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Does anyone know definition of Levi-Civita connection map that defined as $K: TTM\to TM$. and how to prove the following theorem:

Theorem: If $X\in\mathfrak{X}(M)$ be a vector field over $M$ and $K:TTM\to TM$ Levi-Civita connection map then $$K\circ dX=\nabla X$$ where $dX$ is derivation of vector field $X$ as a map.

Thanks.

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After the comment of Jez, here is the corrected answer:

There is the well-defined vertical bundle $VTM\subset TTM\to TM$ given as the kernel of the (differential of the) projection. Moreover, for any $v\in TM,$ there is a natural identification $\Phi$ of $V_vTM$ with $T_\pi(v)M.$

Using the Levi-Civita connection (or ans other connection), you can construct a horizontal bundle $HTM\subset TTM\to TM$ with the properties: (1) $$HTM\oplus VTM=TTM,$$ $d\pi\colon H_vTM\to T_{\pi(v)}M$ is an isomorphism for any $v\in TM$ and (3) for any vector fields X and Y we have $$\Phi^{-1} \nabla_XY= \pi^{VTM} dY(X),$$ where $\pi^{VTM}$ is the projection to $VTM$ corresponding to the splitting (1). The proof is the same as the proof of existence of the geodesic spray. From these properties, you obtain that $K=\Phi\circ \pi^{VTM}$ is satisfying your equation.

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    $\begingroup$ Isn't $\mathrm{d}\pi \circ \pi^{HTM}$ just $\mathrm{d}\pi$? I think instead you should project onto $VTM$ along $HTM$, then use the identification of $VT_vM$ with $T_{\pi(v)}M$. $\endgroup$
    – user81684
    Commented May 18, 2016 at 9:18
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    $\begingroup$ Dear Jez, you are right, I was mixing up things. $\endgroup$
    – Sebastian
    Commented May 18, 2016 at 9:21
  • $\begingroup$ @Sebastian Thanks for your answer. but you don't defined the connection map and what is it properties. $\endgroup$
    – C.F.G
    Commented May 18, 2016 at 10:01
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    $\begingroup$ Dear C.F.G., the connection map is given by the projection onto the vertical bundle composed with its identification with the (pull-back of the) tangent bundle of $M.$ The idea is that one does not need to know the horizontal part of the derivative of a section (as it is basically the identity), so only the vertical part maters. In order to compute the vertical part, one needs a complementary bundle, the horizontal one. A vector $\hat X$ in TTM lies in $HTM$ if and only if it is the germ of a parallel section. $\endgroup$
    – Sebastian
    Commented May 19, 2016 at 7:49

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