## connection of the unit sphere

 Quote by Dick It really doesn't matter what initial conditions you take. But if you want to follow the instructions of the problem to the letter, then (I've just reread it) it says to take the initial unit vector parallel to the phi=0 circle. I guess this would mean we actually should have had the initial conditions to be v^phi(0)=0 and v^theta(0)=1.
I would think that it does matter what the initial conditions are but we'll see. So, with v^phi(0)=0 and v^theta(0)=1, I get:

A = -B = i/2 csc theta_0

 Recognitions: Homework Help Science Advisor Looks ok to me.
 Plugging in 2 pi for theta gives v^phi (phi = 2pi) = 0, right? So, now I need to solve the DE for v^theta and hopefully get v^theta (phi = 2pi) = 1, right?
 Recognitions: Homework Help Science Advisor Nooo. I get v^phi=constant*sin(cos(theta)*phi). Putting in phi=2pi doesn't give you zero for a general value of theta. And you should expect this - the problem asks you to show that after transport the direction v is not generally the same, but the length is the same.
 Recognitions: Homework Help Science Advisor Sure, but it's e^(i*cos(theta)*2pi), not e^(i*2pi).
 Sorry, I deleted a post in between there. Anyway, e^(i*cos theta 2pi) = (e^(i*2pi))^cos(theta) = 1^{cos theta} = 1, right?
 Recognitions: Homework Help Science Advisor Wrong. That rule only works for positive real numbers. i=exp(i*2pi*(1/4))=exp(i*2pi)^(1/4)=1^(1/4)=1. What went wrong there? It's true that i and 1 are both fourth roots of 1 but they aren't equal.

 Quote by Dick Wrong. That rule only works for positive real numbers. i=exp(i*2pi*(1/4))=exp(i*2pi)^(1/4)=1^(1/4)=1. What went wrong there? It's true that i and 1 are both fourth roots of 1 but they aren't equal.
Interesting!

I would need to see a counterexample for negative real numbers, though.

Yes, so then I get:

$$v^{\phi}(\phi = 2 \pi) = \sin ( \cos (\theta) 2 \pi) \csc \theta$$ .

I am not really sure how to simplify that.

So, now I need to solve the DE for v^theta and hopefully get v^theta (phi = 2pi) = sqrt( 1- sin^2 ( cos (theta) 2 pi) csc^2 theta ), right?

 Recognitions: Homework Help Science Advisor Something like that. Just find v^phi and v^theta and put them into the metric. BTW ((-1)^2)^(1/2)=1 which is NOT EQUAL to (-1)^(2*(1/2))=(-1)^1=-1. In the real case you can allow the exponents to be negative, but not the base (or whatever you call the -1 part).
 OK. I plugged the expression for v^phi into the DE for v^theta and integrated and plugged in 2 pi for the phi parameter and got: $$v^{\theta}( \phi = 2 \pi) = \cos^2 \theta \sin (\cos (\theta) 2 \pi)$$ When you say plug it into the metric, do you mean find g_ab v^a v_b?
 Recognitions: Homework Help Science Advisor I don't get the cos(theta)^2 and do I get that its a cos function of the argument. Check your derivation.
 Do you get: $$v^{\theta}( \phi = 2 \pi) = -\cos (\cos (\theta) 2 \pi)$$ No way, that actually does give 1 when we plug into the metric. I think that completes the problem.
 Recognitions: Homework Help Science Advisor If you plug phi=0 in you get v^phi(0)=0, v^theta(0)=(-1) (we wanted +1). I think you have another sign error lurking around. But, yeah, that's basically it, once you correct your derivation of v^theta.