What is the relationship between spin connections and tetrads on a 2-sphere?

[exponent137]

I wish to imagine Spin connection on a 2-sphere.

Now, I am reading Carroll's http://arxiv.org/abs/gr-qc/9712019. He begins to write about spin connections in page 88. Calculations of $\Gamma$ on sphere are evident in page 84.
Can you help me, how spin connection on a 2-sphere look like and how suitable tetrads look like?

 

[fzero]

The 2-sphere would provide an excellent example for you to work out on your own. To get you started, an obvious choice of orthonormal 1-forms (3.122) is

$$\begin{split} & \hat{\theta}^1 = d\theta, \\
&\hat{\theta}^2 = \sin\theta d\phi.\end{split} $$

From these you can read off the $e^a_\mu$ and invert as a matrix to find the $e^\mu_a$.

 

[exponent137]

I calculated $e^a_\mu$ as $e^1_1=dx/d\theta$ and $e^2_2=\frac{dy/d\phi}{\sin(\theta)}$. and use of (3.132) gave me, for instance $\omega^1_{1 1}=-\frac{d\theta}{dx}\partial_\theta(\frac{1}{\sin(\theta)}\frac{dx}{d\theta})$

I do not feel that these examples are correct or finished? (I assumed sphere radius as 1.)

Can I insert $dx/d\theta=1$ and $dy/d\phi=\sin(\theta)$ and I can calculate further? But this means that
$e^1_1=1$ and $e^2_2=\sin(\theta)$, etc? Thus, the the above $\omega^1_{1 1}$ needs correction?

 

[exponent137]

Can I insert $dx/d\theta=1$ and $dy/d\phi=\sin(\theta)$ and I can calculate further? But this means that
$e^1_1=1$ and $e^2_2=\sin(\theta)$, etc? Thus, the the above $\omega^1_{1 1}$ needs correction? Thus $\omega^1_{1 1}=-\frac{d\theta}{dx}\partial_\theta(\frac{dx}{d\theta})=0$ and $\omega^2_{1 2}=\cot(\theta)-\cot(\theta)=0$. And $\omega^2_{1 1}=\omega^1_{1 2}=0$.
$\omega^1_{2 2}=-\sin(\theta)\cos(\theta)$ and $\omega^2_{2 1}=\cot(\theta)$ Other $\omega$'s are zero.

But, when I repeated once again, the results were: $\omega^1_{2 2}=-\cos(\theta)$ and $\omega^2_{2 1}=\cos(\theta)$, other $\omega$'s were zero.

Are these calculations now correct?

Now a question about (3.134) appears, because $\Lambda^{a'}_c$ at the end is independent of $\theta$ and thus this last part equals zero. Is this correct? This also means that comparison with (3.146) cannot be done properly visually for this example.

 

[fzero]

First off, the notation is very confusing since we can't distinguish between $e^1_1$ and the inverse element with the same labels, so let's write

$$e^1_\theta = 1, ~~ e^2_\phi = \sin\theta,$$

so the non-vanishing inverse elements are

$$e^\theta_1 = 1, ~~ e^\phi_2 = \frac{1}{\sin\theta}.$$

The non-vanishing Christoffel symbols are

$$\Gamma^\theta_{\phi\phi}= -\sin\theta \cos\theta,~~~ \Gamma^\phi_{\theta\phi} = \cot\theta.$$

Using (3.132) I find the only nonvanishing components of the spin connection are

$$ {{\omega_\phi}^1}_2 = -\cos \theta, ~~~~{{\omega_\phi}^2}_1 = \cos \theta,$$

which agrees with your result when we penetrate the notation.

I don't agree that ${\Lambda^{a'}}_c$ is necessarily independent of $\theta$. This is a local Lorentz transformation so it is a 2x2 orthgonal matrix that leaves the identity matrix invariant, $\Lambda \Lambda^T =1$. We could certainly choose something like

$$\Lambda = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}.$$

The whole point is that $\Lambda = \Lambda(\theta,\phi)$ which means an LLT is a prescription to define a collection of 2x2 orthogonal matrices at each point of the space(time). Calling it local means that we necessarily want to consider the case that the matrix at the point $(\theta_1,\phi_1)$ is different from that at $(\theta_2,\phi_2)$.

 

[exponent137]

Now it is important also imagination of this spin connection.

The common connection can be imagined as formula 3.65 and the figure above it, and formula 3.1 on page 56.

Thus let us rewrite ${\Gamma^\phi_\theta}_\phi$ also as ${\Gamma^1_2}_3$, where numbers are locations of indeces. Let us imagine that in every point of the sphere it is a vector, which shows in $\phi$ direction. Let us assume that all those vectors are the same.
Index 3 means direction of these vectors. Index 2 means direction of covariant derivation. Index 1 means direction of vector, which arises after covariant derivation. So the above ${\Gamma^\phi_\theta}_\phi=\cot\theta$ means: if we have above mentioned longitudinal field (index 3 or direction $\phi$) and we look its derivation in direction 2, this means direction $\theta$, at the end we obtain a result of covariant derivation proportional to the direction 1, (direction $\phi$), this means proportional to $\cot{\theta}$.

According to similarities of equations 3.67 and 3.138 this interpretation is possible also for spin connection? But covariant derivative of spin connection is not explicitly written in the book, except the equation between 3.141 and 3.142?

Another visualization of the common connection $\Gamma$ is also eq. 3.47. It means, if we move in directions $\rho$ and $\sigma$ with uniform speed, we feel additional 'force' in direction $\mu$. (Imagine this for sphere) This means that we move diagonally in this example. Is this correct visualization? Can we use something like 3.47 for spin connection?

 

[fzero]

The spinor covariant derivative takes the form

$$ \nabla_\mu \phi = \left( \partial_\mu - {\omega_\mu}^{ab} \Sigma_{ab} \right) \psi,$$

where up to numerical factors the $\Sigma_{ab}$ are the appropriate Dirac matrices.

The spin connection does lead to some sort of picture of parallel transport like you're referring to. However, the introduction of the local frames through the vierbeins makes the picture a little bit complicated. For ${\Gamma^\mu}_{\rho\sigma}$ we have coordinate indices everywhere, so the figure on p 56 makes sense. When we use the vierbein $e^a_\mu$, we are assigning some flat coordinate $\sigma^a$ at every point $x^\mu$ of our manifold. If you try to draw an analogous picture, you have to be clear that you are doing this assignment at every point, so you need to map quantities appropriately back and forth. I think morally you can argue that the picture is the same.

I don't really see the point of trying to modify 3.47, which is the geodesic equation. A geodesic naturally lives on the manifold and I don't know what a generalization of it would be that would require the spin connection as a fundamental object in that equation. The straightforward thing to do would be to just use 3.131 to write the Christoffel in terms of the spin connection.