Why does null covariant derivative not imply null second covariant derivative? I am troubled by this. 
How come when a vector $V$ is parallel transported around a loop (say a parallelogram of vectors $a$ and $b$), its second covariant derivatives $\triangledown^{2}_{a,b}V$ and $\triangledown^{2}_{b,a}V$ aren't zero, since by definition $\triangledown^{}_aV=0$ and $\triangledown^{}_bV=0$ ?
And again, if we take $a$ and $b$ to be $\partial x^i$ and $\partial x^j$, and parallel transport $V$ in those directions, we also have $\triangledown^{}_{[a,b]}=0$, so what's left for the Riemann tensor ?
Assuming $\triangledown$ is the Levi-Civita connexion.
 A: When you're parallel transporting a vector around a parallelogram, what you end up with is a vector field $V$ along the parallelogram that satisfies $\nabla_a V = 0$ only along the sides of the parallelogram in the $a$-drection, and $\nabla_b V = 0$ only along the sides in the $b$-direction. To compute $\nabla^2_{a,b}V$, you'd have to differentiate $\nabla_b V$ in the $a$-direction; but there's no guarantee that $\nabla_b V=0$ anywhere except along the $b$-sides.  
Think about the simplest case of a unit square in $\mathbb R^2$, with sides $[0,1]\times \{0\}$, $\{1\}\times [0,1]$, $[0,1]\times \{1\}$, and $\{0\}\times [0,1]$.  If $V$ is parallel along the first side, all we know is that its $x$-derivative is zero when $y=0$ and $0\le x \le 1$.  We have no information about that derivative when $y\ne 0$.
A: For the same reason that a first derivative of zero does not imply a second derivative of zero.
The operator ($\nabla_b$) is an operator, so it doesn't evaluate to a number in general, so it is perfectly fine to apply multiple derivative operators to a vector. Vectors are defined in a region around a particular point, and $\nabla$ provides a connection between different nearby points.
