\right)(\Omega m^2+3\Psi )=0.$$

Now the choice $ \frac{1}{s\alpha\rho}+\mu_0\frac{b^2}{s\alpha}+\frac{\mu_1}{\alpha}=0$ gives the ODE $\rho-s\rho_1-s^2\phi\phi''=0$ which has the solution $\phi=k s$. This solution is just again the same  background Riemannian metric $\alpha$ up to some constants. So, we should have $\Phi+m^2\Psi=0$ and $\Omega m^2+3\Psi =0.$

Conversely, if the conditions (a), (b) and (c) are satisfied, then the result is obviously obtained.
\end{proof}

\section{Some ODEs}
In this section, we  focus our study  on the T-condition and $\sigma$T-condition. By solving some ODEs, we find explicit formulas for $(\alpha,\beta)$-metrics that satisfy the T-condition and $\sigma$T-condition.

We define a function $Q(s)$ as follows
$$Q(s):=\frac{\phi'}{\phi-s\phi'} .$$
The function $Q(s)$ simplifies and helps to solve the ODEs that will be treated in this section. Moreover, $\phi$ is given by
\begin{equation}\label{Q}
\phi(s)=exp\left(\int \frac{Q}{1+sQ}ds\right).
\end{equation}
\begin{theorem}\label{Theorem_T-condition}
An $(\alpha,\beta)$-metric with $n\geq 3$ satisfies the T-condition; that is, the T-tensor  vanishes identically, if and only if it is Riemannian or  $\phi$ is given by 
\begin{equation}\label{berwald}
\phi(s)=
c_3s^{\frac{c b^2-1}{cb^2}}(cb^2-cs^2)^{\frac{1}{2cb^2}}.
\end{equation}
\end{theorem}

\begin{proof}
By Theorem \ref{T-condition}, any  $(\alpha,\beta)$-metric satisfies the T-condition if and only if $\Phi=0$.
So,  taking the fact that $\phi-s\phi\neq 0$ into account, the ODE $s+\alpha k_1m^2=0$ can be rewritten as follows
$$Q'+\left(\frac{1}{s}+\frac{2s}{m^2}\right)Q=-\frac{2}{m^2}.$$ 
This is a first order linear differential equation and has the solution
$$Q=\frac{c\,b^2-1}{s}-cs=\frac{c\,(b^2-s^2)-1}{s}, \quad c \,\,\text{is a constant}.$$
Hence, 
$$1+sQ=c\,b^2-cs^2, \quad \frac{Q}{1+sQ}=\frac{1}{s}-\frac{1}{c s(b^2-s^2)}.$$
By using \eqref{Q}, $\phi(s)$ is given by \eqref{berwald}. Plugging  $\phi(s)$ in  $\Psi$ and $\Omega$, we have $\Psi=0$ and $\Omega=0$.
\end{proof}


\begin{theorem}
An $(\alpha,\beta)$-metric with $n\geq 3$ satisfies the $\sigma$T-condition if and only if it satisfies the T-condition or $\phi$ is given by 
\begin{equation}\label{Landsberg}
\phi(s)=c_3 \,exp\left(\int \frac{c_1\sqrt{b^2-s^2}+c_2 s}{s(c_1\sqrt{b^2-s^2}+c_2 s)+1}ds\right). 
\end{equation}
\end{theorem}
\begin{proof}
First we should write $\Phi$ and $\Psi$ in terms of $Q(s)$ and its derivations with respect to $s$, as  follows
$$\Phi=-\frac{\phi (\phi-s\phi')^2(Q-sQ')(sm^2 \phi' Q'+(2s\phi+m^2\phi')Q)}{4 \alpha(m^2 \phi' Q'+\phi Q)},$$
$$\Psi=-\frac{\phi (\phi-s\phi')^2Q''(sm^2 \phi' Q'+(2s\phi+m^2\phi')Q)}{4 \alpha(m^2 \phi' Q'+\phi Q)}.$$
Now, making use of the condition \eqref{phi_conditions}, Remark \ref{denomaintor} and the fact that $\phi-s\phi'\neq 0$, the condition $\Phi+m^2\Psi=0$ gives the following two possible ODEs
\begin{equation}\label{nontrivial_case}
(b^2-s^2)Q''-sQ'+Q=0
\end{equation}
or
\begin{equation}\label{trivial_case}
sm^2\phi'Q'+2s\phi Q+m^2\phi'Q=0
\end{equation}
The ODE \eqref{trivial_case} can be given in the form 
$$Q'+\left(\frac{1}{s}+\frac{2s}{m^2}\right)Q=-\frac{2}{m^2}$$ 
which gives the trivial case; that is, the T-tensor vanishes. The ODE \eqref{nontrivial_case} has the solution
$$Q(s)=c_1s+c_2\sqrt{b^2-s^2}.$$
By using \eqref{Q}, $\phi(s)$ is given by  \eqref{Landsberg}.
\end{proof}

\section{Concluding remarks}

$\bullet$ By the following special choice   $c_2:=-c$ and $c_1:=cb^2-1$ ($b^2$ is constant), the class \eqref{berwald} becomes
$$\phi(s)=c_3s^{\frac{c_1}{c_1+1}}(1+c_1+c_2s^2)^{\frac{1}{2(c_1+1)}}.$$
which is exactly as same as the one obtained by \cite{Shen_example} ( (7.4) in Theorem 7.2). Moreover,    this metric is  positively almost regular Berwaldian. 

It should be noted that this   irregularity is studied by Z. Shen \cite{Shen_example}. Here we confirm that this metric is not regular Finsler metric because it  has vanishing T-tensor. This because of Szabo Zoltan's result; that is, positive definite Finsler metric with vanishing T-tensor is Riemannian.

\bigskip

$\bullet$ If $b(x)=b_0$, then \eqref{Landsberg} can be rewritten an follows 
$$\phi(s)=c_3 \,exp\left(\int \frac{c_2'\sqrt{1-(s/b_0)^2}+c_1 s}{s(c_2'\sqrt{1-(s/b_0)^2}+c_1 s)+1}ds\right), \quad c_2':=c_2/b_0. $$
We notice that the above formulae for $\phi$ is as same as  the one obtained in \cite{Shen_example} ( (1.3) in Theorem 1.2). Which, under some restrictions on  $\beta$, represents a class of Landsberg non Berwaldian Finsler spaces. Also, with a special choice of the constants, $b_0=1$ and $c_1=0$, we obtain
$$\phi(s)=c_3 \,exp\left(\int \frac{c_2'\sqrt{1-s^2}}{c_2's\sqrt{1-s^2}+1}ds\right), $$
which is obtained by Asanov \cite{Asanov}.


   \bigskip

  
  $\bullet$ As an example, putting $c_2=0$ in the class \eqref{sigmaT-condition}, $\phi(s)$ is given by
  $$\phi(s)=c_3\sqrt{1+c_1b^2\sqrt{b^2-s^2}}\,exp\left(  \frac{c_1b^2}{\sqrt{4-c_1^2b^4}}\arctan \left( \frac{c_1b^2\sqrt{b^2-s^2}+2s}{\sqrt{4-c_1^2b^4}\sqrt{b^2-s^2}} \right)\right).$$
  Here there are no restrictions on $\beta$, so it can be chosen such that $b<1$ and since $\phi$ satisfies \eqref{phi_conditions}, then the Finsler metric $F=\alpha \phi(s)$ is regular Finsler metric admitting functions $\sigma_h(x)$ such that $\sigma_hT^h_{ijk}=0$. However, $F$ is not a solution for the Landsberg unicorn's problem. We have to put some restrictions on $\beta$ (see \cite{Shen_example}) so that we obtain such a solution. Unfortunately, these restrictions turn $F$ into almost regular Finsler metric.
  
$\bullet$     Sumerizing above,the class of $(\alpha,\beta)$-metrics that satisfies the $\sigma$T-condition is as same as the class  which is obtained by Z. Shen. This confirms our previous claim   in  \cite{Elgendi_BL} that  the long existing problem of regular Landsberg  non Berwaldian spaces  is (closely) related to   the question:\\
 
 \textit{Is there  any  Finsler space admitting functions $\sigma_r(x)$ such that $\sigma_rT^r_{ijk}=0$?}
{"config": "arxiv", "file": "1806.02620.tex"}