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# Some Properties of Diagonal Lifts in Semi-Cotangent Bundles

###### Accepted: 26 Sep 2019
Journal Details
Format
Journal
First Published
01 Jan 2016
Publication timeframe
2 times per year
Languages
English

We analyze some properties of diagonal lift of tensor fields of type (1,1) in semi-cotangent bundles with the help of adapted frames.

#### MSC 2010

Lifts of Vector Fields on a Cross-Section in the Semi-Cotangent Bundle

Defining some structure on the tangent bundles and cotangant bundles to obtain subtle information about the topology and geometry of the manifold is the main way for mathematicians. Due to this feature, many authors have been systematically worked on them [1, 2, 9, 13, 14]. One of these studies is analyzing some properties of diagonal lift of tensor fields of type (1,1) in semi-cotangent bundles with the help of adapted frames

Let Mn be an n-dimensional differentiable manifold of class C and T (Mn) the tangent bundle determined by a natural projection (submersion) π1 : T (Mn) → Mn. We use the notation (xi) = (xα̅,xα), where the indices i, j,... run from 1 to 2n, the indices α,β,... from 1 to n and the indices α̅,β̅ ,... from n + 1 to 2n, xα are coordinates in Mn, xα̅ = yα are fibre coordinates of the tangent bundle T (Mn). If $(xi′)=(xα¯′,xα′)$ ({x^{{i^\prime}}}) = ({x^{{{\overline \alpha }^\prime}}},{x^{{\alpha ^\prime}}}) is another system of local adapted coordinates in the tangent bundle T (Mn), then we have ${xα¯′=∂xα′∂xβyβ,xα′=xα′(xβ).$ \left\{ {\matrix{ {{x^{{{\overline \alpha }^\prime}}} = {{\partial {x^{{\alpha ^\prime}}}} \over {\partial {x^\beta }}}{y^\beta },} \hfill & \cr {{x^{{\alpha ^\prime}}} = {x^{{\alpha ^\prime}}}\left( {{x^\beta }} \right).} \hfill & \cr } } \right.

The Jacobian of (1) has components $(Aji′)=(∂xi′∂xj)=(Aβα′Aβεα′yε0Aβα′),$ (A_j^{{i^\prime}}) = \left( {{{\partial {x^{{i^\prime}}}} \over {\partial {x^j}}}} \right) = \left( {\matrix{ {A_\beta ^{{\alpha ^\prime}}} & {A_{\beta \varepsilon }^{{\alpha ^\prime}}{y^\varepsilon }} \cr 0 & {A_\beta ^{{\alpha ^\prime}}} \cr } } \right), where $Aβα′=∂xα′∂xβ$ A_\beta ^{{\alpha ^\prime}} = {{\partial {x^{{\alpha ^\prime}}}} \over {\partial {x^\beta }}} , $Aβεα′=∂2xα′∂xβ∂xε$ A_{\beta \varepsilon }^{{\alpha ^\prime}} = {{{\partial ^2}{x^{{\alpha ^\prime}}}} \over {\partial {x^\beta }\partial {x^\varepsilon }}} . Let $Tx∗(Mn)(x=π1(x˜),x˜=(xα¯,xα)∈T(Mn))$ T_x^ * ({M_n})(x = {\pi _1}(\widetilde x),\widetilde x = ({x^{\overline \alpha }},{x^\alpha }) \in T\left( {{M_n}} \right)) be the cotangent space at a point x of Mn. If pα are components of $p∈Tx∗(Mn)$ p \in T_x^ * ({M_n}) with respect to the natural coframe {dxα}, i.e. p = pi dxi, then by definition the set t*(Mn) of all points $(xI)=(xα¯,xα,xα¯¯)$ \left( {{x^I}} \right) = ({x^{\overline \alpha }},{x^\alpha },{x^{\overline {\overline \alpha } }}) , $xα¯¯=pα$ {x^{\overline {\overline \alpha } }} = {p_\alpha } ; I,J,... = 1,...,3n with projection π2 : t*(Mn) → T (Mn) (i.e. $π2:(xα¯,xα,xα¯¯)→(xα¯,xα))$ {\pi _2}:({x^{\overline \alpha }},{x^\alpha },{x^{\overline {\overline \alpha } }}) \to ({x^{\overline \alpha }},{x^\alpha })) ) is a semi-cotangent (pull-back [12]) bundle of the cotangent bundle by submersion π1 : T (Mn) → Mn (For definition of the pull-back bundle, see for example [3], [5], [6], [7]). It is remarkable fact that the semi-cotangent (pull-back) bundle has a degenerate symplectic structure [12] $ω=(ωAB)=dp=(00000−δβα0δαβ0).$ \omega = ({\omega _{AB}}) = dp = \left( {\matrix{ 0 & 0 & 0 \cr 0 & 0 & { - \delta _\beta ^\alpha } \cr 0 & {\delta _\alpha ^\beta } & 0 \cr } } \right).

It is clear that the pull-back bundle t*(Mn) of the cotangent bundle T* (Mn) also has the natural bundle structure over Mn, its bundle projection π: t*(Mn) → Mn being defined by $π:(xα¯,xα,xα¯¯)→(xα)$ \pi :({x^{\overline \alpha }},{x^\alpha },{x^{\overline {\overline \alpha } }}) \to ({x^\alpha }) , and hence π = π1π2. Thus (t*(Mn),π1π2) is the composite bundle [ [14], p.9] or step-like bundle [15].

The main purpose of the present paper is to study complete lift of vector fields and tensor fields of type (1,1) from tangent bundle T (Mn) to semi-cotangent (pull-back) bundle (t*(Mn),π2).

We denote by $ℑqp(T(Mn))$ \Im _q^p(T({M_n})) and $ℑqp(Mn)$ \Im _q^p({M_n}) the modules over F (T (Mn)) and F (Mn) of all tensor fields of type (p,q) on T (Mn) and Mn respectively, where F (T (Mn)) and F (Mn) denote the rings of real-valued C−functions on T (Mn) and Mn, respectively.

To a transformation (1) of local coordinates of T (Mn), there corresponds on t*(Mn) the coordinate transformation [10]: ${xα¯′=∂xα′∂xβyβ,xα′=xα′(xβ),xα¯¯′=∂xβ∂xα′pβ.$ \left\{ {\matrix{ {{x^{{{\overline \alpha }^\prime}}} = {{\partial {x^{{\alpha ^\prime}}}} \over {\partial {x^\beta }}}{y^\beta },} \cr {\;{x^{{\alpha ^\prime}}} = {x^{{\alpha ^\prime}}}\left( {{x^\beta }} \right),} \cr {{x^{{{\overline {\overline \alpha } }^\prime}}} = {{\partial {x^\beta }} \over {\partial {x^{{\alpha ^\prime}}}}}{p_\beta }.} \cr } } \right.

The Jacobian of (2) has components [10]: $A¯=(AJI′)=(Aβα′Aβεα′yε00Aβα′00pσAββ′Aβ′α′σAα′β),$ \overline A = (A_J^{{I^\prime}}) = \left( {\matrix{ {A_\beta ^{{\alpha ^\prime}}} & {A_{\beta \varepsilon }^{{\alpha ^\prime}}{y^\varepsilon }} & 0 \cr 0 & {A_\beta ^{{\alpha ^\prime}}} & 0 \cr 0 & {{p_\sigma }A_\beta ^{{\beta ^\prime}}A_{{\beta ^\prime}{\alpha ^\prime}}^\sigma } & {A_{{\alpha ^\prime}}^\beta } \cr } } \right), where $Aβεα′=∂2xα′∂xβ∂xε, Aβ′α′α=∂2xα∂xβ′∂xα′.$ A_{\beta \varepsilon }^{{\alpha ^\prime}} = {{{\partial ^2}{x^{{\alpha ^\prime}}}} \over {\partial {x^\beta }\partial {x^\varepsilon }}},\quad A_{{\beta ^\prime}{\alpha ^\prime}}^\alpha = {{{\partial ^2}{x^\alpha }} \over {\partial {x^{{\beta ^\prime}}}\partial {x^{{\alpha ^\prime}}}}}.

We denote by $ℑqp(T(Mn))$ \Im _q^p(T({M_n})) and $ℑqp(Mn)$ \Im _q^p({M_n}) the modules over F (T (Mn)) and F (Mn) of all tensor fields of type (p,q) on T (Mn) and Mn, respectively, where F (T (Mn)) and F (Mn) denote the rings of real-valued C −functions on T (Mn) and Mn, respectively.

Let θ be a covector field on T (Mn). Then the transformation pθp, θp being the value of θ at pT (Mn), determines a cross-section βθ of semi-cotangent bundle. Thus if σ : MnT* (Mn) is a cross-section of (T* (Mn), π̃,Mn), such that π̃σ = I(Mn), an associated cross-section βθ : T (Mn) → t*(Mn) of semi-cotangent (pull-back) bundle (t*(Mn),π2,T (Mn)) of cotangent bundle by using projection (submersion) of the tangent bundle T (Mn) defined by [ [4], p. 217–218], [ [9], p. 301]: $βθ(xα¯,xα)=(xα¯,xα,σ∘π1(xα¯,xα))=(xα¯,xα,σ(xα))=(xα¯,xα,θα(xβ)).$ {\beta _\theta }\left( {{x^{\overline \alpha }},{x^\alpha }} \right) = \left( {{x^{\overline \alpha }},{x^\alpha },\sigma \circ {\pi _1}\left( {{x^{\overline \alpha }},{x^\alpha }} \right)} \right) = \left( {{x^{\overline \alpha }},{x^\alpha },\sigma \left( {{x^\alpha }} \right)} \right) = \left( {{x^{\overline \alpha }},{x^\alpha },{\theta _\alpha }\left( {{x^\beta }} \right)} \right).

If the covector field θ has the local components θα (xβ), the cross-section βθ (T (Mn)) of t*(Mn) is locally expressed by $xα¯=yα=Vα(xβ), xα=xα, xα¯¯=pα=θα(xβ)$ {x^{\overline \alpha }} = {y^\alpha } = {V^\alpha }\left( {{x^\beta }} \right),\quad {x^\alpha } = {x^\alpha },\quad {x^{\overline {\overline \alpha } }} = {p_\alpha } = {\theta _\alpha }\left( {{x^\beta }} \right) with respect to the coordinates $xA=(xα¯,xα,xα¯¯)$ {x^A} = ({x^{\overline \alpha }},{x^\alpha },{x^{\overline {\overline \alpha } }}) in t* (Mn). xα̅ = yα being considered as parameters. Differentiating (4) by xα̅ = yα, we have vector fields B(β̅) (β̅ = 1,...,n) with components $B(β¯)=∂xA∂xβ¯=∂β¯xA=(∂β¯Vα∂β¯xα∂β¯θα),$ {B_{\left( {\overline \beta } \right)}} = {{\partial {x^A}} \over {\partial {x^{\overline \beta }}}} = {\partial _{\overline \beta }}{x^A} = \left( {\matrix{ {{\partial _{\overline \beta }}{V^\alpha }} \hfill \cr {{\partial _{\overline \beta }}{x^\alpha }} \hfill \cr {{\partial _{\overline \beta }}{\theta _\alpha }} \hfill \cr } } \right), which are tangent to the cross-section βθ (T (Mn)) [10].

Thus B(β̅) have components $B(β¯):(B(β¯)A)=(δβ¯α00)$ {B_{\left( {\overline \beta } \right)}}:\left( {B_{\left( {\overline \beta } \right)}^A} \right) = \left( {\matrix{ {\delta _{\overline \beta }^\alpha } \hfill \cr 0 \hfill \cr 0 \hfill \cr } } \right) with respect to the coordinates $(xα¯,xα,xα¯¯)$ ({x^{\overline \alpha }},{x^\alpha },{x^{\overline {\overline \alpha } }}) in t*(Mn), where $δβ¯α=Aβ¯α=∂xα∂xβ¯.$ \delta _{\overline \beta }^\alpha = A_{\overline \beta }^\alpha = {{\partial {x^\alpha }} \over {\partial {x^{\overline \beta }}}}.

Let $X∈ℑ01(T(Mn))$ X \in \Im _0^1\left( {T({M_n})} \right) , i.e. X = Xαα. We denote by BX the vector field with local components $BX:(B(β¯)AXβ¯)=(δβ¯αXβ¯00)=(Aβ¯αXβ¯00)=(Xα00)$ BX:\left( {B_{\left( {\overline \beta } \right)}^A{X^{\overline \beta }}} \right) = \left( {\matrix{ {\delta _{\overline \beta }^\alpha {X^{\overline \beta }}} \hfill \cr 0 \hfill \cr 0 \hfill \cr } } \right) = \left( {\matrix{ {A_{\overline \beta }^\alpha {X^{\overline \beta }}} \hfill \cr 0 \hfill \cr 0 \hfill \cr } } \right) = \left( {\matrix{ {{X^\alpha }} \hfill \cr 0 \hfill \cr 0 \hfill \cr } } \right) with respect to the coordinates $(xα¯,xα,xα¯¯)$ ({x^{\overline \alpha }},{x^\alpha },{x^{\overline {\overline \alpha } }}) in t*(Mn), which is defined globally along βθ (T (Mn)). Then a mapping $B:ℑ01(T(Mn))→ℑ01(βθ(T(Mn)))$ B:\Im _0^1(T({M_n})) \to \Im _0^1({\beta _\theta }\left( {T({M_n})} \right)) is defined by (5). The mapping B is the differential of βθ : T (Mn) → t* (Mn) and so an isomorphism of $ℑ01(T(Mn))$ \Im _0^1(T({M_n})) onto $ℑ01(βθ(T(Mn)))$ \Im _0^1({\beta _\theta }\left( {T({M_n})} \right)) [10].

Since a cross-section is locally expressed by xα̅ = yα = const., $xα¯¯=pα=const.$ {x^{\overline {\overline \alpha } }} = {p_\alpha } = const. , xα = xα, xα being considered as parameters. Differentiating (4) by xα, we have vector fields C(β) (β = n + 1,...,2n) with components $C(β)=∂xA∂xβ=∂βxA=(∂βVα∂βxα∂βθα),$ {C_{\left( \beta \right)}} = {{\partial {x^A}} \over {\partial {x^\beta }}} = {\partial _\beta }{x^A} = \left( {\matrix{ {{\partial _\beta }{V^\alpha }} \hfill \cr {{\partial _\beta }{x^\alpha }} \hfill \cr {{\partial _\beta }{\theta _\alpha }} \hfill \cr } } \right), which are tangent to the cross-section βθ (T (Mn)).

Thus C(β) have components $C(β):(C(β)A)=(∂βVαδβα∂βθα)$ {C_{\left( \beta \right)}}:\left( {C_{\left( \beta \right)}^A} \right) = \left( {\matrix{ {{\partial _\beta }{V^\alpha }} \hfill \cr {\delta _\beta ^\alpha } \hfill \cr {{\partial _\beta }{\theta _\alpha }} \hfill \cr } } \right) with respect to the coordinates $(xα¯,xα,xα¯¯)$ ({x^{\overline \alpha }},{x^\alpha },{x^{\overline {\overline \alpha } }}) in t* (Mn), where $δβα=Aβα=∂xα∂xβ.$ \delta _\beta ^\alpha = A_\beta ^\alpha = {{\partial {x^\alpha }} \over {\partial {x^\beta }}}.

Let $X∈ℑ01(T(Mn))$ X \in \Im _0^1\left( {T({M_n})} \right) . Then we denote by CX the vector field with local components $CX:(C(β)AXβ)=(Xβ∂βVαXαXβ∂βθα)$ CX:\left( {C_{\left( \beta \right)}^A{X^\beta }} \right) = \left( {\matrix{ {{X^\beta }{\partial _\beta }{V^\alpha }} \hfill \cr {{X^\alpha }} \hfill \cr {{X^\beta }{\partial _\beta }{\theta _\alpha }} \hfill \cr } } \right) with respect to the coordinates $(xα¯,xα,xα¯¯)$ ({x^{\overline \alpha }},{x^\alpha },{x^{\overline {\overline \alpha } }}) in t* (Mn), which is defined globally along βθ (T (Mn)). Then a mapping $C:ℑ01(T(Mn))→ℑ01(βθ(T(Mn)))$ C:\Im _0^1(T({M_n})) \to \Im _0^1({\beta _\theta }\left( {T({M_n})} \right)) is defined by (6). The mapping C is the differential of βθ : T (Mn) → t* (Mn) and so an isomorphism of $ℑ01(T(Mn))$ \Im _0^1(T({M_n})) onto $ℑ01(βθ(T(Mn)))$ \Im _0^1({\beta _\theta }\left( {T({M_n})} \right)) [10].

Now, consider $ω∈ℑ10(Mn)$ \omega \in \Im _1^0({M_n}) and vector field $X∈ℑ01(T(Mn))$ X \in \Im _0^1\left( {T({M_n})} \right) , then vvω (vertical lift), ccX (complete lift) and HHX (horizontal lift) have respectively, components on the semi-cotangent bundle t*(Mn) [11]: $vvω=(00ωα), ccX=(yε∂εXαXα−pσ(∂αXσ)), HHX=(−ΓβαXβXαXβΓβα)$ ^{^{vv}}\omega = \left( {\matrix{ 0 \hfill \cr 0 \hfill \cr {{\omega _\alpha }} \hfill \cr } } \right),{\quad ^{cc}}X = \left( {\matrix{ {{y^\varepsilon }{\partial _\varepsilon }{X^\alpha }} \hfill \cr {{X^\alpha }} \hfill \cr { - {p_\sigma }({\partial _\alpha }{X^\sigma })} \hfill \cr } } \right),{\quad ^{HH}}X = \left( {\matrix{ { - \Gamma _\beta ^\alpha {X^\beta }} \hfill \cr {{X^\alpha }} \hfill \cr {{X^\beta }{\Gamma _\beta }_\alpha } \hfill \cr } } \right) with respect to the coordinates $(xα¯,xα,xα¯¯)$ ({x^{\overline \alpha }},{x^\alpha },{x^{\overline {\overline \alpha } }}) , where $Γβα=VεΓεαβ, Γβα=θεΓβεα.$ \Gamma _\beta ^\alpha = {V^\varepsilon }{{\Gamma_\varepsilon^\alpha}_\beta},\quad {\Gamma _{\beta\, \alpha}} = {\theta _\varepsilon }{{\Gamma_{\beta \alpha}^\varepsilon}}.

On the other hand, the fibre is locally represented by $xα¯=yα=const., xα=const., xα¯=pα=pα,$ {x^{\overline \alpha }} = {y^\alpha } = const.,\quad {x^\alpha } = const.,\quad x\overline {^\alpha } = {p_\alpha } = {p_\alpha }, pα being considered as parameters. Thus, on differentiating with respect to pα, we easily see that the vector fields $E(β¯¯)=vv(dxβ) (β¯¯=2n+1,...,3n)$ {E_{\left( {\overline {\overline \beta } } \right)}}{ = ^{vv}}\left( {d{x^\beta }} \right)\;(\overline {\overline \beta } = 2n + 1,...,3n) with components $E(β¯¯):(E(β¯¯)A)=∂(β¯¯)xA=(∂β¯¯yα∂β¯¯xα∂β¯¯pα)=(00δαβ)$ {E_{\left( {\overline {\overline \beta } } \right)}}:\left( {E_{\left( {\overline {\overline \beta } } \right)}^A} \right) = {\partial _{\left( {\overline {\overline \beta } } \right)}}{x^A} = \left( {\matrix{ {{\partial _{\overline {\overline \beta } }}{y^\alpha }} \hfill \cr {{\partial _{\overline {\overline \beta } }}{x^\alpha }} \hfill \cr {{\partial _{\overline {\overline \beta } }}{p_\alpha }} \hfill \cr } } \right) = \left( {\matrix{ 0 \hfill \cr 0 \hfill \cr {\delta _\alpha ^\beta } \hfill \cr } } \right) is tangent to the fibre, where $δαβ=Aαβ=∂xβ∂xα.$ \delta _\alpha ^\beta = A_\alpha ^\beta = {{\partial {x^\beta }} \over {\partial {x^\alpha }}}.

Let ω be an 1-form with local components ωα on Mn, so that ω is a 1-form with local expression ω = ωαdxα. We denote by the vector field with local components $Eω:(E(β¯¯)Aωβ)=(00ωα),$ E\omega :\left( {E_{\left( {\overline {\overline \beta } } \right)}^A{\omega _\beta }} \right) = \left( {\matrix{ 0 \hfill \cr 0 \hfill \cr {{\omega _\alpha }} \hfill \cr } } \right), which is tangent to the fibre. Then a mapping $E:ℑ10(Mn)→ℑ01(t∗(Mn))$ E:\Im _1^0({M_n}) \to \Im _0^1({t^ * }({M_n})) is defined by (8) and so an isomorphism of $ℑ10(Mn)$ \Im _1^0({M_n}) in to $ℑ01(t∗(Mn))$ \Im _0^1({t^ * }({M_n})) [10].

We consider in π1 (U) 3n local vector fields B(β̅), C(β) and $E(β¯¯)$ {E_{\left( {\overline {\overline \beta } } \right)}} along βθ (T (Mn)), which are respectively represented by $B(β¯)=B∂∂xβ¯, C(β)=C∂∂xβ, E(β¯¯)=Edxβ.$ {B_{\left( {\overline \beta } \right)}} = B{\partial \over {\partial {x^{\overline \beta }}}},\quad {C_{\left( \beta \right)}} = C{\partial \over {\partial {x^\beta }}},\quad {E_{\left( {\overline {\overline \beta } } \right)}} = Ed{x^\beta }.

Theorem 1

Let X be a vector field on T (Mn). We have along βθ (T (Mn)) the formula $ccX=CX+B(LVX)+E( −LXθ),$ ^{cc}X = CX + B\left( {{L_V}X} \right) + E\left( {\; - {L_X}\theta } \right), where LV X denotes the Lie derivative of X with respect to V, and LX θ denotes the Lie derivative of θ with respect to X [10].

On the other hand, on putting $C(β¯¯)=E(β¯¯)$ {C_{\left( {\overline {\overline \beta } } \right)}} = {E_{\left( {\overline {\overline \beta } } \right)}} , we write the adapted frame of βθ (T (Mn)) as ${B(β¯),C(β),C(β¯¯)}$ \left\{ {{B_{\left( {\overline \beta } \right)}},{C_{\left( \beta \right)}},{C_{\left( {\overline {\overline \beta } } \right)}}} \right\} . The adapted frame ${B(β¯),C(β),C(β¯¯)}$ \left\{ {{B_{\left( {\overline \beta } \right)}},{C_{\left( \beta \right)}},{C_{\left( {\overline {\overline \beta } } \right)}}} \right\} of βθ (T (Mn)) is given by the matrix $A˜=(A˜BA)=(δβα∂βVα00δβα00∂βθαδαβ).$ \widetilde A = \left( {\widetilde A_B^A} \right) = \left( {\matrix{ {\delta _\beta ^\alpha } & {{\partial _\beta }{V^\alpha }} & 0 \cr 0 & {\delta _\beta ^\alpha } & 0 \cr 0 & {{\partial _\beta }{\theta _\alpha }} & {\delta _\alpha ^\beta } \cr } } \right).

Since the matrix Ã in (9) is non-singular, it has the inverse. Denoting this inverse by (Ã)−1, we have $(A˜)−1=(A˜CB)−1=(δθβ−∂θVβ00δθβ00−∂θθβδβθ),$ {\left( {\widetilde A} \right)^{ - 1}} = {\left( {\widetilde A_C^B} \right)^{ - 1}} = \left( {\matrix{ {\delta _\theta ^\beta } & { - {\partial _\theta }{V^\beta }} & 0 \cr 0 & {\delta _\theta ^\beta } & 0 \cr 0 & { - {\partial _\theta }{\theta _\beta }} & {\delta _\beta ^\theta } \cr } } \right), where $A˜(A˜)−1=(A˜BA)(A˜CB)−1=δCA=I˜$ \widetilde A{\left( {\widetilde A} \right)^{ - 1}} = (\widetilde A_B^A){\left( {\widetilde A_C^B} \right)^{ - 1}} = \delta _C^A = \widetilde I , where $A=(α¯,α,α¯¯)$ A = \left( {\overline \alpha ,\alpha ,\overline {\overline \alpha } } \right) , $B=(β¯,β,β¯¯)$ B = \left( {\overline \beta ,\beta ,\overline {\overline \beta } } \right) , $C=(θ¯,θ,θ¯¯)$ C = \left( {\overline \theta ,\theta ,\overline {\overline \theta } } \right) .

Then we see from Theorem 1 that the complete lift ccX of a vector field $X∈ℑ01(T(Mn))$ X \in \Im _0^1(T({M_n})) has along βθ (T (Mn)) components of the form $ccX:(LVXαXα−LXθα)$ ^{cc}X:\left( {\matrix{ {{L_V}{X^\alpha }} \hfill \cr {{X^\alpha }} \hfill \cr { - {L_X}{\theta _\alpha }} \hfill \cr } } \right) with respect to the adapted frame ${B(β¯),C(β),C(β¯¯)}$ \left\{ {{B_{\left( {\overline \beta } \right)}},{C_{\left( \beta \right)}},{C_{\left( {\overline {\overline \beta } } \right)}}} \right\} [10].

Theorem 2

The complete lift ccX of a vector field X in Mn to t* (Mn) is tangent to the cross-section βθ (T (Mn)) determined by a 1form θ and vector field V in Mn if and only if $LXθ=0,LVX=0,$ {L_X}\theta = 0,{L_V}X = 0, where LV X denotes the Lie derivative of X with respect to V, and LX θ denotes the Lie derivative of θ with respect to X.

BX, CX and also have components: $BX=(Xα00), CX=(0Xα0), Eω=(00ωα)$ BX = \left( {\matrix{ {{X^\alpha }} \hfill \cr 0 \hfill \cr 0 \hfill \cr } } \right),\quad CX = \left( {\matrix{ 0 \hfill \cr {{X^\alpha }} \hfill \cr 0 \hfill \cr } } \right),\quad E\omega = \left( {\matrix{ 0 \hfill \cr 0 \hfill \cr {{\omega _\alpha }} \hfill \cr } } \right) respectively, with respect to the adapted frame ${B(β¯),C(β),C(β¯¯)}$ \left\{ {{B_{\left( {\overline \beta } \right)}},{C_{\left( \beta \right)}},{C_{\left( {\overline {\overline \beta } } \right)}}} \right\} of the cross-section βθ (T (Mn)) determined by a 1-form θ on T (Mn) [10].

Complete Lift of Tensor Fields of Type (1,1) on a Cross-Section in Semi-Cotangent Bundle

Suppose now that $F∈ℑ11(T(Mn))$ F \in \Im _1^1(T({M_n})) and F has local components $Fβα$ F_\beta ^\alpha in a neighborhood U of Mn, $F=Fβα∂α⊗dxβ$ F = F_\beta ^\alpha {\partial _\alpha } \otimes d{x^\beta } . Then the semi-cotangent (pull-back) bundle t* (Mn) of cotangent bundle T* (Mn) by using projection of the tangent bundle T (Mn) admits the complete lift ccF of F with components [11]: $ccF=( ccFJI)=(Fβαyε∂εFβα00Fβα00pσ(∂βFασ−∂αFβσ)Fαβ),$ ^{cc}F = ({\;^{^{cc}}}F_J^I) = \left( {\matrix{ {F_\beta ^\alpha } & {{y^\varepsilon }{\partial _\varepsilon }F_\beta ^\alpha } & 0 \cr 0 & {F_\beta ^\alpha } & 0 \cr 0 & {{p_\sigma }({\partial _\beta }F_\alpha ^\sigma - {\partial _\alpha }F_\beta ^\sigma )} & {F_\alpha ^\beta } \cr } } \right), with respect to the coordinates $(xα¯,xα,xα¯¯)$ ({x^{\overline \alpha }},{x^\alpha },{x^{\overline {\overline \alpha } }}) on t* (Mn). Then ccF has components $FBA$ F_B^A given by $ccF=(ccFBA)=(FβαLVFβα00Fβα00φFθFαβ)$ ^{cc}F{ = (^{^{cc}}}F_B^A) = \left( {\matrix{ {F_\beta ^\alpha } & {{L_V}F_\beta ^\alpha } & 0 \cr 0 & {F_\beta ^\alpha } & 0 \cr 0 & {{\varphi _F}\theta } & {F_\alpha ^\beta } \cr } } \right) with respect to the adapted frame ${B(β¯),C(β),C(β¯¯)}$ \left\{ {{B_{\left( {\overline \beta } \right)}},{C_{\left( \beta \right)}},{C_{\left( {\overline {\overline \beta } } \right)}}} \right\} of the cross-section βθ (T (Mn)) determined by a 1-form θ in T (Mn), where $A=(α¯,α,α¯¯)$ A = \left( {\overline \alpha ,\alpha ,\overline {\overline \alpha } } \right) , $B=(β¯,β,β¯¯)$ B = \left( {\overline \beta ,\beta ,\overline {\overline \beta } } \right) [10]. Also, the component $ccFβα¯¯$ ^{^{cc}}F_\beta ^{\overline {\overline \alpha } } of $ccFBA$ ^{^{cc}}F_B^A is defined as Tachibana operator φFθ of F, i.e., $ccFβα¯¯=ϕFθ=(∂βFασ−∂αFβσ)θσ−Fβγ∂γθα+Fαγ∂βθγ,$ {\;^{^{cc}}}F_\beta ^{\overline {\overline \alpha } } = {\phi _F}\theta = ({\partial _\beta }F_\alpha ^\sigma - {\partial _\alpha }F_\beta ^\sigma ){\theta _\sigma } - F_\beta ^\gamma {\partial _\gamma }{\theta _\alpha } + F_\alpha ^\gamma {\partial _\beta }{\theta _\gamma }, and $LVFβα$ {L_V}F_\beta ^\alpha denotes the Lie derivative of $Fβα$ F_\beta ^\alpha with respect to V, i.e., $LVFβα=Vγ∂γFβα+Fγα∂βVγ−Fβγ∂γVα.$ {L_V}F_\beta ^\alpha = {V^\gamma }{\partial _\gamma }F_\beta ^\alpha + F_\gamma ^\alpha {\partial _\beta }{V^\gamma } - F_\beta ^\gamma {\partial _\gamma }{V^\alpha }.

Adapted Frames and Diagonal Lifts of Affinor Fields

Let ∇ be a symmetric affine connection in Mn. In each coordinate neighborhood {U,xα} of Mn, we put $X(α)=∂∂xα, θ(α)=dxα.$ {X_{\left( \alpha \right)}} = {\partial \over {\partial {x^\alpha }}},\quad {\theta ^{\left( \alpha \right)}} = d{x^\alpha }.

Then 3n local vector fields Y(α), HHX(α) and vvθ (α) have respectively components of the form $Y(α):(δαβ00), HHX(α):(−ΓβαδαβΓβα), vvθ(α):(00δβα)$ {Y_{\left( \alpha \right)}}:\left( {\matrix{ {\delta _\alpha ^\beta } \hfill \cr 0 \hfill \cr 0 \hfill \cr } } \right),{\quad ^{HH}}{X_{\left( \alpha \right)}}:\left( {\matrix{ { - \Gamma _\beta ^\alpha } \hfill \cr {\delta _\alpha ^\beta } \hfill \cr {{\Gamma _\beta }_\alpha } \hfill \cr } } \right),{\quad ^{vv}}{\theta ^{\left( \alpha \right)}}:\left( {\matrix{ 0 \hfill \cr 0 \hfill \cr {\delta _\beta ^\alpha } \hfill \cr } } \right) with respect to the induced coordinates $(xα¯,xα,xα¯¯)$ ({x^{\overline \alpha }},{x^\alpha },{x^{\overline {\overline \alpha } }}) in π−1 (U), where we have used (7). We call the set {Y(α),HH X(α),vv θ (α)} the frame adapted to the symmetric affine connection ∇ in π−1 (U). On putting $e^(α¯)=Y(α), e^(α)=HHX(α), e^(α¯¯)=vvθ(α)$ {\widehat e_{\left( {\overline \alpha } \right)}} = {Y_{\left( \alpha \right)}},\quad {\widehat e_{\left( \alpha \right)}}{ = ^{HH}}{X_{\left( \alpha \right)}},\quad {\widehat e_{\left( {\overline {\overline \alpha } } \right)}}{ = ^{vv}}{\theta ^{\left( \alpha \right)}} we write the adapted frame as ${e^(B)}={e^(α¯),e^(α),e^(α¯¯)}.$ \left\{ {{{\widehat e}_{\left( B \right)}}} \right\} = \left\{ {{{\widehat e}_{\left( {\overline \alpha } \right)}},{{\widehat e}_{\left( \alpha \right)}},{{\widehat e}_{\left( {\overline {\overline \alpha } } \right)}}} \right\}.

The adapted frame ${e^(B)}={e^(α¯),e^(α),e^(α¯¯)}$ \left\{ {{{\widehat e}_{\left( B \right)}}} \right\} = \left\{ {{{\widehat e}_{\left( {\overline \alpha } \right)}},{{\widehat e}_{\left( \alpha \right)}},{{\widehat e}_{\left( {\overline {\overline \alpha } } \right)}}} \right\} is given by the matrix $A^=(A^BA)=(δβα−Γβα00δβα00Γβαδαβ).$ \widehat A = \left( {\widehat A_B^A} \right) = \left( {\matrix{ {\delta _\beta ^\alpha } & { - \Gamma _\beta ^\alpha } & 0 \cr 0 & {\delta _\beta ^\alpha } & 0 \cr 0 & {{\Gamma _\beta }_\alpha } & {\delta _\alpha ^\beta } \cr } } \right).

Since the matrix $A^$ \widehat A in (17) is non-singular, it has the inverse. Denoting this inverse by $(A^)−1$ {\left( {\widehat A} \right)^{ - 1}} , we have $(A^)−1=(A^CB)−1=(δθβΓθβ00δθβ00−Γθβδβθ),$ {\left( {\widehat A} \right)^{ - 1}} = {\left( {\widehat A_C^B} \right)^{ - 1}} = \left( {\matrix{ {\delta _\theta ^\beta } & {\Gamma _\theta ^\beta } & 0 \cr 0 & {\delta _\theta ^\beta } & 0 \cr 0 & { - {\Gamma _\theta }_\beta } & {\delta _\beta ^\theta } \cr } } \right), where $A^(A^)−1=(A^BA)(A^CB)−1=δCA=I˜$ \widehat A{\left( {\widehat A} \right)^{ - 1}} = (\widehat A_B^A){\left( {\widehat A_C^B} \right)^{ - 1}} = \delta _C^A = \widetilde I , where $A=(α¯,α,α¯¯)$ A = \left( {\overline \alpha ,\alpha ,\overline {\overline \alpha } } \right) , $B=(β¯,β,β¯¯)$ B = \left( {\overline \beta ,\beta ,\overline {\overline \beta } } \right) , $C=(θ¯,θ,θ¯¯)$ C = \left( {\overline \theta ,\theta ,\overline {\overline \theta } } \right) .

If we take account of (16), we see that the diagonal lift DDF of $F∈ℑ11(T(Mn))$ F \in \Im _1^1(T({M_n})) has components [10]: $DDF=(DDFJI)=(−Fβα−ΓεαFβε−ΓβεFεα00Fβα00ΓβσFασ+ΓασFβσ−Fαβ),$ ^{^{DD}}F{ = (^{^{DD}}}F_J^I) = \left( {\matrix{ { - F_\beta ^\alpha } & { - \Gamma _\varepsilon ^\alpha F_\beta ^\varepsilon - \Gamma _\beta ^\varepsilon F_\varepsilon ^\alpha } & 0 \cr 0 & {F_\beta ^\alpha } & 0 \cr 0 & {{\Gamma _\beta }_\sigma F_\alpha ^\sigma + {\Gamma _\alpha }_\sigma F_\beta ^\sigma } & { - F_\alpha ^\beta } \cr } } \right), with respect to the coordinates $(xα¯,xα,xα¯¯)$ ({x^{\overline \alpha }},{x^\alpha },{x^{\overline {\overline \alpha } }}) on t* (Mn), where $Γεα=yγΓγαε, Γασ=pγΓαγσ.$ {\Gamma _\varepsilon ^\alpha} = {y^\gamma }{{\Gamma_\gamma^\alpha}_\varepsilon},\quad {\Gamma _{\alpha\sigma} } = {p_\gamma }{{\Gamma_\alpha^\gamma}_\sigma}. which proves (19).

We now see, from (16), that the diagonal lift DDF of $F∈ℑ11(T(Mn))$ F \in \Im _1^1(T({M_n})) has components of the form $DDF=(DDFBA)=(−Fβα000Fβα000−Fαβ)$ ^{^{DD}}F{ = (^{^{DD}}}F_B^A) = \left( {\matrix{ { - F_\beta ^\alpha } & 0 & 0 \cr 0 & {F_\beta ^\alpha } & 0 \cr 0 & 0 & { - F_\alpha ^\beta } \cr } } \right) with respect to the adapted frame ${e^(B)}$ \left\{ {{{\widehat e}_{\left( B \right)}}} \right\} in t* (Mn).

We now obtain from (19) that the diagonal lift DDF of an affinor field $F∈ℑ11(T(Mn))$ F \in \Im _1^1(T({M_n})) has along βθ (T (Mn)) components of the form [10]: $DDF:(−Fβα−(∇εVα)Fβε−(∇βVε)Fεα00Fβα00−(∇βθσ)Fασ−(∇αθσ)Fβσ−Fαβ),$ ^{^{DD}}F:\left( {\matrix{ { - F_\beta ^\alpha } & { - \left( {{\nabla _\varepsilon }{V^\alpha }} \right)F_\beta ^\varepsilon - \left( {{\nabla _\beta }{V^\varepsilon }} \right)F_\varepsilon ^\alpha } & 0 \cr 0 & {F_\beta ^\alpha } & 0 \cr 0 & { - \left( {{\nabla _\beta }{\theta _\sigma }} \right)F_\alpha ^\sigma - \left( {{\nabla _\alpha }{\theta _\sigma }} \right)F_\beta ^\sigma } & { - F_\alpha ^\beta } \cr } } \right), with respect to the adapted frame ${B(β¯),C(β),C(β¯¯)}$ \left\{ {{B_{\left( {\overline \beta } \right)}},{C_{\left( \beta \right)}},{C_{\left( {\overline {\overline \beta } } \right)}}} \right\} .

Then we see from (7) that the horizontal lift HHX of a vector field $X∈ℑ01(T(Mn))$ X \in \Im _0^1\left( {T({M_n})} \right) has along βθ (T (Mn)) components of the form $HHX:(−Xβ(∇βVα)Xα−(∇βθα)Xβ)$ ^{HH}X:\left( {\matrix{ { - {X^\beta }\left( {{\nabla _\beta }{V^\alpha }} \right)} \hfill \cr {{X^\alpha }} \hfill \cr { - \left( {{\nabla _\beta }{\theta _\alpha }} \right){X^\beta }} \hfill \cr } } \right) with respect to the adapted frame ${B(β¯),C(β),C(β¯¯)}$ \left\{ {{B_{\left( {\overline \beta } \right)}},{C_{\left( \beta \right)}},{C_{\left( {\overline {\overline \beta } } \right)}}} \right\} [10].

Using (7), (20) and (21), we have along βθ (T (Mn)):

Theorem 3

If F and X are affinor and vector fields on T (Mn), and $ω∈ℑ10(Mn)$ \omega \in \Im _1^0({M_n}) , then with respect to a symetric affine connectionin Mn, we have

(i) DDF (HHX) = HH (FX),

(ii) DDF (vvω) = −vv (ωF) [10].

Theorem 4

If $F,G∈ℑ11(Mn)$ F,G \in \Im _1^1({M_n}) , then with respect to a symetric affine connectionin Mn, we have $DDFDDG+DDGDDF=HH(FG+GF).$ ^{DD}{F^{DD}}G{ + ^{DD}}{G^{DD}}F{ = ^{HH}}\left( {FG + GF} \right).

Proof

If $ω∈ℑ10(Mn)$ \omega \in \Im _1^0({M_n}) and $X∈ℑ01(T(Mn))$ X \in \Im _0^1\left( {T({M_n})} \right) , then by Theorem 3 and ([11], HHFHHX =HH (FX)), we have $(DDFDDG+DDGDDF)HHX=DDF(HHGX)+DDG(HHFX) =HH(FGX)+HH(GFX) =HH((FG+GF)X) =HH(FG+GF)HHX.$ \matrix{ {{{{(^{DD}}{F^{DD}}G{ + ^{DD}}{G^{DD}}F)}^{HH}}X ={ ^{DD}}F{(^{HH}}GX){ + ^{DD}}G{(^{HH}}FX)} \hfill \cr {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = {^{HH}}(FGX){ + ^{HH}}(GFX)} \hfill \cr {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = {^{HH}}(\left( {FG + GF} \right)X)} \hfill \cr {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; ={ ^{HH}}{{\left( {FG + GF} \right)}^{HH}}X.} \hfill \cr }

Thus, we have DDFDDG +DD GDDF = HH (FG + GF).

Theorem 5

If $F,G∈ℑ11(Mn)$ F,G \in \Im _1^1({M_n}) , then with respect to a symetric affine connectionin Mn, we have $DDFHHG+DDGHHF=HHFDDG+HHGDDF=DD(FG+GF).$ ^{DD}{F^{HH}}G{ + ^{DD}}{G^{HH}}F{ = ^{HH}}{F^{DD}}G{ + ^{HH}}{G^{DD}}F{ = ^{DD}}\left( {FG + GF} \right).

Proof

If $ω∈ℑ10(Mn)$ \omega \in \Im _1^0({M_n}) and $F,G∈ℑ11(T(Mn))$ F,G \in \Im _1^1(T({M_n})) , then by Theorem 3 and ([11], HHFvvω =vv (ωF) we have $(DDFDDG+DDGDDF)vvω=−DDF(vv(ω∘G))−DDG(vv(ω∘F)) =vv(ω∘GF)−vv(ω∘FG) =vv(ω∘(GF+FG)) =HH(GF+FG)vvω.$ \matrix{ {{{{(^{DD}}{F^{DD}}G{ + ^{DD}}{G^{DD}}F)}^{vv}}\omega = { - ^{DD}}F{(^{vv}}\left( {\omega \circ G} \right)){ - ^{DD}}G{(^{vv}}\left( {\omega \circ F} \right))} \hfill \cr {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = {^{vv}}\left( {\omega \circ GF} \right){ - ^{vv}}\left( {\omega \circ FG} \right)} \hfill \cr {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = {^{vv}}\left( {\omega \circ (GF + FG} \right))} \hfill \cr {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = {^{HH}}{{(GF + FG)}^{vv}}\omega .} \hfill \cr }

Thus, we have Theorem 5.

Putting F = G in Theorem 4 and Theorem 5, we have $HHFDDF=DDFHHF=DD(F2)(DDF)2p=HH(F2p),(DDF)2p+1=DD(F2p+1),(p=1,2,...)$ \matrix{ {^{HH}{F^{DD}}F{ = ^{DD}}{F^{HH}}F{ = ^{DD}}({F^2})} \hfill \cr {{{{(^{DD}}F)}^{2p}}{ = ^{HH}}({F^{2p}}{{),(}^{DD}}F{)^{2p + 1}}{ = ^{DD}}({F^{2p + 1}}),(p = 1,2,...)} \hfill \cr } for any $F∈ℑ11(T(Mn))$ F \in \Im _1^1(T({M_n})) .

Theorem 6

The diagonal lift $J^$ \widehat J of the identity tensor field I of type (1,1) has the components $J^:(−δβα2Γβα00δβα002Γβα−δαβ).$ \widehat J:\left( {\matrix{ { - \delta _\beta ^\alpha } & {2\Gamma _\beta ^\alpha } & 0 \cr 0 & {\delta _\beta ^\alpha } & 0 \cr 0 & {2{\Gamma _\beta }_\alpha } & { - \delta _\alpha ^\beta } \cr } } \right).

Proof

Since the matrix $J^$ \widehat J in (22) is non-singular, it has the inverse. Denoting this inverse by $(J^)−1$ {\left( {\widehat J} \right)^{ - 1}} , we have $(J^)−1=(J^CB)−1=(−δθβ2Γθβ00δθβ002Γθβ−δβθ),$ {\left( {\widehat J} \right)^{ - 1}} = {\left( {\widehat J_C^B} \right)^{ - 1}} = \left( {\matrix{ { - \delta _\theta ^\beta } & {2\Gamma _\theta ^\beta } & 0 \cr 0 & {\delta _\theta ^\beta } & 0 \cr 0 & {2{\Gamma _\theta }_\beta } & { - \delta _\beta ^\theta } \cr } } \right), where $J^(J^)−1=(J^BA)(J^CB)−1=δCA=I˜$ \widehat J{\left( {\widehat J} \right)^{ - 1}} = (\widehat J_B^A){\left( {\widehat J_C^B} \right)^{ - 1}} = \delta _C^A = \widetilde I , where $A=(α¯,α,α¯¯)$ A = \left( {\overline \alpha ,\alpha ,\overline {\overline \alpha } } \right) , $B=(β¯,β,β¯¯)$ B = \left( {\overline \beta ,\beta ,\overline {\overline \beta } } \right) , $C=(θ¯,θ,θ¯¯)$ C = \left( {\overline \theta ,\theta ,\overline {\overline \theta } } \right) . In fact, from (22) and (23), we easily see that $J^(J^)−1=(J^BA)(J^CB)−1 =(−δβα2Γβα00δβα002Γβα−δαβ)(−δθβ2Γθβ00δθβ002Γθβ−δβθ) =(δθα2Γθα−2Γθα00δθα002Γθα−2Γθαδαθ) =(δθα000δθα000δαθ) =δCA =I^.$ \matrix{ {\widehat J{{\left( {\widehat J} \right)}^{ - 1}} = (\widehat J_B^A){{\left( {\widehat J_C^B} \right)}^{ - 1}}} \hfill \cr {\;\;\;\;\;\;\;\;\;\;\;\;\; = \left( {\matrix{ { - \delta _\beta ^\alpha } & {2\Gamma _\beta ^\alpha } & 0 \cr 0 & {\delta _\beta ^\alpha } & 0 \cr 0 & {2{\Gamma _\beta }_\alpha } & { - \delta _\alpha ^\beta } \cr } } \right)\left( {\matrix{ { - \delta _\theta ^\beta } & {2\Gamma _\theta ^\beta } & 0 \cr 0 & {\delta _\theta ^\beta } & 0 \cr 0 & {2{\Gamma _\theta }_\beta } & { - \delta _\beta ^\theta } \cr } } \right)} \hfill \cr {\;\;\;\;\;\;\;\;\;\;\;\;\; = \left( {\matrix{ {\delta _\theta ^\alpha } & {2\Gamma _\theta ^\alpha - 2\Gamma _\theta ^\alpha } & 0 \cr 0 & {\delta _\theta ^\alpha } & 0 \cr 0 & {2{\Gamma _\theta }_\alpha - 2{\Gamma _\theta }_\alpha } & {\delta _\alpha ^\theta } \cr } } \right)} \hfill \cr {\;\;\;\;\;\;\;\;\;\;\;\;\; = \left( {\matrix{ {\delta _\theta ^\alpha } & 0 & 0 \cr 0 & {\delta _\theta ^\alpha } & 0 \cr 0 & 0 & {\delta _\alpha ^\theta } \cr } } \right)} \hfill \cr {\;\;\;\;\;\;\;\;\;\;\;\;\; = \delta _C^A} \hfill \cr {\;\;\;\;\;\;\;\;\;\;\;\;\; = \widehat I.} \hfill \cr }

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