# Connections and Projection Maps

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.

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.