Can someone explain me the following statement about the covariant derivatives? Let $S$ be an n-dimensional manifold and $M$ be an m-dimensional submanifold of $S$.Let $[\xi^i]$ and $[u^a]$ be coordinate system for $S$ and $M$ respectively,and let $X$ and $Y$ be two vector fields on $M$,we may write $\nabla_{X_p}Y$,"the directional derivative of $Y$ along $X_p$",and $\nabla_{X_p}Y$ is a  tangent vector of $S$ but is not necessarily a tangent vector of $M$.Why it is not a tangent vector in $M$?
 A: Let $S=\Bbb R^2$ endowed with the usual euclidean metric, $M=\Bbb S^1$.
Recall that for $p\in \Bbb S^1$, $T_p\Bbb S^1 = \{p\}^{\perp}$.
The Levi-Civita connection is just the directional derivative of the coefficients:
$$
\nabla_X(f \partial _x + g \partial_y) = (Xf)\partial_x + (Xg)\partial_y.
$$
Consider $X(x,y)=Y(x,y)=y\partial_x-x\partial_y$.
Then:

*

*$X$ and $Y$ are tangent to $\Bbb S^1$, but

*$\nabla_XY$ isn't: indeed
\begin{align}
\nabla_XY &= \nabla_{y\partial_x - x\partial_y}(y\partial_x - x\partial_y)\\
&= y \nabla_{\partial_x}(y\partial_x - x\partial_y) - x \nabla_{\partial_y}(y\partial_x - x\partial_y)\\
&= -x\partial_x - y\partial_y, 
\end{align}
so that for $p\in \Bbb S^1$, $\left(\nabla_XY\right)(p) \perp T_p\Bbb S^1$.

Note: this example does not come out of nowhere.
The integral curves of these vector fields are concentric circles.
They can be interpreted as the velocity vector $\vec{v}$ of a particle that would move circularly with constant speed (given by $\sqrt{x^2+y^2}$).
It turns out that $\nabla_{\vec{v}}{\vec{v}}$ is the acceleration of such a particle, and it is physically clear that it is directed toward the origin, and thus cannot be tangent to the trajectory.
