Consider the Rimannian manifold $(M,g)$, for it's tangent bundle $TM$ we can consider a local curve inside a coordiante neiborhoof for example $$\gamma(t) = (p(t) = p_0,t\frac{\partial}{\partial x^i})\subset TM$$
for some $i$.Which is a curve on the fiber.
Then consider $p(t)$ as a constant curve we can take covariant derivative of the vector field $t\frac{\partial}{\partial x^i}$ along the curve:
$$D_t (t\frac{\partial}{\partial x^i}) = \frac{\partial}{\partial x^i} + t D_t(\frac{\partial}{\partial x^i}) = \frac{\partial}{\partial x^i}\tag{1}$$ by the product rule correct. But if we do it alternatively :
$$D_t(t\frac{\partial}{\partial x^i}) = \nabla_{\dot{p}} (t\frac{\partial}{\partial x^i}) = \nabla_0(t\frac{\partial}{\partial x^i}) = 0 \tag{2}$$
From my perspective (1) should be the correct solution (2) is incorrect, but I can't see where goes wrong.
[This question comes from something about the Sasaki metric on the tangent bundle which is defined to be :
$$\langle V, W\rangle_{(p_0, v_0)}=\langle d \pi(V), d \pi(W)\rangle_{p_0}+\left\langle\frac{D v}{d t}(0), \frac{D w}{d s}(0)\right\rangle_{p_0}$$
For $V,W \in T_{(p_0,v_0)}TM$, with the associate curve $\alpha:t \mapsto (p(t),v(t))$ and $\beta:s\mapsto (q(s),w(s))$ where $p(0) = q(0) = p_0, v(0) = w(0) = v_0$. With $\alpha'(0) = V, \beta'(0) = W$,and $\frac{D v}{d t}$ denote the covariant derivative of $v(t)$ along the curve $\alpha$.
Best Answer
Oh I see the problem,
Note that $t\mapsto \partial_{x^i} \in $ is extenable vector field,while $t\mapsto t\partial_{x^i}$ is not a extenable vector field along the curve $p(t) = p_0$.
So If we compute $$D_t(t \partial_{x^i}) \ne \nabla_{\dot{p}}(t\partial_{x^i})$$.
While the Leibniz rule holds for $D_t$ therefore:
$$D_t(t\partial{x^i}) = \partial x^i + tD_t\partial x^i$$ in this step we can apply $$D_t \partial x^i = \nabla_{\dot{p}}\partial x^i = 0$$
$$D_t(t \partial x^i) = \partial x^i$$.
As a direct consequence of this result we can prove :
Proof if part can be checked directly, for the only if part.
For horizontal $V$ (with associative curve $t\mapsto (q(t),w(t))$), it means for every $W \in \ker d\pi$, we have $$\tilde{g} (V,W) = 0$$
In particular consider the vector field $W$ induced by the vertical curve $t\mapsto (p_0,t\partial x^i)$ (easy to check $d\pi (W) = 0$) so:
$$\tilde{g}(V,W) = g(d\pi(V),d\pi (W)) + g(D_t w(0),D_t(t\partial x^i)) = g(d\pi(V),0) + g(D_t w(0),\partial x^i) = 0 $$
Since $i$ can be arbitary so $D_t w(0) = 0$, which means $w$ is parallel to the curve $q(t)$.