Lifting Dual Connections with the Riemann Extension

: Le ( M , g ) be a Riemannian manifold equipped with a pair of dual connections 1 ( ∇ , ∇ ∗ ) .Such a structure is known as a statistical manifold since it was deﬁned in the context 2 of information geometry. This paper aims at deﬁning the complete lift of such a structure to the 3 cotangent bundle T ∗ M using the Riemannian extension of the Levi-Civita connection of M .In the ﬁrst 4 section, common tensors associated with pairs of dual connections, emphasizing the cyclic symmetry 5 property of the so-called skewness tensor. In a second section, the complete lift of this tensor is 6 obtained, allowing the deﬁnition of dual connections on TT ∗ M with respect to the Riemannian 7 extension. 8

Version July 31, 2020 submitted to Mathematics 2 of 13 with: T ijk = E p θ ∂ θ i log p(x, θ)∂ θ j log p(x, θ)∂ θ k log p(x, θ) As a generalization, a smooth Riemannian manifold (M, g) equipped with a pair (∇, ∇ * ) of 20 torsionless dual connections is called a statistical manifold. It can be defined equivalently by (M, g, T) 21 where T is a fully symmetric (0, 3)-tensor. It turns out [5] that any statistical manifold can be embedded 22 as a statistical model, i.e. one related to a parameterized family of densities. 23 For a Riemannian manifold (M, g), lifting geometric objects to the tangent bundle TM (resp. 24 cotangent bundle T * M) is a classical problem [6][7][8]  defined to be the Levi-Civita one with respect to its Riemann extension. Complete and vertical lifts of 33 different kind of tensors are also presented in [6].
The torsion of a connection ∇ is the tensor T defined as: next well known proposition relates the torsion tensors of dual connections.

46
Proposition 1. Let ∇, ∇ * be dual connections. Let T (resp. T * ) be the torsion tensor of ∇ (resp. ∇ * ). Then, Proof. For any triple (X, Y, Z) of vector fields: As a particular, but important case, if the torsion of T vanishes, so does the torsion of T * .

55
In the case of dual connections with vanishing torsion, the commutation defect of the divergence 56 is related to the mutual curvature of the connections.

57
Definition 3. Let (∇ 1 , ∇ 2 ) be a pair of connections. Their mutual curvature is the tensor (1, 3)-tensor: As in the case of the curvature, it is often useful to introduce the (0, 4)-tensor: The curvature and the mutual curvature of dual connections enjoy symmetry properties.

58
Proposition 3. Let (∇, ∇ * ) be a pair of dual connections. Then, for any vector fields X, Y, Z, U; Proof. The proof of the first property is found in, e.g. [4]. For the second, the definition of R ∇∇ * is written as: Using the duality property: Using duality once again: In the case of dual connections without torsion, the definition of D(X, Y) simplifies to ∇ X Y − 60 ∇ * X Y . Letting D X : Y → D(X, y), the next proposition relates the commutation defect to the 61 curvatures.

62
Proposition 4. For any vector fields X, Y, Z: Proof. By simple computation: and the claims follows by identification of the terms.

63
Proposition 5. Let ∇, ∇ * be dual affine connections on TM. Then, for any triple X, Y, Z of vector fields: where ∇ lc is the Levi-Civita connection.

64
Proof. Since the two connections are dual: Using the definition of D ∇,∇ * it comes: Using then an alternating sum over the cyclic permutations of (X, Y, Z) and the Koszul formula: yields the result.

65
Remark 2. Prop. 5 is the analogue of the Kozsul formula for dual connections. It is a defining property 66 given D ∇∇ * .

67
Notation 1. The (0, 3)-tensor: is the skewness tensor associated the connections ∇ 1 , ∇ 2 . When no confusion is possible in the case of dual connections, the subscripts will be dropped so that U(X, Y, Z) stands for U ∇,∇ * (X, Y, Z) 69 Remark 3. The formula of prop. 5 can be rewritten to give the expression of ∇ * : Proposition 6. For any triple (X, Y, Z): where T is the torsion of ∇.
Proof. Using the definition: and the fact that the Levi-Civita has vanishing torsion: and so: Proposition 7. The tensor U has the cyclic symmetry propery, that is for any triple (X, Y, Z) of vector fields: Proof. Using the symmetry of the Riemann metric, the same derivation as in prop. 5 but applied to the terms X(g(Z, Y), Y(g(X, Z), Z(g(Y, X) yields: By identification it comes: Proposition 8. Let U be a tensor with cyclic symmetry, then the connections defined by: are dual.
Proof. For any triple (X, Y, Z) of vector fields: Under the assumption of eq. 10, it comes: and since U has cyclic symmetry: Proposition 9. Let ∇ 1 , ∇ 2 be a pair of affine connections. For any triple (X, Y, Z) of vector fields: Proof. Direct computation from the definition of U.

75
Remark 4. Prop. 9 shows that the mutual torsion of a pair of dual connections is uniquely defined by a 76 cyclic symmetric tensor. Conversely, for a pair ∇ 1 , ∇ 2 of connections, the cyclic symmetry defect of dual. Please note also that the torsion for a pair of dual connections can be seen as the obstruction for 79 the tensor U to be totally symmetric.
Complementary to it, there is an horizontal distribution spanned by the vectors: with: These basis vectors are conveniently put into a matrix form, following the convention of [11]: where Γ is the matrix with entries: Definition 4. The Riemannian extension of a torsion-free affine connection ∇ on TM is the symmetric (0, 2)-tensor with component matrix: where Γ is the matrix defined in 16.
which is equal to: −2Γ + Γ + Γ t Id Id 0 using the assumption that ∇ is torsion-free, Γ t = Γ and the claim follows.

89
Definition 5. The Levi-Civita connection with respect to Riemannian extension, denoted by ∇ c , is 90 called the complete lift of the connection ∇. .

91
Proposition 11. The Christoffel symbols of the complete lift ∇ c are given by: When ∇ = ∇ lc , the torsion-free assumption is automatically satisfied, so that in an adapted frame 92 the Riemannian extension reduces to the one of prop. 10 93 Proposition 12. Let (∇, ∇ * ) be a pair of dual affine connections on TM. Then, with respect to the Riemannian extension ∇ R of ∇ lc , the following relations hold: whereD is the matrix with entries:D ji = p k D k ji and L (resp. L * ) is the component matrix of the adapted frame to ∇ (resp. ∇ * ).

94
Proof. In the case of dual connections, eq. 12 yields: and so: where D t (X, Y) = D(Y, X). From 20 (resp. 21), it comes: When then have: and: The other equations are proved the same way.

95
The above relations show that the horizontal subspaces of ∇ and ∇ * are related by the Riemannian extension in a very simple way. Let X, Y be a vector in T x,p T * M with decomposition X = X V + X H (resp. Y = Y V * + Y H * ) according to the horizontal subspace of ∇ (resp. ∇ * ), then: with ·, · the euclidean inner product.

96
Another interesting fact is that with respect to the adapted frames of ∇ (resp. ∇), the Riemannian  Since duality is related to metric, it is not so obvious how to lift a pair of mutually dual connections 102 in a canonical way since the complete lifts of ∇ and ∇ * involve different Riemannian extensions.

103
The preferred approach will be thus to lift the mutual torsion D to a (0, 3)-tensor, what can be done 104 extending the approach of [6], and to exploit the fact that it has the cyclic symmetry property.

105
In the sequel, the symmetric (resp. anti-symmetric) part with respect to the contravariant indices of the (1, 2)-tensor D will be denoted by s D (resp. a D), i.e.: Proposition 13. The expression: defines a 2-form on TT * M. Its exterior derivative dσ is given by: Rearranging the terms, the form dσ can be rewritten as: It turns out that the above tensor has cyclic symmetry since it is (0, 3) and skew-symmetric. This 106 can made more explicit by first noticing that the first line in the right hand side has obviously this 107 property. In the second line, considering as an example the first term a D k ij dp k ∧ dx i ∧ dx j , a cyclic 108 permutation of the arguments yields a D k ij dx j ∧ dp k ∧ dx i . Now, the indices change j → k, k → i, i → j 109 gives a D i jk dx k ∧ dp i ∧ dx j , which is exactly the original second term. The remaining terms can be 110 worked the same way.

111
Considering now the symmetric part of D, a similar procedure can applied to obtain a fully symmetric (0, 3)-tensor. Let us denote by the symmetric tensor product, that is: . From s D, a symmetric tensor on TT * M can be defined as: ∂x i dx i dx j dx k + s D k ij dp k dx i dx j + s D i jk dx k dp i dx j + s D j ki dx k dx i dp j Gathering things together, both the symmetric and the anti-symmetric part of D can be lifted to a 113 cyclic symmetric (0, 3)-tensor. In the sequel, the notation of [6] is adopted: Latin letters i, j, . . . refer to 114 x components, overlined letters i, j, . . . refers to p components and capital letters can be used for both.

115
As an example, dx i = dp i , δ i = ∂ i .

116
Definition 7. The cyclic symmetric complete lift of the (1, 2)-tensor D, denoted U c , is the (0, 3)-tensor with components u c ABC dx A ⊗ dx B ⊗ dx C : From U c , the complete lift of D can be defined as the (1, 2)-tensor D c such that for any triple of vector fields: Given the matrix form of the Riemannian extension: its inverse is readily obtained as: The components of D c in coordinates can be obtained by composing the matrix A, yielding: with the notation Γ i l = Γ il . Please note that the above relations are different from the one given in [6] 117 for the complete lift of a skew-symmetric (1, 2)-tensor since here the Riemann extension is used in 118 place of the canonical (1, 1)-tensor and only the cyclic symmetry is assumed. This last fact can be 119 noticed in the third and forth lines of eq. (28).

120
The next definitions are recalled for the sake of completeness.

121
Definition 8. Let ω = ω i dx i be a degree 1 differential form. Its vertical lift to TT * M is the vector field: Vector fields admit both a vertical and a complete lift. Only the later will be used here.

122
Definition 9. Let X = X i ∂ i be a vector field on M. Its complete lift to TT * M is the vector field: Finally (1, 1)-tensors can be lifted in a quite obvious way:

123
Definition 10. Let F be a (1, 1)-tensor field. Its vertical lift to TT * M is the vector field: The action of D c on vertical and complete lift can now be obtained.

Proof.
Let ω = ω i dx i , θ = θ j dx j . Then D(ω v , θ V ) = ω i θ j D c A ij = 0. Let X be vector field and X c its complete lift. By linearity: Since D c A ik = 0, the second term in the right hand side vanishes. For the fist one, only D c k ij = D i jk is non-zero, so that: The tensor D X has expression D X (Y) = D k ij X i Y j ∂ k , so that ωD X is the form ωD X = ω k X i D k ij dx j , whose 126 vertical lift is ω k X i D k ij ∂ k .
Putting the expression in eq. (31) yields: Let K be a (1, 1) tensor K. Its Lie derivative can be written [14, p. 32, prop. 35.]: It thus comes: Which can be written in coordinates: Plugging it into eq. (35) finally gives the reduced expression: The pair (∇,∇ * ) defines the complete lift of the original statistical structure to the pseudo-Riemannian 131 manifold (T * M, ∇ R ). When ∇ is without torsion, then D is symmetric. Using eq. (38) and the fact that 132 in such a case D X = D X show that D c is itself symmetric, proving that∇ has vanishing torsion.

133
Conflicts of Interest: 'The author declare no conflict of interest 134