## connection of the unit sphere

1. The problem statement, all variables and given/known data
I am confused about this question and probably the connection in general. When we only have two independent coordinates here what do the three indices in $$\Gamma ^a_{bc}$$ stand for?

I found the metric for this space, but every formula for Gamma has indices a,b,c and I do not know what to plug in for them?

2. Relevant equations

3. The attempt at a solution
Attached Thumbnails

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 Recognitions: Homework Help Science Advisor a,b and c stand for either phi or theta. I can say this without even seeing your attachment.
 So, when it asks me to calculate the coefficients, i need to calculate for (a,b,c) = (theta, theta, theta) (a,b,c) = (theta, theta, phi) (a,b,c) = (theta, phi, theta) (a,b,c) = (theta, phi, phi) and then four more with a = phi ?

Recognitions:
Homework Help

## connection of the unit sphere

Yes. but you know case (a,b,c)=(a,c,b) right? And didn't you just do an exercise about diagonal metrics that would come in handy here?
 So, to do the next part, should I just plug the connection into 3.41 and integrate from 0 to 2 pi? Attached Thumbnails
 Recognitions: Homework Help Science Advisor Your attachments still haven't been approved, but you should plug the connection into the parallel transport equations, if that's what you mean. And then solve them.
 Which equations do you mean? I know the connection and I think that I know the initial vector, so which equation relates these and the rotation angle to the final vector?

Recognitions:
Homework Help
 Quote by ehrenfest Which equations do you mean? I know the connection and I think that I know the initial vector, so which equation relates these and the rotation angle to the final vector?
Look up the 'parallel transport' operation mentioned in the problem.
 What to you mean look up the parallel transport operation? I already looked it up and I even posted that entire section from my book. Did you mean look it up elsewhere? Because I have tried that also. I just want help figuring out which equations to plug the connection and the initial vector into and get the final vector.
 Recognitions: Homework Help Science Advisor Equation (3.42). Start trying to use it and then people can start trying to help you. Did you read what you posted? It's a differential equation for the rate of change of the transported vector components in terms of the tangent vector to the curve and the connection coefficients. You have to solve this differential equation.
 Here is my attempt: $$\Gamma^{\phi}_{\phi \theta} = \Gamma^{\phi}_{\theta \phi} = \cot \theta$$ $$\Gamma^{\theta}_{\phi \phi} = - \sin \theta \cos \theta$$ The rest of the connection vanishes. I plug this into the equation along with the initial position and the theta equation vanishes and the phi equation is $$\frac{ dv^\phi}{du} = - \theta_0 \cot \theta \frac{dx^\phi}{du}$$ So, is the next step finding dx^phi/du with some parametrization?
 Recognitions: Homework Help Science Advisor The theta equation does not vanish. And why theta_0? Shouldn't that be v^theta? And use the obvious parametrization. Use phi as the path parameter.
 You're right. I was using the initial conditions for v^theta and v^phi. That is why I thought the theta equation vanishes. So, the two equations are: $$\frac{ dv^\theta}{du} = \sin \theta \cos \phi v^{\theta} \frac{d\phi}{du}$$ $$\frac{ dv^\phi}{du} = - v^{\theta} \cot \theta \frac{d\phi}{du}$$ So when I replace u with theta as my parameter, why don't both of these equation vanish since they both would have a d\phi/d\theta and theta and phi are supposed to be independent?

Recognitions:
Homework Help
 Quote by ehrenfest You're right. I was using the initial conditions for v^theta and v^phi. That is why I thought the theta equation vanishes. So, the two equations are: $$\frac{ dv^\theta}{du} = \sin \theta \cos \phi v^{\theta} \frac{d\phi}{du}$$ $$\frac{ dv^\phi}{du} = - v^{\theta} \cot \theta \frac{d\phi}{du}$$ So when I replace u with theta as my parameter, why don't both of these equation vanish since they both would have a d\phi/d\theta and theta and phi are supposed to be independent?
 You're right, the equations should be: $$\frac{ dv^\theta}{du} = \sin \theta \cos \theta v^{\phi} \frac{d\phi}{du}$$ $$\frac{ dv^\phi}{du} = - v^{\theta} \cot \theta \frac{d\phi}{du}$$ And when I apply the phi parametrization (I don't know what I was thinking with the theta parametrization), I get $$\frac{ dv^\theta}{d\phi} = \sin \theta \cos \theta v^{\phi}$$ $$\frac{ dv^\phi}{d\phi} = - v^{\theta} \cot \theta$$ So if we integrate both of these equations w.r.t. phi, we get: $$\Delta v^\theta = \sin \theta_0 \cos \theta_0 \int_{0}^{2\pi}v^{\phi}d\phi$$ $$\Delta v^\phi = - \cot\theta_0 \int_{0}^{2\pi} v^{\theta} d\phi$$ How do I evaluate the integrals on the RHS?
 I get $$\frac{d^2v^{\phi}}{d\theta} = - \cos \theta v^{\phi}$$ Since $$\theta = \theta_0$$ is a constant all along this parametrization, this DE has solutions: $$v^{\phi} = A e^{i \cos \theta_0 \phi} + B e^{-i \cos \theta_0 \phi}$$ So, now I need to plug this into the other DE, find A and B, and plug in 2 pi for phi to get the final value of v^phi and v^theta, right? EDIT: the first equation should be $$\frac{d^2v^{\phi}}{d\phi^2} = - \cos^2 \theta v^{\phi}$$