3 Fixed error

Let us write the isomorphism

$T_{(x,v)}TM = H_{(x,v)}TM \oplus V_{(x,v)})TM \cong T_xM \oplus T_xM$

by

$\xi \simeq (\xi^h,\xi^v)$,

so that $\xi^h \in T_xM$ and $\xi^v \in T_xM$.

Here the identification $H_{(x,v)}TM \cong T_xM$ is given by the restriction of $d_{(x,v)}\pi$ to $H_{(x,v)}TM$ (where $\pi:TM \rightarrow M$ is the projection), and the isomorphism $V_{(x,v)}TM \cong T_xM$ is canonical.

The key point is that under these identifications, if $z$ is a curve on $TM$, say $z(t)=(\gamma(t),u(t))$ then

$\dot{z}(0) \simeq (\dot{\gamma}(0),(\nabla_tu)(0))$.

So suppose $x \in M$ and $v,w,y \in T_xM$. Let $\gamma$ be a curve in $M$ such that $\gamma(0)=x$ and $\dot{\gamma}(0)=w$, and let $u$ be a vector field along $\gamma$ such that $u(0)=v$ and $(\nabla_tu)(0)=y$. Let $z(t)=(\gamma(t),u(t))$.

Think of $A$ as a map $TM \rightarrow TM$, so that the differential $dA$ is a map

$d_{(x,v)}A:T_{(x,v)}TM \rightarrow T_{(x,v)}TM$.

Then given $w\in T_xM$, if $\xi :\xi_w$ is the unique vector whose horizontal component is $w$ and whose vertical component is zero (i.e. $\xi^h = \dot{z}(0)$ w$and$\xi^v = 0$), then we define$(\nabla_xA)(x,v)(w):=(d_{(x,v)}A(\xi))^h$,(\nabla_xA)(x,v)(w):=(d_{(x,v)}A(\xi_w))^v$,

and similarly if $(\nabla_vA)(x,v)(y):=(d_{(x,v)}A(\xi))^v$,

that \zeta_w$is the unique vector whose horizontal component is zero and whose vertical component is$w$(i.e.$\zeta^h = 0$and$\zeta^v = w$), then we define$(\nabla_vA)(x,v)(y):=(d_{(x,v)}A(\zeta_w))^v$. Then it follows that$d_{(x,v)}A(\xi) \simeq ((\nabla_xA)(x,v)(w),(\nabla_vA)(x,v)(y))$, and these two maps have the properties you're looking for. Post Deleted by Will Merry 2 added 253 characters in body Let us write the isomorphism$T_{(x,v)}TM = H_{(x,v)}TM \oplus V_{(x,v)})TM \cong T_xM \oplus T_xM$by$\xi \simeq (\xi^h,\xi^v)$, so that$\xi^h \in T_xM$and$\xi^v \in T_xM$. Here the identification$H_{(x,v)}TM \cong T_xM$is given by the restriction of$d_{(x,v)}\pi$to$H_{(x,v)}TM$(where$\pi:TM \rightarrow M$is the projection), and the isomorphism$V_{(x,v)}TM \cong T_xM$is canonical. The key point is that under these identifications, if$z$is a curve on$TM$, say$z(t)=(\gamma(t),u(t))$then$\dot{z}(0) \simeq (\dot{\gamma}(0),(\nabla_tu)(0))$. So suppose$x \in M$and$v,w,y \in T_xM$. Let$\gamma$be a curve in$M$such that$\gamma(0)=x$and$\dot{\gamma}(0)=w$, and let$u$be a vector field along$\gamma$such that$u(0)=v$and$(\nabla_tu)(0)=y$. Let$z(t)=(\gamma(t),u(t))$. Think of$A$as a map$TM \rightarrow TM$, so that the differential$dA$is a map$d_{(x,v)}A:T_{(x,v)}TM \rightarrow T_{(x,v)}TM$. Then if$\xi := \dot{z}(0)$then we define$(\nabla_xA)(x,v)(w):=(d_{(x,v)}A(\xi))^h$,$(\nabla_vA)(x,v)(y):=(d_{(x,v)}A(\xi))^v$, that is,$d_{(x,v)}A(\xi) \simeq ((\nabla_xA)(x,v)(w),(\nabla_vA)(x,v)(y))\$,

and these two maps have the properties you're looking for.

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