I am trying to gain a better feel for Covariant Differentiation.
I am thinking of the circle , parametrized as $ C(t):=(\cos t, \sin t) ; 0\leq t \leq 2\pi)$.
Is it correct to say that it's derivative vector field $C'(t)$ , given by $ C'(t):=(-\sin t, \cos t)$, is purely
Tangential, in that , for each $t_0$, $(\cos t, \sin t)$ lives in the tangent space at $(\cos t_0, \sin t_0)$?
I know that an informal way to test for it being tangent is that it's perpendicular to the normal
through the segment$(0, t_0)$. Is this rigorous?
Is it accurate too, to say the Normal vector field assigning a normal vector at each point, has
no tangential component, so that it projects to $0$ in the tangent space, and it's covariant derivative is thus $0$? ( The 0 vector).
Can anyone provide an example of a vector field $X$ ; on the circle or otherwise, whose
derivative has both Tangential and Non-Tangential components?
EDIT: I am aware of one way of doing this decomposition: Say the circle is
embedded in $\mathbb R^2 $. Then we do a vector-space decomposition of
$\mathbb R^2$, as $T_p S^1 + N $ , with N normal to the circle, and express tangent vectors accordingly.
EDIT: I realized some careless issues here:
We differentiate tangent vector fields along a curve. (Cost, Sint); at least as I am using it, is not a vector field, but the chosen parametrization of the curve. Since the derivative vector field (-Sint, Cost) is indeed a tangent vector field, it can be differentiated. We find the standard Euclidean derivative , assuming the vectors are vectors in the tangent space to $\mathbb R^3 $ and project the results to $T_p S^1$ at each point.
The normal vector field N(x,y)=(x,y) is not a tangent vector field in the unit circle.
Thank you