How to solve the following:
Let $f : (M,\nabla)\rightarrow (\overline{M},\overline{\nabla})$ be a diffeomorphism of manifolds with torsion-free connections.
a) For reparametrisation $\alpha$ of geodesic line on M holds $\nabla_{T_\alpha}T_{\alpha}= \rho(t)T_\alpha$, and conversely, if tangent field of curve satisfies this condition, then exits its reparametrisation that is geodesic. Prove. (These curves we can call general geodesics.)
b) Diffeomorfism $f$ is a geodesic map if image of general geodesic on manifold M is general geodesic on $\overline{M}$. By identifying $f$ connected vector-fields $M$ and $\overline{M}$: $P(X,Y)=\overline{\nabla}_{X}Y-\nabla_{X}Y$. Prove that $P$ is symmetric and $F$-bilinear map. Conclude that $P(X,Y)= \psi(Y)X+\psi(X)Y$, where $\psi$ is a form. By using vector-fields of coordinate frame, find $\psi$.
For a torsionless connection we have: $\nabla_{X}Y-\nabla_{Y}X=[X,Y]$. $P$ is a $F$-bilinear map, since it is $F$-linear by first coordinate: \begin{align*}P(X,f_1 Y_1+f_2 Y_2)&=\nabla_{X}(f_1 Y_1+f_2 Y_2)-\overline\nabla_{X}(f_1 Y_1+f_2 Y_2)\\ &=\nabla_{X}f_1 Y_1+\nabla_{X} f_2 Y_2-\overline\nabla_{X}f_1 Y_1-\overline\nabla_{X}f_2 Y_2\\ &=X(f_1)Y_1+f_1\nabla_{X}Y_1+X(f_2)Y_2+f_2\nabla_{X}Y_2\\ &\quad -X(f_1)Y_1-f_1\overline\nabla_{X} Y_1-X(f_2)Y_2-f_2\overline\nabla_{X} Y_2\\ &=f_1(\nabla_{X}Y_1-\overline\nabla_{X}Y_1)+f_2 (\nabla_{X}Y_2-\overline\nabla_{X}Y_2)\\ &=f_1 P(X,Y_1)+f_2 P(X,Y_2),\end{align*} and similarly, it is $F$-linear by second coordinate.
How to deduce what is $\psi$? Also, how to prove a statement for general geodesics?
Detailed explanations are welcome. Thanks in advance.
