Show that a geodesic remains of the same type at all points by first proving that $v^a w_a$ remains constant when parallel-transported The title basically says it all. Given $\boldsymbol{v}$ and $\boldsymbol{w}$ with contravariant components $v^a$ and $w^a$, I'm asked to show that $v^a w_a$ after being parallel-transported along a curve, and using this result, prove that a timelike (or spacelike or null) geodesic will remain such at all points.
I've tried taking the intrinsic derivative of both the contravariant components of $\boldsymbol{v}$ and the covariant components of $\boldsymbol{w}$ and takind the product between those two, but it turns out really difficult to calculate and I'm not even sure I'm doing it right... Any help will be much appreciated!
 A: Firstly, a vector $v$ (or any arbitrary tensor) is parallel transported along a curve $X^{\mu}(\lambda)$ if its absolute derivative vanishes along the curve,
$$ \tag{1}
\frac{D}{D \lambda}v^{\mu} = \nabla_{u} v^{\mu} \equiv u^{\nu} \nabla_{\nu} v^{\mu} = 0 \ ,
$$
where $u^{\nu} = \frac{d X^{\nu}(\lambda)}{d \lambda}$ is the tangent vector along the curve. Note that the definition of parallel transport makes no reference to geodesics. (A geodesic$^1$ is an example of a curve where the tangent vector is parallel transported with respect to itself.)
Applying this to the norm of two vectors $v$ $w$ we have
$$\tag{2}
\frac{D}{D \lambda}(v^{\mu}w_{\mu}) = u^{\rho} \nabla_{\rho}(g_{\mu \nu} v^{\mu} w^{\nu}) = g_{\mu \nu}  w^{\nu}( u^{\rho}\nabla_{\rho}v^{\mu}) + g_{\mu \nu} v^{\mu} ( u^{\rho}\nabla_{\rho}w^{\nu}) = 0
$$
where we've used $\nabla_{\rho}g_{\mu \nu}=0$ and both remaining terms vanish due to (1).
As previously mentioned, a geodesic is a special case where the vector we're parallel transporting is the tangent vector to the curve itself
$u = \frac{d X}{d \lambda}$. Recall the geodesic equation$^{1}$ is just
$$
\nabla_{u}u^{\rho}=u^{\sigma}\nabla_{\sigma}u^{\rho}=0 \ ,
$$
and that we also classify geodesics (and vectors in general) as timelike/null/spacelike by looking at $u^{\nu}u_{\nu}$. In (2) we saw that when parallel transported along a curve, the norm of two vectors is conserved. The expression $u^{\nu}u_{\nu}$ is just a specific example of (2), and hence geodesics that are timelike/null/spacelike must remain so.

$^1$Note that for all the geodesics paths $X^{\mu}(\lambda)$ we've assumed $\lambda$ to be an affine parameter, which can always be found.
