Question about covariant derivate on a manifold

Question

Let $S$ be an n-dimensional manifold and $M$ be an m-dimensional submanifold of $S$. Let $[\xi^i]$ and $[u^a]$ be coordinate systems for $S$ and $M$, respectively, and let $\partial_i=\frac{\partial}{\partial \xi^i}$ and $\partial_a=\frac{\partial}{\partial u^a}$ Suppose also that $\nabla$ is an affine connection on $S$ and that $\Gamma_{ij}^{k}$ are the connection coefficients of $\nabla$ with respect to $[\xi^i]$ Now letting $X=X^a\partial_a$ and $Y=Y^a\partial_a$ be two vector fields in $M$ then we define $\nabla_{X_p}Y$ "the directional derivative of $Y$ along $X_p$" ,In general $\nabla_{X_p}Y$ is a tangent vector in $S$ but not in $M$.
Can someone give an example, I am finding it difficult to visualize why $\nabla_{X_p}Y$ is not tangent vector in a submanifold $M$.Thanks

Answer 1

Consider a smooth curve $ u:(-1,1) \longrightarrow \mathbb{R}$ with $u_x, u_{xx} \neq0 $ on its domain. Then, $X= \frac{\partial}{\partial x} + u_x \frac{\partial}{\partial y}$ is tangent to the curve at every point. We have that $D_XX= u_{xx}\frac{\partial}{\partial y}$. Since $N =(u_x\frac{\partial}{\partial x}-\frac{\partial}{\partial y})$ is perpendicular to the curve at every point and $$\langle D_XX,N \rangle= -u_{xx} \neq 0,$$ the normal component of $D_XX$ is not zero. This means $D_XX$ is not in the tangent space of the curve at any point.

Note that because $u_x$ is the slope of the tangent line on curve at every point, $y=u_x x $ is parallel to the tangent line, in turn, $(1,u_x) \in T_xM$. (One can also consider parametrization $ X :(-1,1) \longrightarrow \mathbb{R}^2$ by $X(x) =(x,u(x))$. Therefore $ d X (\frac{\partial}{\partial x}) = (1, u_x)$ is a basis for $T_xM$). And since $\langle N,X \rangle =0$, $N\perp T_xM$.