Given a closed Riemannian manifold $(X,g)$ and let $p\colon TX\to X$ be the usual projection, the paper I'm reading asserts that the Levi-Civita connection induces a splitting $T(TX)= H(TX)\oplus V(TX)\cong p^\* TX\oplus p^\*TX$, so that
$$
\begin{aligned}\Omega^\bullet(TX) &= \{ \Gamma(H(TX))\to C^\infty(TX)\}\otimes\_{C^\infty(TX)} \{\Gamma(V(TX))\to C^\infty(TX) \} \\
&\cong p^\*\Omega^\bullet (X)\otimes\_{C^\infty(TX)} p^\*\Omega^\bullet(X).
\end{aligned}
 $$


I want to investigate this isomorphism explicitly, in the sense that I can compute the images of the local coordinate functions on $X$ under the two maps


$$ \phi\_1\colon C^\infty(X)\xrightarrow{d}\Omega^1(X)\xrightarrow{p^\*} p^\*\Omega^\bullet(X) \cong \{\Gamma(V(TX))\to C^\infty(TX) \} \hookrightarrow \Omega^\bullet(TX),$$


$$\phi\_2\colon C^\infty(X)\xrightarrow{d}\Omega^1(X)\xrightarrow{p^\*} p^\*\Omega^\bullet(X) \cong \{\Gamma(H(TX))\to C^\infty(TX) \} \hookrightarrow \Omega^\bullet(TX). $$


But I failed finding any material that could tell me how the horizontal vector bundle is splitted out using the Levi-Civita connection, without which I cannot do any computation.


Also, I found that $V(TX) = \ker ( d p\colon T(TX)\to TX)$ on Wikipedia, but I'm not quite sure how the isomorphism $V(TX)\cong p^\*TX$ is given. The isomorphism that I came up with is the following:



> 
> Let $x\_1,\cdots,x\_n$ be the coordinate functions of a coordinate
> neighborhood $U$ of $X$, then $x\_1,\cdots,x\_n,y\_1:=d x\_1,\cdots, y\_n:=dx\_n$
> are the coordinates of $p^{-1}(U)\cong U\times\mathbb{R}^n$
> of $TX$. Then the fibres of $V(TX)|\_{p^{-1}(U)}$ are exactly the spans of $\frac{\partial}{\partial y\_i}$'s. Since the fibres of $p^\*TX|\_{p^{-1}(U)}$ are the spans of $\frac{\partial}{\partial x\_i}$'s, the isomorphism is given by $\frac{\partial}{\partial x\_i}\mapsto \frac{\partial}{\partial y\_i} $ along each fibre.
> 
> 
> 


So my questions are:


1. How the horizontal vector bundle is constructed using the Levi-Civita connection?
2. Is the above isomorphism the correct one?
3. What are the images of the local coordinate functions on $X$ under the maps $\phi\_1$ and $\phi\_2$ given above? (Probably I will be able to do this by myself with (1) and (2) solved, but I'm putting it here anyway, since it is the ultimate goal of mine)


To be honest, I'm not familiar to Riemannian geometry, so it would be the best if there exists a self-contained material that could provide a throughout treatment to this matter.


Thanks in advance for any help.




---


EDIT: I thought it over and a new question (4) arises: See, a choice of coordinates $x\_1,\cdots,x\_n$ of a neighborhood $U$ of $X$ induces coordinates $x\_1,\cdots,x\_n,y\_1:=d x\_1,\cdots, y\_n:=dx\_n$ of $p^{-1}(U)$ of $TX$, which gives a basis $\frac{\partial }{\partial x\_1},\cdots,\frac{\partial }{\partial x\_n},\frac{\partial }{\partial y\_1},\cdots,\frac{\partial }{\partial y\_n}$ of $T(TX)|\_{p^{-1}(U)}$. While the span of $\frac{\partial }{\partial y\_i}$’s gives $V(TX)| \_{p^{-1}(U)} $, why isn’t $H(TX)|\_{p^{-1}(U)}$ simply the span of $\frac{\partial}{\partial x\_i}$’s? I don’t see any trouble gluing these choices of $H(TX)|\_{p^{-1}(U)}$‘s up as $U$ varies, which seems to have no difference with $p^\*TX$. Am I missing something in this construction, or is there any other reason to define $H(TX)$ differently? This approach looks canonical — not even a Riemannian metric is required.



Explicit computation of the vertical and horizontal vector bundles
I am just answering to (1) and (2): The horizontal bundle is defined as follows. Consider a point $v\in T\_pX\subset TX,$ and $w\in T\_pm$. Consider a curve $\gamma$ through $p$ with $\gamma'(0)=w.$ Consider the parallel transport $V$ through $v$ along $\gamma,$ i.e. the curve $V\colon (-\epsilon, \epsilon)\to TX$ satisfying $$\nabla\_\gamma' V=0\quad\text{and}\quad V(0)=v.$$ Then, $V'(0)\in T\_vTX$ is per Definition horizontal, i.e.
$$V'(0)\in H\_v(TX).$$ Of course, one has to show that this gives a well-defined bundle, which is isomorphic to $p^\*TX.$ The later isomorphism is given by $$w\mapsto V'(0),$$ with inverse $$V'(0)\mapsto d\_vp(V'(0)).$$


(2) seems correct. There is the following invariant way describing the isomorphism: Take $v\in T\_pX\subset TX,$ and $w\in T\_pX.$ Then, $$t\in(-\epsilon,\epsilon)\mapsto v+t w\in T\_pX$$ is a vertical curve $\gamma\_w$. The isomorphism then is $$w\in T\_{p(v)} X\mapsto \gamma\_w'(0)\in V\_vV.$$


