TrickyDicky said:
Hmm, maybe it's just me but if perpendicularity is meaningless without a metric(wich probably is), I don't know how parallelism is any more meaningful, I mean one seems to need something resembling Euclidean space locally (actually its affine generalization) to have the notion of parallelism between lines.
To see the problem with perpendicularity, let's take a look at a manifold having nothing to do with spatial distances. Imagine that your manifold represents thermodynamic states of some system (say, a certain quantity of a gas). We can label the states by a pair of numbers (P,V) representing the pressure and the volume.
Suppose I have four states: S_1 = (P_1,V_1), S_2 = (P_2,V_2), S_3 = (P_3,V_3) and S_4 = (P_4,V_4)
Define:
\delta(P)_{12} = P_2 - P_1
\delta(V)_{12} = V_2 - V_1
\delta(P)_{34} = P_4 - P_3
\delta(V)_{34} = V_4 - V_3
To say that the line from S_1 to S_2 is parallel to the line from S_3 to S_4 is to say that there is some nonzero real number \lambda such that:
\delta(P)_{34} = \lambda \delta(P)_{12}
\delta(V)_{34} = \lambda \delta(V)_{12}
So parallel displacement vectors are defined for this space. On the other hand, how would you define
perpendicularity for displacement vectors? What does it mean to say that the line from S_1 to S_2 is perpendicular to the line from S_3 to S_4?
If the points were points in Euclidean space, and the coordinates were Cartesian, then we could say that the displacements are perpendicular if
\delta(P)_{12} \delta(P)_{34} + \delta(V)_{12}\delta(V)_{34} = 0
But that equation doesn't even make any sense for pressures and volumes. You can't add a square pressure to a square volume. In order to make sense of adding squared pressures and squared volumes, you need a conversion factor that relates pressure to volume.
So for a general abstract manifold, you can always make sense of parallel lines, but you can't always make sense of perpendicular lines.