parallel vector field I was wondering about the following:
I know that a vector field along a geodesic that is parallel has a constant angle to the tangent vector of the curve and constant length. Now, is the converse also true? Can we say that something is a parallel vector field if these two properties hold?
If anything is unclear, please let me know.
 A: As a couple of commenters have suggested, the statement is true in the $2$-dimensional case, and false in all higher dimensions. 
The proofs of both claims rely on the following lemma:

Lemma: Let $M$ be a Riemannian manifold of dimension $n\ge 2$. Given any unit-speed geodesic $\gamma\colon J\to M$, there exist parallel vector fields $E_2,\dots,E_n$ along $\gamma$ such that $(\gamma'(t),E_2(t),\dots,E_n(t))$ form an orthonormal basis for $T_{\gamma(t)}M$ for each $t\in J$.

To prove the lemma, just choose any $t_0\in J$, extend $\gamma'(t_0)$ to an orthonormal basis $(\gamma'(t_0),E_2(t_0),\dots,E_n(t_0))$ for $T_{\gamma(t_0)}M$, and parallel transport these vectors along $\gamma$. Because parallel transport preserves inner products, the resulting frame is orthonormal at each point. 
Now suppose $n=2$ and $\gamma$ is any geodesic in $M$. By reparametrizing, we may assume it's unit speed. Let $E_2$ be a vector field along $\gamma$ such that $(\gamma(t),E_2(t))$ is orthonormal for each $t$. Suppose $V$ is a vector field along $\gamma$ whose length and angle with $\gamma'(t)$ are both constant. We can write $V(t) = a(t)\gamma'(t)+b(t)E_2(t)$, where $a(t) = \langle V(t), \gamma'(t)\rangle$ and $b(t) = \langle V(t), E_2(t)\rangle$, and the hypotheses imply that both $a(t)$ and $b(t)$ are constant. Since any linear combination of parallel vector fields with constant coefficients is parallel, it follows that $V$ is parallel.
On the other hand, when $n>2$, if $\gamma$ is any geodesic, we can construct an orthonormal frame $(\gamma'(t),E_2(t),\dots,E_n(t))$ as in the lemma, and let $V(t) = (\cos t) E_2(t) + (\sin t) E_3(t)$. Then $V(t)$ is a unit vector for all $t$, and its angle with $\gamma'(t)$ is constant ($90^\circ$). But it's not parallel along $\gamma$.
