On the α-connections and the α-conformal equivalence on statistical manifolds

Purpose

In this paper, we give some properties of the α-connections on statistical manifolds and we study the α-conformal equivalence where we develop an expression of curvature $\bar{R}$ for $\bar{\nabla}$ in relation to those for ∇ and $\hat{\nabla}$ ⁠.

Design/methodology/approach

In the first section of this paper, we prove some results about the α-connections of a statistical manifold where we give some properties of the difference tensor K and we determine a relation between the curvature tensors; this relation is a generalization of the results obtained in [1]. In the second section, we introduce the notion of α-conformal equivalence of statistical manifolds treated in [1, 3], and we construct some examples.

Findings

We give some properties of the difference tensor K and we determine a relation between the curvature tensors; this relation is a generalization of the results obtained in [1]. In the second section, we introduce the notion of α-conformal equivalence of statistical manifolds, we give the relations between curvature tensors and we construct some examples.

Originality/value

1. Introduction

Let $(M^{m}, g)$ be a Riemannian manifold and ∇ a torsion free linear connection on M. The triple $(M^{m}, \nabla, g)$ is called a statistical manifold if ∇g is symmetric and the pair $(\nabla, g)$ is called a statistical structure. For a statistical manifold $(M^{m}, \nabla, g)$ ⁠, let ∇* be an affine connection on M such that,

X (g (Y, Z)) = g (\nabla_{X} Y, Z) + g (Y, \overset{*}{\nabla_{X}} Z),

for all X, Y and Z in $Γ (T M)$ ⁠. The affine connection $\overset{*}{\nabla}$ is torsion free, and $\overset{*}{\nabla} g$ symmetric. Then $\overset{*}{\nabla}$ is called the dual connection of ∇, the triple $(M^{m}, \overset{*}{\nabla}, g)$ is called the dual statistical manifold of $(M^{m}, \nabla, g)$ and $(\nabla, \overset{*}{\nabla}, g)$ is the dualistic structure. Denoted by $\hat{\nabla}$ the Levi-Civita connection associated with g, the difference tensor K is the $(1,2)$ -tensor defined by (see [1]).

K (X, Y) = \overset{*}{\nabla_{X}} Y - \nabla_{X} Y .

The difference tensor K satisfies for any vector fields X, Y, Z and any smooth function f on M the following properties:

\nabla_{X} Y = {\hat{\nabla}}_{X} Y - \frac{1}{2} K (X, Y), \overset{*}{\nabla_{X}} Y = {\hat{\nabla}}_{X} Y + \frac{1}{2} K (X, Y),

\begin{array}{l} K (X, Y) = K (Y, X), \\ K (X, Y + Z) = K (X, Y) + K (X, Z) \end{array}

and

K (f X, Y) = K (X, f Y) = f K (X, Y) .

Moreover, we have,

g (K (X, Y), Z) = g (K (X, Z), Y) .

A statistical structure is called trace-free if ∇v_g = 0 where v_g is the volume form determined by g. This condition is equivalent to the condition Tr_gK = 0. A statistical structure $(\nabla, g)$ is said to be of constant curvature $k \in R$ if the curvature tensor field R of ∇ satisfies,

R (X, Y) Z = k \{g (Y, Z) X - g (X, Z) Y\} .

If k = 0, $(\nabla, g)$ is called a Hessian structure. The concept of α-conformally equivalence was treated in [1] where the author develops an expression of the curvature $R^{(α)}$ ⁠. In [2], the authors studied a 1-conformally flat statistical submanifold of a flat statistical manifold; they proved that a 1-conformally flat statistical manifold with a Riemannian metric can be locally realized as a submanifold of a flat statistical manifold. The author in [3] gives a procedure to realize a statistical manifold, which is α-conformally equivalent to a manifold with an α-transitively flat connection, as a statistical submanifold and in [4], he describe a method to obtain α-conformally equivalent connections from the relation between tensors and the symmetric cubic form. In [5], the authors studied the statistical hypersurfaces of some types of the statistical manifolds, which enable to construct a structure of a constant curvature. The divergence of 1-conformally flat statistical manifolds is studied in [6] where the authors prove that the generalized Pythagorean theorem holds if the statistical manifold has a constant curvature. In the first section of this paper, we prove some results about the α-connections of a statistical manifolds where we give some properties of the difference tensor K and we determine a relation between the curvature tensors $R^{(α)}$ and $R^{(β)}$ ⁠; this relation is a generalization of the results obtained in [1]. In the second section, we introduce the notion of α-conformal equivalence of statistical manifolds treated in [1, 3], and we give the relations between $\bar{R}$ ⁠, R and $\hat{R}$ and we construct some examples.

2. Some results on the α-connections of a statistical manifolds

Let $(M^{m}, \nabla, g)$ a statistical manifold with a dualistic structure $(\nabla, \overset{*}{\nabla}, g)$ ⁠. For $α \in R$ ⁠, we define a family of torsion-free connections $\nabla^{(α)}$ by,

\nabla^{(α)} = \frac{1 + α}{2} \nabla + \frac{1 - α}{2} \overset{*}{\nabla} .

$\nabla^{(α)}$ is called an α-connection of $(M^{m}, \nabla, g)$ ⁠. The triple $(M^{m}, \nabla^{(α)}, g)$ is also a statistical manifold, and $\nabla^{(- α)}$ is the dual connection of $\nabla^{(α)}$ ⁠. In particular,

\nabla^{(0)} = \hat{\nabla}, \nabla^{(1)} = \nabla, \nabla^{(- 1)} = \overset{*}{\nabla} .

Moreover, we have the following equality,

\nabla_{X}^{(α)} Y = {\hat{\nabla}}_{X} Y - \frac{α}{2} K (X, Y) .

In general, for any $α, β \in R$ ⁠, it is easy to see that,

\nabla_{X}^{(α)} Y = \nabla_{X}^{(β)} Y - \frac{α - β}{2} K (X, Y)

(1)

Proposition 1.

Let $(M^{m}, \nabla, g)$ a statistical manifold with a dualistic structure $(\nabla, \overset{*}{\nabla}, g)$ ⁠. For all vector fields X, Y, Z on M, we have,

\begin{aligned} (\nabla_{X}^{(α)} K) (Y, Z) & = (\nabla_{X}^{(β)} K) (Y, Z) - \frac{α - β}{2} K (X, K (Y, Z)) \\ + \frac{α - β}{2} K (K (X, Y), Z) + \frac{α - β}{2} K (Y, K (X, Z)) . \end{aligned}

(2)

Proof of Proposition 1. Let $X, Y, Z \in Γ (T M)$ ⁠, by definition, we have,

(\nabla_{X}^{(α)} K) (Y, Z) = \nabla_{X}^{(α)} K (Y, Z) - K (\nabla_{X}^{(α)} Y, Z) - K (Y, \nabla_{X}^{(α)} Z) .

The properties of the difference tensor K gives us,

\nabla_{X}^{(α)} K (Y, Z) = \nabla_{X}^{(β)} K (Y, Z) - \frac{α - β}{2} K (X, K (Y, Z)),

K (\nabla_{X}^{(α)} Y, Z) = K (\nabla_{X}^{(β)} Y, Z) - \frac{α - β}{2} K (K (X, Y), Z)

and

K (Y, \nabla_{X}^{(α)} Z) = K (Y, \nabla_{X}^{(β)} Z) - \frac{α - β}{2} K (Y, K (X, Z)),

then

(\nabla_{X}^{(α)} K) (Y, Z) = \nabla_{X}^{(β)} K (Y, Z) - \frac{α - β}{2} K (X, K (Y, Z)) - K (\nabla_{X}^{(β)} Y, Z) + \frac{α - β}{2} K (K (X, Y), Z) - K (Y, \nabla_{X}^{(β)} Z) + K (Y, K (X, Z))

Finally, using the fact that,

(\nabla_{X}^{(β)} K) (Y, Z) = \nabla_{X}^{(β)} K (Y, Z) - K (\nabla_{X}^{(β)} Y, Z) - K (Y, \nabla_{X}^{(β)} Z),

we deduce that,

(\nabla_{X}^{(α)} K) (Y, Z) = (\nabla_{X}^{(β)} K) (Y, Z) - \frac{α - β}{2} K (X, K (Y, Z)) + \frac{α - β}{2} K (K (X, Y), Z) + \frac{α - β}{2} K (Y, K (X, Z)) .

Remark 1.

As particular cases of Eqn (2), we have

(\nabla_{X}^{(α)} K) (Y, Z) = ({\hat{\nabla}}_{X} K) (Y, Z) - \frac{α}{2} K (X, K (Y, Z)) + \frac{α}{2} K (K (X, Y), Z) + \frac{α}{2} K (Y, K (X, Z)) = (\nabla_{X} K) (Y, Z) - \frac{α - 1}{2} K (X, K (Y, Z)) + \frac{α - 1}{2} K (K (X, Y), Z) + \frac{α - 1}{2} K (Y, K (X, Z)) = (\overset{*}{\nabla_{X}} K) (Y, Z) - \frac{α + 1}{2} K (X, K (Y, Z)) + \frac{α + 1}{2} K (K (X, Y), Z) + \frac{α + 1}{2} K (Y, K (X, Z))

For a statistical structure $(\nabla, \overset{*}{\nabla}, g)$ ⁠, we denote R, R*, $\hat{R}$ the curvature tensors for ∇, $\overset{*}{\nabla}$ ⁠, $\hat{\nabla}$ , respectively, and $R^{(α)}$ the curvature tensor for $\nabla^{(α)}$ ⁠. In the first results, we give the relation between $R^{(α)}$ and $R^{(β)}$ for any $α, β \in R$ ⁠.

Theorem 1.

Let $(M^{m}, \nabla, g)$ a statistical manifold. The relation between $R^{(α)}$ and $R^{(β)}$ is given by,

\begin{aligned} R^{(α)} (X, Y) Z & = R^{(β)} (X, Y) Z + \frac{β - α}{2} (\nabla_{X}^{(β)} K) (Y, Z) - \frac{β - α}{2} (\nabla_{Y}^{(β)} K) (X, Z) \\ + \frac{{(β - α)}^{2}}{4} K (X, K (Y, Z)) - \frac{{(β - α)}^{2}}{4} K (Y, K (X, Z)), \end{aligned}

(3)

for all $X, Y, Z \in Γ (T M)$ ⁠.

Proof of Theorem 1. Let $X, Y, Z \in Γ (T M)$ ⁠, By definition we have,

R^{(α)} (X, Y) Z = \nabla_{X}^{(α)} \nabla_{Y}^{(α)} Z - \nabla_{X}^{(α)} \nabla_{Y}^{(α)} Z - \nabla_{[X, Y]}^{(α)} Z .

(4)

For the first term $\nabla_{X}^{(α)} \nabla_{Y}^{(α)} Z$ ⁠, we have,

\nabla_{Y}^{(α)} Z = \nabla_{Y}^{(β)} Z + \frac{β - α}{2} K (Y, Z),

then

\nabla_{X}^{(α)} \nabla_{Y}^{(α)} Z = \nabla_{X}^{(α)} {\hat{\nabla}}_{Y} Z - \frac{α}{2} \nabla_{X}^{(α)} K (Y, Z) .

It is simple to see that,

\nabla_{X}^{(α)} \nabla_{Y}^{(β)} Z = \nabla_{X}^{(β)} \nabla_{Y}^{(β)} Z + \frac{β - α}{2} K (X, \nabla_{Y}^{(β)} Z)

and

\nabla_{X}^{(α)} K (Y, Z) = \nabla_{X}^{(β)} K (Y, Z) + \frac{β - α}{2} K (X, K (Y, Z)),

which gives us

\begin{aligned} \nabla_{X}^{(α)} \nabla_{Y}^{(α)} Z & = \nabla_{X}^{(β)} \nabla_{Y}^{(β)} Z + \frac{β - α}{2} K (X, \nabla_{Y}^{(β)} Z) \\ + \frac{β - α}{2} \nabla_{X}^{(β)} K (Y, Z) + \frac{{(β - α)}^{2}}{4} K (X, K (Y, Z)) . \end{aligned}

(5)

A similar calculation gives,

\begin{aligned} \nabla_{Y}^{(α)} \nabla_{X}^{(α)} Z & = \nabla_{Y}^{(β)} \nabla_{X}^{(β)} Z + \frac{β - α}{2} K (Y, \nabla_{X}^{(β)} Z) \\ + \frac{β - α}{2} \nabla_{Y}^{(β)} K (X, Z) + \frac{{(β - α)}^{2}}{4} K (Y, K (X, Z)) . \end{aligned}

(6)

Finally, we have,

\nabla_{[X, Y]}^{(α)} Z = \nabla_{[X, Y]}^{(β)} Z + \frac{β - α}{2} K ([X, Y], Z)

(7)

If we replace (5), (6) and (7) in (4), we deduce that,

\begin{aligned} R^{(α)} (X, Y) Z & = R^{(β)} (X, Y) Z + \frac{β - α}{2} \nabla_{X}^{(β)} K (Y, Z) - \frac{β - α}{2} \nabla_{Y}^{(β)} K (X, Z) \\ + \frac{β - α}{2} K (X, \nabla_{Y}^{(β)} Z) - \frac{β - α}{2} K (Y, \nabla_{X}^{(β)} Z) - \frac{β - α}{2} K ([X, Y], Z) \\ + \frac{{(β - α)}^{2}}{4} K (X, K (Y, Z)) - \frac{{(β - α)}^{2}}{4} K (Y, K (X, Z)) \end{aligned}

Using the fact that,

\nabla_{X}^{(β)} K (Y, Z) = (\nabla_{X}^{(β)} K) (Y, Z) + K (\nabla_{X}^{(β)} Y, Z) + K (Y, \nabla_{X}^{(β)} Z),

\nabla_{Y}^{(β)} K (X, Z) = (\nabla_{Y}^{(β)} K) (X, Z) + K (\nabla_{Y}^{(β)} X, Z) + K (X, \nabla_{Y}^{(β)} Z)

and

K ([X, Y], Z) = K (\nabla_{X}^{(β)} Y, Z) - K (\nabla_{Y}^{(β)} X, Z),

we get

\begin{aligned} R^{(α)} (X, Y) Z & = R^{(β)} (X, Y) Z + \frac{β - α}{2} (\nabla_{X}^{(β)} K) (Y, Z) - \frac{β - α}{2} (\nabla_{Y}^{(β)} K) (X, Z) \\ + \frac{{(β - α)}^{2}}{4} K (X, K (Y, Z)) - \frac{{(β - α)}^{2}}{4} K (Y, K (X, Z)) . \end{aligned}

As particular cases of Theorem 1, we get the following Corollary:

Corollary 1.

Let $(M^{m}, \nabla, g)$ a statistical manifold with a dualistic structure $(\nabla, \overset{*}{\nabla}, g)$ ⁠. The relations between $R^{(α)}$ ⁠, $\hat{R}$ , R and R* are given by

\begin{aligned} R^{(α)} (X, Y) Z & = \hat{R} (X, Y) Z - \frac{α}{2} ({\hat{\nabla}}_{X} K) (Y, Z) + \frac{α}{2} ({\hat{\nabla}}_{Y} K) (X, Z) \\ + \frac{α^{2}}{4} K (X, K (Y, Z)) - \frac{α^{2}}{4} K (Y, K (X, Z)), \end{aligned}

(8)

\begin{aligned} R^{(α)} (X, Y) Z & = R (X, Y) Z + \frac{1 - α}{2} (\nabla_{X} K) (Y, Z) - \frac{1 - α}{2} (\nabla_{Y} K) (X, Z) \\ + \frac{{(1 - α)}^{2}}{4} K (X, K (Y, Z)) - \frac{{(1 - α)}^{2}}{4} K (Y, K (X, Z)) \end{aligned}

(9)

and

\begin{aligned} R^{(α)} (X, Y) Z & = R^{*} (X, Y) Z - \frac{1 + α}{2} (\overset{*}{\nabla_{X}} K) (Y, Z) + \frac{1 + α}{2} (\overset{*}{\nabla_{Y}} K) (X, Z) \\ + \frac{{(1 + α)}^{2}}{4} K (X, K (Y, Z)) - \frac{{(1 + α)}^{2}}{4} K (Y, K (X, Z)), \end{aligned}

(10)

for all

X, Y, Z \in Γ (T M)

⁠.

Remark 2.

From Theorem 1, we can give other relations:

The relation between $R^{(α)}$ and $R^{(- α)}$ is given by (see [1]).
$R^{(α)} (X, Y) Z = R^{(- α)} (X, Y) Z + α (R (X, Y) Z - R^{*} (X, Y) Z) .$
The relation between $R^{(α)}$ ⁠, R and R* is given by (see [1]).
$\begin{aligned} R^{(α)} (X, Y) Z & = \frac{1 + α}{2} R (X, Y) Z + \frac{1 - α}{2} R^{*} (X, Y) Z \\ - \frac{1 - α^{2}}{4} K (X, K (Y, Z)) + \frac{1 - α^{2}}{4} K (Y, K (X, Z)) \end{aligned}$

Corollary 2.

Let ${\{e_{i}\}}_{i = 1}^{m}$ be a local orthonormal frame field on $(M^{m}, g)$ ⁠, for a statistical structure $(\nabla, \overset{*}{\nabla}, g)$ ⁠, if we denote

R i c c i^{(α)} (X) = T r_{g} R^{(α)} (X, \cdot) \cdot = R^{(α)} (X, e_{i}) e_{i}

and

R i c c i^{(β)} (X) = T r_{g} R^{(β)} (X, \cdot) \cdot = R^{(β)} (X, e_{i}) e_{i},

for any

X \in Γ (T M)

⁠, then the relation between

R i c c i^{(α)} (X)

and

R i c c i^{(β)} (X)

is given by the following formula :

\begin{aligned} R i c c i^{(α)} (X) & = R i c c i^{(β)} (X) + \frac{{(β - α)}^{2}}{4} K (X, E) + \frac{β - α}{2} T r_{g} (\nabla_{X}^{(β)} K) (\cdot, \cdot) \\ - \frac{β - α}{2} T r_{g} (\nabla_{\cdot}^{(β)} K) (X, \cdot) - \frac{{(β - α)}^{2}}{4} T r_{g} K (K (X, \cdot), \cdot), \end{aligned}

where

T r_{g} (\nabla_{X}^{(β)} K) (\cdot, \cdot) = (\nabla_{X}^{(β)} K) (e_{i}, e_{i}),

T r_{g} (\nabla_{\cdot}^{(β)} K) (X, \cdot) = (\nabla_{e_{i}}^{(β)} K) (X, e_{i}),

T r_{g} K (K (X, \cdot), \cdot) = K (K (X, e_{i}), e_{i}),

and

E = T r_{g} K = K (e_{i}, e_{i}) .

In particular for $β \in \{- 1,0,1\}$ ⁠, we obtain,

\begin{aligned} R i c c i^{(α)} (X) & = \hat{R i c c i} (X) + \frac{α^{2}}{4} K (X, E) - \frac{α}{2} T r_{g} ({\hat{\nabla}}_{X} K) (\cdot, \cdot) \\ + \frac{α}{2} T r_{g} ({\hat{\nabla}}_{\cdot} K) (X, \cdot) - \frac{α^{2}}{4} T r_{g} K (K (X, \cdot), \cdot), \end{aligned}

\begin{aligned} R i c c i^{(α)} (X) & = R i c c i (X) + \frac{{(1 - α)}^{2}}{4} K (X, E) + \frac{1 - α}{2} T r_{g} (\nabla_{X} K) (\cdot, \cdot) \\ - \frac{1 - α}{2} T r_{g} (\nabla_{\cdot} K) (X, \cdot) - \frac{{(1 - α)}^{2}}{4} T r_{g} K (K (X, \cdot), \cdot) \end{aligned}

and

\begin{aligned} R i c c i^{(α)} (X) & = R i c c i^{*} (X) + \frac{{(1 + α)}^{2}}{4} K (X, E) - \frac{1 + α}{2} T r_{g} (\overset{*}{\nabla_{X}} K) (\cdot, \cdot) \\ + \frac{1 + α}{2} T r_{g} (\overset{*}{\nabla} \cdot K) (X, \cdot) - \frac{{(1 + α)}^{2}}{4} T r_{g} K (K (X, \cdot), \cdot) \end{aligned}

Example 1.

Let $(R^{2}, g)$ be a statistical manifold with Riemannian metric g = dx² + dy² and ∇ an affine connection defined by

\nabla_{e_{1}} e_{1} = e_{2}, \nabla_{e_{2}} e_{2} = 0, \nabla_{e_{1}} e_{2} = \nabla_{e_{2}} e_{1} = e_{1}

where

\{e_{1} = \frac{\partial}{\partial x}, e_{2} = \frac{\partial}{\partial y}\}

is an orthonormal frame field. A simple calculation gives,

{\overset{*}{\nabla}}_{e_{1}} e_{1} = - e_{2}, {\overset{*}{\nabla}}_{e_{2}} e_{2} = 0, {\overset{*}{\nabla}}_{e_{1}} e_{2} = {\overset{*}{\nabla}}_{e_{2}} e_{1} = - e_{1} .

We deduce that,

K (e_{1}, e_{1}) = - 2 e_{2}, K (e_{2}, e_{2}) = 0, K (e_{1}, e_{2}) = K (e_{2}, e_{1}) = - 2 e_{1},

then,

E = T r_{g} K = K (e_{1}, e_{1}) + K (e_{2}, e_{2}) = - 2 e_{2} .

In this case, we have,

\nabla_{e_{1}}^{(α)} e_{1} = α e_{2}, \nabla_{e_{2}}^{(α)} e_{2} = 0, \nabla_{e_{1}}^{(α)} e_{2} = \nabla_{e_{2}}^{(α)} e_{1} = α e_{1},

R^{(α)} (e_{1}, e_{2}) e_{2} = - α^{2} e_{1}, R^{(α)} (e_{2}, e_{1}) e_{1} = - α^{2} e_{2} .

and

R i c c i^{(α)} (X) = - α^{2} X, R i c^{(α)} (X, Y) = - α^{2} g (X, Y), S_{g}^{(α)} = - 2 α^{2} .

Then $(R^{2}, \nabla^{(α)}, g)$ is a statistical manifold of constant curvature − α² and it is a Hessian structure if and only if α = 0.

Example 2.

Let $(H^{2} = \{(x, y) \in R^{2}, y ≻ 0\}, g)$ be a statistical manifold with Riemannian metric $g = \frac{1}{y^{2}} (d x^{2} + d y^{2})$ and ∇ an affine connection defined by

\nabla_{e_{1}} e_{1} = \nabla_{e_{2}} e_{2} = 2 e_{2}, \nabla_{e_{1}} e_{2} = 0, \nabla_{e_{2}} e_{1} = e_{1},

where

\{e_{1} = y \frac{\partial}{\partial x}, e_{2} = y \frac{\partial}{\partial y}\}

is an orthonormal frame field. A simple calculation gives

{\overset{*}{\nabla}}_{e_{1}} e_{1} = 0, {\overset{*}{\nabla}}_{e_{2}} e_{2} = - 2 e_{2}, {\overset{*}{\nabla}}_{e_{1}} e_{2} = - 2 e_{1}, {\overset{*}{\nabla}}_{e_{2}} e_{1} = - e_{1} .

Then,

\nabla_{e_{1}}^{(α)} e_{1} = (1 + α) e_{2}, \nabla_{e_{2}}^{(α)} e_{2} = 2 α e_{2}, \nabla_{e_{1}}^{(α)} e_{2} = - (1 - α) e_{1}, \nabla_{e_{2}}^{(α)} e_{1} = α e_{1} .

We deduce that,

R^{(α)} (e_{1}, e_{2}) e_{2} = (α^{2} - 1) e_{1}, R^{(α)} (e_{2}, e_{1}) e_{1} = (α^{2} - 1) e_{2},

it follows that,

R i c c i^{(α)} (X) = (α^{2} - 1) X, R i c^{(α)} (X, Y) = (α^{2} - 1) g (X, Y), S_{g}^{(α)} = 2 (α^{2} - 1) .

In this case, $(H^{2}, \nabla^{(α)}, g)$ is a statistical manifold of constant curvature α² − 1 and it is a Hessian structure if and only if α = ±1.

3. The α-conformal equivalence

For a real number α, statistical manifolds $(M^{m}, \nabla, g)$ and $(M^{m}, \bar{\nabla}, \bar{g})$ are said to be α-conformally equivalent if there exists a function γ on M such that the Riemannian metrics $\bar{g}$ and g and h are related by the following relation,

\bar{g} (X, Y) = e^{2 γ} g (X, Y)

(11)

and the connection $\bar{\nabla}$ is given by,

{\bar{\nabla}}_{X} Y = \nabla_{X} Y + (1 - α) Y (γ) X + (1 - α) X (γ) Y - (1 + α) g (X, Y) g r a d γ,

(12)

for all $X, Y, Z \in Γ (T M)$ ⁠. Using the fact that $\nabla_{X} Y = {\hat{\nabla}}_{X} Y - \frac{1}{2} K (X, Y)$ ⁠, we obtain,

{\bar{\nabla}}_{X} Y = {\hat{\nabla}}_{X} Y + (1 - α) Y (γ) X + (1 - α) X (γ) Y - (1 + α) g (X, Y) g r a d γ - \frac{1}{2} K (X, Y) .

(13)

Theorem 2.

\begin{aligned} \bar{R} (X, Y) Z & = \hat{R} (X, Y) Z - (1 + α) g (Y, Z) {\hat{\nabla}}_{X} g r a d γ + (1 + α) g (X, Z) {\hat{\nabla}}_{Y} g r a d γ \\ + {(1 + α)}^{2} g (Y, Z) X (γ) g r a d γ - {(1 + α)}^{2} g (X, Z) Y (γ) g r a d γ + {(1 - α)}^{2} Y (γ) Z (γ) X \\ - (1 - α^{2}) g (Y, Z) {|g r a d γ|}^{2} X - \frac{1 - α}{2} K (Y, Z) (γ) X - (1 - α) g ({\hat{\nabla}}_{Y} g r a d γ, Z) X \\ - {(1 - α)}^{2} X (γ) Z (γ) Y + (1 - α) g ({\hat{\nabla}}_{X} g r a d γ, Z) Y + (1 - α^{2}) g (X, Z) {|g r a d γ|}^{2} Y \\ + \frac{1 - α}{2} K (X, Z) (γ) Y - \frac{1}{2} ({\hat{\nabla}}_{X} K) (Y, Z) + \frac{1}{2} ({\hat{\nabla}}_{Y} K) (X, Z) \\ + \frac{1 + α}{2} g (Y, Z) K (X, g r a d γ) - \frac{1 + α}{2} g (X, Z) K (Y, g r a d γ) \\ + \frac{1}{4} K (X, K (Y, Z)) - \frac{1}{4} K (Y, K (X, Z)) . \end{aligned}

(14)

Proof of Theorem 2. By definition, we have,

\bar{R} (X, Y) Z = {\bar{\nabla}}_{X} {\bar{\nabla}}_{Y} Z - {\bar{\nabla}}_{Y} {\bar{\nabla}}_{X} Z - {\bar{\nabla}}_{[X, Y]} Z .

(15)

We will study the right side of this equation term by term. By (13), we obtain,

{\bar{\nabla}}_{Y} Z = {\hat{\nabla}}_{Y} Z + (1 - α) Z (γ) Y + (1 - α) Y (γ) Z - (1 + α) g (Y, Z) g r a d γ - \frac{1}{2} K (Y, Z),

which gives us,

\begin{aligned} {\bar{\nabla}}_{X} {\bar{\nabla}}_{Y} Z & = {\bar{\nabla}}_{X} {\hat{\nabla}}_{Y} Z + (1 - α) {\bar{\nabla}}_{X} Z (γ) Y + (1 - α) {\bar{\nabla}}_{X} Y (γ) Z \\ - (1 + α) {\bar{\nabla}}_{X} g (Y, Z) g r a d γ - \frac{1}{2} {\bar{\nabla}}_{X} K (Y, Z) . \end{aligned}

Using Eqn (13), we deduce that,

\begin{aligned} {\bar{\nabla}}_{X} {\hat{\nabla}}_{Y} Z & = {\hat{\nabla}}_{X} {\hat{\nabla}}_{Y} Z + (1 - α) ({\hat{\nabla}}_{Y} Z) (γ) X + (1 - α) X (γ) {\hat{\nabla}}_{Y} Z \\ - (1 + α) g (X, {\hat{\nabla}}_{Y} Z) g r a d γ - \frac{1}{2} K (X, {\hat{\nabla}}_{Y} Z), \end{aligned}

{\bar{\nabla}}_{X} Z (γ) Y = Z (γ) {\hat{\nabla}}_{X} Y + (1 - α) Y (γ) Z (γ) X + (1 - α) X (γ) Z (γ) Y - (1 + α) g (X, Y) Z (γ) g r a d γ - \frac{1}{2} Z (γ) K (X, Y) + g ({\hat{\nabla}}_{X} g r a d γ, Z) Y + g (g r a d γ, {\hat{\nabla}}_{X} Z) Y,

{\bar{\nabla}}_{X} Y (γ) Z = Y (γ) {\hat{\nabla}}_{X} Z + (1 - α) Y (γ) Z (γ) X + (1 - α) X (γ) Y (γ) Z - (1 + α) g (X, Z) Y (γ) g r a d γ - \frac{1}{2} Y (γ) K (X, Z) + g ({\hat{\nabla}}_{X} g r a d γ, Y) Z + g (g r a d γ, {\hat{\nabla}}_{X} Y) Z,

\begin{aligned} {\bar{\nabla}}_{X} g (Y, Z) g r a d γ & = g (Y, Z) {\hat{\nabla}}_{X} g r a d γ + (1 - α) g (Y, Z) {|g r a d γ|}^{2} X - 2 α g (Y, Z) X (γ) g r a d γ \\ - \frac{1}{2} g (Y, Z) K (X, g r a d γ) + g ({\hat{\nabla}}_{X} Y, Z) g r a d γ + g (Y, {\hat{\nabla}}_{X} Z) g r a d γ \end{aligned}

and

\begin{aligned} {\bar{\nabla}}_{X} K (Y, Z) & = {\hat{\nabla}}_{X} K (Y, Z) + (1 - α) K (Y, Z) (γ) X + (1 - α) X (γ) K (Y, Z) \\ - (1 + α) g (X, K (Y, Z)) g r a d γ - \frac{1}{2} K (X, K (Y, Z)) . \end{aligned}

It follows that,

{\bar{\nabla}}_{X} {\bar{\nabla}}_{Y} Z = {\hat{\nabla}}_{X} {\hat{\nabla}}_{Y} Z - (1 + α) g (Y, Z) {\hat{\nabla}}_{X} g r a d γ + \frac{1 + α}{2} g (X, K (Y, Z)) g r a d γ - (1 - α^{2}) g (X, Y) Z (γ) g r a d γ - (1 - α^{2}) g (X, Z) Y (γ) g r a d γ + 2 α (1 + α) g (Y, Z) X (γ) g r a d γ - (1 + α) g (X, {\hat{\nabla}}_{Y} Z) g r a d γ - (1 + α) g ({\hat{\nabla}}_{X} Y, Z) g r a d γ - (1 + α) g (Y, {\hat{\nabla}}_{X} Z) g r a d γ - (1 - α^{2}) g (Y, Z) {|g r a d γ|}^{2} X + 2 {(1 - α)}^{2} Y (γ) Z (γ) X + (1 - α) ({\hat{\nabla}}_{Y} Z) (γ) X + {(1 - α)}^{2} X (γ) Z (γ) Y + (1 - α) g ({\hat{\nabla}}_{X} g r a d γ, Z) Y + (1 - α) g (g r a d γ, {\hat{\nabla}}_{X} Z) Y + {(1 - α)}^{2} X (γ) Y (γ) Z + (1 - α) g ({\hat{\nabla}}_{X} g r a d γ, Y) Z + (1 - α) g (g r a d γ, {\hat{\nabla}}_{X} Y) Z + (1 - α) X (γ) {\hat{\nabla}}_{Y} Z - \frac{1}{2} K (X, {\hat{\nabla}}_{Y} Z) + (1 - α) Z (γ) {\hat{\nabla}}_{X} Y + (1 - α) Y (γ) {\hat{\nabla}}_{X} Z - \frac{1 - α}{2} Y (γ) K (X, Z) - \frac{1 - α}{2} Z (γ) K (X, Y) + \frac{1 + α}{2} g (Y, Z) K (X, g r a d γ) - \frac{1}{2} {\hat{\nabla}}_{X} K (Y, Z) - \frac{1 - α}{2} K (Y, Z) (γ) X - \frac{1 - α}{2} X (γ) K (Y, Z) + \frac{1}{4} K (X, K (Y, Z)) .

(16)

A similar calculation gives us,

{\bar{\nabla}}_{Y} {\bar{\nabla}}_{X} Z = {\hat{\nabla}}_{Y} {\hat{\nabla}}_{X} Z - (1 + α) g (X, Z) {\hat{\nabla}}_{Y} g r a d γ + \frac{1 + α}{2} g (Y, K (X, Z)) g r a d γ - (1 - α^{2}) g (X, Y) Z (γ) g r a d γ - (1 - α^{2}) g (Y, Z) X (γ) g r a d γ + 2 α (1 + α) g (X, Z) Y (γ) g r a d γ - (1 + α) g (Y, {\hat{\nabla}}_{X} Z) g r a d γ - (1 + α) g ({\hat{\nabla}}_{Y} X, Z) g r a d γ - (1 + α) g (X, {\hat{\nabla}}_{Y} Z) g r a d γ - (1 - α^{2}) g (X, Z) {|g r a d γ|}^{2} Y + 2 {(1 - α)}^{2} X (γ) Z (γ) Y + (1 - α) ({\hat{\nabla}}_{X} Z) (γ) Y + {(1 - α)}^{2} Y (γ) Z (γ) X + (1 - α) g ({\hat{\nabla}}_{Y} g r a d γ, Z) X + (1 - α) g (g r a d γ, {\hat{\nabla}}_{Y} Z) X + {(1 - α)}^{2} X (γ) Y (γ) Z + (1 - α) g ({\hat{\nabla}}_{X} g r a d γ, Y) Z + (1 - α) g (g r a d γ, {\hat{\nabla}}_{Y} X) Z + (1 - α) Y (γ) {\hat{\nabla}}_{X} Z - \frac{1}{2} K (Y, {\hat{\nabla}}_{X} Z) + (1 - α) Z (γ) {\hat{\nabla}}_{Y} X + (1 - α) X (γ) {\hat{\nabla}}_{Y} Z - \frac{1 - α}{2} X (γ) K (Y, Z) - \frac{1 - α}{2} Z (γ) K (X, Y) + \frac{1 + α}{2} g (X, Z) K (Y, g r a d γ) - \frac{1}{2} {\hat{\nabla}}_{Y} K (X, Z) - \frac{1 - α}{2} K (X, Z) (γ) Y - \frac{1 - α}{2} Y (γ) K (X, Z) + \frac{1}{4} K (Y, K (X, Z)) .

(17)

Finally, it is easy to see that,

{\bar{\nabla}}_{[X, Y]} Z = {\hat{\nabla}}_{[X, Y]} Z - (1 + α) g ({\hat{\nabla}}_{X} Y, Z) g r a d γ + (1 + α) g ({\hat{\nabla}}_{Y} X, Z) g r a d γ + (1 - α) ({\hat{\nabla}}_{X} Y) (γ) Z - (1 - α) ({\hat{\nabla}}_{Y} X) (γ) Z + (1 - α) Z (γ) {\hat{\nabla}}_{X} Y - (1 - α) Z (γ) {\hat{\nabla}}_{Y} X - \frac{1}{2} K ({\hat{\nabla}}_{X} Y, Z) + \frac{1}{2} K ({\hat{\nabla}}_{Y} X, Z) .

(18)

By replacing (16), (17) and (18) in (15), we conclude that,

\bar{R} (X, Y) Z = \hat{R} (X, Y) Z - (1 + α) g (Y, Z) {\hat{\nabla}}_{X} g r a d γ + (1 + α) g (X, Z) {\hat{\nabla}}_{Y} g r a d γ + {(1 + α)}^{2} g (Y, Z) X (γ) g r a d γ - {(1 + α)}^{2} g (X, Z) Y (γ) g r a d γ + {(1 - α)}^{2} Y (γ) Z (γ) X - (1 - α^{2}) g (Y, Z) {|g r a d γ|}^{2} X - \frac{1 - α}{2} K (Y, Z) (γ) X - (1 - α) g ({\hat{\nabla}}_{Y} g r a d γ, Z) X - {(1 - α)}^{2} X (γ) Z (γ) Y + (1 - α) g ({\hat{\nabla}}_{X} g r a d γ, Z) Y + (1 - α^{2}) g (X, Z) {|g r a d γ|}^{2} Y + \frac{1 - α}{2} K (X, Z) (γ) Y - \frac{1}{2} ({\hat{\nabla}}_{X} K) (Y, Z) + \frac{1}{2} ({\hat{\nabla}}_{Y} K) (X, Z) + \frac{1 + α}{2} g (Y, Z) K (X, g r a d γ) - \frac{1 + α}{2} g (X, Z) K (Y, g r a d γ) + \frac{1}{4} K (X, K (Y, Z)) - \frac{1}{4} K (Y, K (X, Z)) .

The same method of calculation used in Theorem 2 and the following equations,

{\hat{\nabla}}_{X} Y = \nabla_{X} Y + \frac{1}{2} K (X, Y),

\begin{aligned} \hat{R} (X, Y) Z & = R (X, Y) Z + \frac{1}{2} (\nabla_{X} K) (Y, Z) - \frac{1}{2} (\nabla_{Y} K) (X, Z) \\ + \frac{1}{4} K (X, K (Y, Z)) - \frac{1}{4} K (Y, K (X, Z)) \end{aligned}

gives us the following theorem

Theorem 3.

\begin{aligned} \bar{R} (X, Y) Z & = R (X, Y) Z - (1 + α) g (Y, Z) \nabla_{X} g r a d γ + (1 + α) g (X, Z) \nabla_{Y} g r a d γ \\ - {(1 + α)}^{2} g (X, Z) Y (γ) g r a d γ + {(1 + α)}^{2} g (Y, Z) X (γ) g r a d γ \\ - (1 - α^{2}) g (Y, Z) {|g r a d γ|}^{2} X + (1 - α^{2}) g (X, Z) {|g r a d γ|}^{2} Y \\ - (1 - α) g (\nabla_{Y} g r a d γ, Z) X + (1 - α) g (\nabla_{X} g r a d γ, Z) Y \\ + {(1 - α)}^{2} Y (γ) Z (γ) X - {(1 - α)}^{2} X (γ) Z (γ) Y \\ - (1 - α) g (K (Y, Z), g r a d γ) X + (1 - α) g (K (X, Z), g r a d γ) Y \end{aligned}

(19)

Corollary 3.

Let us choose ${\{e_{i}\}}_{1 \leq i \leq m}$ to be an orthonormal frame on $(M^{m}, \nabla, g)$ ⁠, an orthonormal frame on $(M^{m}, \bar{\nabla}, \bar{g} = e^{2 γ} g)$ is given by ${\{{\bar{e}}_{i} = e^{- γ} e_{i}\}}_{1 \leq i \leq m}$ ⁠. For any $X, Y \in Γ (T M)$ ⁠, we define

R i c c i (X) = T r_{g} R (X, \cdot) \cdot = R (X, e_{i}) e_{i}, \bar{R i c c i} (X) = T r_{\bar{g}} \bar{R} (X, \cdot) \cdot = \bar{R} (X, \bar{e_{i}}) \bar{e_{i}},

R i c (X, Y) = g (R i c c i (X), Y), \bar{R i c} (X, Y) = \bar{g} (\bar{R i c c i} (X), Y)

and

S_{g} = T r_{g} R i c = R i c (e_{i}, e_{i}), S_{\bar{g}} = T r_{\bar{g}} \bar{R i c} = \bar{R i c} (\bar{e_{i}}, \bar{e_{i}}) .

Using Theorem 3, we obtain the following relations,

\begin{aligned} \bar{R i c c i} (X) & = e^{- 2 γ} R i c c i (X) + ((m - 2) α^{2} + 2 m α + m - 2) e^{- 2 γ} X (γ) g r a d γ \\ - (m α + m - 2) e^{- 2 γ} \nabla_{X} g r a d γ + (m α^{2} - 2 α - m + 2) e^{- 2 γ} {|g r a d γ|}^{2} X \\ - (1 - α) e^{- 2 γ} (\hat{Δ} γ) X - \frac{(1 - α)}{2} e^{- 2 γ} E (γ) X + (1 - α) e^{- 2 γ} K (X, g r a d γ), \end{aligned}

\begin{aligned} \bar{R i c} (X, Y) & = R i c (X, Y) + ((m - 2) α^{2} + 2 m α + m - 2) X (γ) Y (γ) \\ + (m α^{2} - 2 α - m + 2) {|g r a d γ|}^{2} g (X, Y) - (1 - α) (\hat{Δ} γ) g (X, Y) \\ - (m α + m - 2) g (\nabla_{X} g r a d γ, Y) - \frac{(1 - α)}{2} E (γ) g (X, Y) \\ + (1 - α) g (K (X, Y), g r a d γ) \end{aligned}

and

\begin{aligned} S_{\bar{g}} & = e^{- 2 γ} S_{g} + (m - 1) ((m + 2) α^{2} - m + 2) e^{- 2 γ} {|g r a d γ|}^{2} \\ - 2 (m - 1) e^{- 2 γ} (\hat{Δ} γ) + (m - 1) α e^{- 2 γ} E (γ) \end{aligned}

Corollary 4.

Theorem 3 and Corollary 3 gives us two particular cases:

If α = 1, we obtain,
$\begin{aligned} \bar{R} (X, Y) Z & = R (X, Y) Z - 2 g (Y, Z) \nabla_{X} g r a d γ + 2 g (X, Z) \nabla_{Y} g r a d γ \\ - 4 g (X, Z) Y (γ) g r a d γ + 4 g (Y, Z) X (γ) g r a d γ, \end{aligned}$
$\bar{R i c c i} (X) = e^{- 2 γ} (R i c c i (X) + 4 (m - 1) X (γ) g r a d γ - 2 (m - 1) \nabla_{X} g r a d γ),$
$\bar{R i c} (X, Y) = R i c (X, Y) + 4 (m - 1) X (γ) Y (γ) - 2 (m - 1) g (\nabla_{X} g r a d γ, Y)$
and
$S_{\bar{g}} = e^{- 2 γ} (S_{g} + 4 (m - 1) {|g r a d γ|}^{2} - 2 (m - 1) (\hat{Δ} γ) + (m - 1) E (γ)) .$
If α = −1, we obtain,
$\begin{aligned} \bar{R} (X, Y) Z & = R (X, Y) Z - 2 g (\nabla_{Y} g r a d γ, Z) X + 2 g (\nabla_{X} g r a d γ, Z) Y \\ + 4 Y (γ) Z (γ) X - 4 X (γ) Z (γ) Y - 2 g (K (Y, Z), g r a d γ) X \\ + 2 g (K (X, Z), g r a d γ) Y, \end{aligned}$
$\begin{aligned} \bar{R i c c i} (X) & = e^{- 2 γ} R i c c i (X) - 4 e^{- 2 γ} X (γ) g r a d γ + 2 e^{- 2 γ} \nabla_{X} g r a d γ \\ + 4 e^{- 2 γ} {|g r a d γ|}^{2} X - 2 e^{- 2 γ} (\hat{Δ} γ) X - e^{- 2 γ} E (γ) X \\ + 2 e^{- 2 γ} K (X, g r a d γ), \end{aligned}$
$\begin{aligned} \bar{R i c} (X, Y) = R i c (X, Y) - 4 X (γ) Y (γ) + 2 g (\nabla_{X} g r a d γ, Y) \\ + 4 {|g r a d γ|}^{2} g (X, Y) - 2 (\hat{Δ} γ) g (X, Y) \\ - E (γ) g (X, Y) + 2 g (K (X, Y), g r a d γ) \end{aligned}$
and
$S_{\bar{g}} = e^{- 2 γ} S_{g} + (m - 1) e^{- 2 γ} (4 {|g r a d γ|}^{2} - 2 (\hat{Δ} γ) + E (γ)),$
where
$\hat{Δ} γ = g ({\hat{\nabla}}_{e_{i}} g r a d γ, e_{i}) = e_{i} (e_{i} (γ)) - ({\hat{\nabla}}_{e_{i}} e_{i}) (γ) .$

Example 3.

Let $(R^{2}, g)$ be a statistical manifold with Riemannian metric g = dx² + dy² and ∇ an affine connection defined by

\nabla_{e_{1}} e_{1} = e_{2}, \nabla_{e_{2}} e_{2} = 0, \nabla_{e_{1}} e_{2} = \nabla_{e_{2}} e_{1} = e_{1}

where

\{e_{1} = \frac{\partial}{\partial x}, e_{2} = \frac{\partial}{\partial y}\}

is an orthonormal frame field. Then

(R^{2}, \nabla, g)

is a statistical manifold of constant curvature − 1 and S_g = −2. We want to determine γ such that

S_{\bar{g}} = 0

⁠. By Corollary\enleadertwodots , we deduce that

S_{\bar{g}}

vanish if and only if

2 (\hat{Δ} γ) - α E (γ) - 4 α^{2} {|g r a d γ|}^{2} + 2 = 0 .

To solve this equation, we will present two cases :

(1)
If we assume that γ depends only on the variable x, then $S_{\bar{g}}$ vanish if and only if.
$γ^{″} - 2 α^{2} {(γ^{'})}^{2} + 1 = 0 .$

Note that if α = 0, the solution of this last equation is,

γ (x) = - \frac{1}{2} x^{2} + a x + b .

In the case where α ≠ 0, a particular solution is given by $γ (x) = \frac{1}{α \sqrt{2}} x + b$ ⁠.

(2)
If the function γ depends only on the variable y, we conclude that $S_{\bar{g}} = 0$ if and only if,
$γ^{″} + α γ^{'} - 2 α^{2} {(γ^{'})}^{2} + 1 = 0 .$

Using the same method, if α = 0, the solution obtained is,

γ (y) = - \frac{1}{2} y^{2} + a y + b

and if we take α ≠ 0, a particular solution is $γ (y) = \frac{1}{α} y + b$ ⁠.

The authors would like to thank the referee for some useful comments and their helpful suggestions that have improved the quality of this paper.

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Khadidja Addad and Seddik Ouakkas

Published in the Arab Journal of Mathematical Sciences. Published by Emerald Publishing Limited. This article is published under the Creative Commons Attribution (CC BY 4.0) licence. Anyone may reproduce, distribute, translate and create derivative works of this article (for both commercial and non-commercial purposes), subject to full attribution to the original publication and authors. The full terms of this licence may be seen at http://creativecommons.org/licences/by/4.0/legalcode

On the α-connections and the α-conformal equivalence on statistical manifolds

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On the α-connections and the α-conformal equivalence on statistical manifolds