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Let $M$ be a manifold and let $\pi: TM\rightarrow M$ be the projection map. Taking the pushforward of $\pi$ we obtain a bundle map $$ \pi_\ast: T(TM)\rightarrow TM. $$ Let $$ V=\ker \pi_\ast $$ be the vertical subbundle of $T(TM)$.

Let $\nabla$ be a covariant derivative on $M$. How does $\nabla$ define a smooth bundle map $K: T(TM)\rightarrow T(TM)$ such that $K(T(TM))=V$ and $K^2=K$? Also, is it true that there is a one to one correspondence between covariant derivatives on $M$ and smooth bundle maps $K:T(TM)\rightarrow T(TM)$ satisfying $K^2=K$ and $K(T(TM))=V$?

Of course, the bundle map $K$ with the above properties defines a decomposition $$ T(TM)=H\oplus V $$ where $H:=\ker K$ is the so-called horizontal distribution.

I have seen similar questions on Stack Exchange but the constructions given in the comments or answers either do not seem well defined or they do not appear to work.

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Let $M$ be a manifold and $\nabla$ a connection on $M$. Let $D/dt$ denote the induced covariant derivative for vector fields on $M$ defined along curves. Let $\sigma: (a,b)\rightarrow M$ be a (smooth) curve and let $X:(a,b)\rightarrow TM$ be a vector field along $c$. In other words, if $\pi: TM\rightarrow M$ is the projection, then $$ c(t)=\pi(X(t)) $$ or $X(t)\in T_{c(t)}M$. Let $t_0\in (a,b)$ and choose a coordinate neihborhood $(U,x_i)$ of $p=c(t_0)$. Let $\varepsilon>0$ be such that $(t_0-\varepsilon, t_0+\varepsilon)\subset (a,b)$ and $$ c(t_0-\varepsilon, t_0+\varepsilon)\subset U. $$ Write $c(t)$ and $X(t)$ for $t\in (t_0-\varepsilon, t_0+\varepsilon)$ in local coordinates: $$ c(t)=(c^i(t)),\hspace{0.1in} X(t)=X^i(t)\frac{\partial}{\partial x^i}\big|_{c(t)} $$ where Einstein notation is employed in the expression for $X(t)$. The covariant derivative of $X(t)$ is then $$ \frac{DX}{dt}(t)=\left(\frac{dX^k}{dt}+\frac{dc^i}{dt}X^j\Gamma^k_{ij}\right)\frac{\partial}{\partial x^k}\big|_{c(t)}\in T_{c(t)}M $$ where $\Gamma^k_{ij}$ are the Chrisoffel symbols of $\nabla$.

The idea now is to use $D/dt$ to define a map $K: T(TM)\rightarrow T(TM)$ with the desired properties.

For convenience, we will denote the tangent vector $v\in T_p M$ as $(p,v)$. Note that $$ V_{(p,v)} = T_{(p,v)}(T_pM) \simeq T_p M $$ and the isomorphism between $V_{(p,v)}$ and $T_pM$ is canonical because $T_pM$ is a vector space. Specifically, given a tangent vector $Y\in T_pM$, we associate it to the vector which in $Y_v\in V_{(p,v)}$ which is specified by the curve $$ \alpha(t)=(p,v+tY)\in T_pM\subset TM. $$ In other words, $$ \alpha(0)=(p,v),\hspace{0.1in} \dot{\alpha}(0)=Y_v\in V_{(p,v)}. $$ We can express $Y_v$ in terms of local coordinates as follows. Writing $$ Y=Y^i\frac{\partial}{\partial x^i}\big|_p, $$ let $(TU,x^i,y^i)$ be the coordinate system induced by $(U,x^i)$. Then one sees that $$ Y_v = Y^i\frac{\partial}{\partial y^i}\big|_{(p,v)} $$ The map $K: TTM\rightarrow TTM$ is then defined as follows. For $\xi\in T_{(p,v)}TM$, let $X:(-\varepsilon,\varepsilon)$ be any curve such that $$ X(0)=(p,v),\hspace{0.1in}\dot{X}(0)=\xi. $$ We then define $$ K\xi = \left(\frac{DX}{dt}(0)\right)_v\in V_{(p,v)}. $$ This definition is independent of the choice of curve $X(t)$. This is clear by expressing $K\xi$ in local coordinates. Write $$ \xi = \xi^i_{(1)} \frac{\partial}{\partial x^i}\big|_{(p,v)}+\xi^i_{(2)} \frac{\partial}{\partial y^i}\big|_{(p,v)} $$ Let $c_X(t)=\pi(X(t))$ and write $$ X(t) = X^i(t)\frac{\partial}{\partial x^i}\big|_{c(t)},\hspace{0.2in} v=v^i\frac{\partial}{\partial x^i}\big|_{p} $$ Since $X(0)=(p,v)$ and $\dot{X}(0)=\xi$, we have $$ X^i(0)=v^i,\hspace{0.1in} \frac{dc_X^i}{dt}(0)=\xi^i_{(1)}, \hspace{0.1in}\frac{dX^i}{dt}(0)=\xi^i_{(2)} $$ So $$ \begin{align} \frac{DX}{dt}(0)&=\left(\frac{dX^k}{dt}(0)+\frac{dc^i}{dt}(0)X^j(0)\Gamma^k_{ij}(p)\right)\frac{\partial}{\partial x^k}\big|_{p}\\ &=\left(\xi^k_{(2)}+\xi^i_{(1)}v^j\Gamma^k_{ij}(p)\right)\frac{\partial}{\partial x^k}\big|_{p} \end{align} $$ $K\xi$ is then $$ K\xi=\left(\frac{DX}{dt}(0)\right)_v=\left(\xi^k_{(2)}+\xi^i_{(1)}v^j\Gamma^k_{ij}(p)\right)\frac{\partial}{\partial y^k}\big|_{(p,v)} $$ The definition of $K\xi$ in local coordinates makes it clear that $K$ is linear on the fibers of $TTM$ and that its smooth. Also, note that if $\xi\in V_{(p,v)}$, then the components $\xi^i_{(1)}=0$ and $$ K\xi=\xi^k_{(2)}\frac{\partial}{\partial y^k}\big|_{(p,v)}=\xi. $$ By construction, the image of $K$ is all of $V$. Also, since $K$ is the identity on $V$, we also have $K^2=K$. $K$ is then the desired smooth bundle map.

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