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  Levi-Civita connection on a sphere in the vielbein formalism

+ 4 like - 0 dislike

I am trying to learn the vielbein formalism and have a question for the example of the Riemann sphere $S^2$. I am afraid my question is rather elementary, as it seems to be a simple sign error. Still, could someone help me figure this out?

On the sphere with coordinates $(x,y,z) = (\cos ϕ \sin θ, \sin ϕ \sin θ, \cos θ)$ and metric $ds^2 = dθ^2 + \sin θ^2 dϕ^2$, we can define the zweibein

$$ e_θ = ∂_θ , \quad e_ϕ = \frac{1}{\sin θ} ∂_ϕ$$

The Levi-Civita connection for the metric is torsion-free, which means

$$ \nabla_{e_ϕ} e_θ - \nabla_{e_θ} e_ϕ = [e_ϕ,e_θ] $$

A separate calculation shows that $\nabla_{e_θ}e_ϕ = 0$, so we can use this formula to quickly calculate the connection form $ω_{ab}$: 

$$ \nabla_{e_ϕ} e_θ = [e_ϕ,e_θ] = -∂_θ(\frac{1}{\sin θ})·∂_ϕ = \cot θ · e_ϕ \equiv  ω_{θ\ \ }^{\ \ ϕ}(e_ϕ) e_ϕ$$

Unfortunately, this calculation seems to be wrong, because it contradicts the statement

$$  ω^{ϕ\ \ }_{\ \ θ}(e_ϕ) = \cot θ $$

that I found in some lecture notes (formula (2.345)). The connection form is antisymmetric, so one of the two values should be $-\cot θ$, but I can't decide which.

Can somebody help me find the source of this sign discrepancy?

asked Feb 9, 2015 in Mathematics by Greg Graviton (775 points) [ revision history ]
Ooh, it appears to be a matter of convention for the connection form $\omega$! It looks like physicists use the notation $\nabla_X e_b = \omega^a_{\ \ b}(X) e_a$ whereas mathematicians tend to use the notation $\nabla_X e_i = \sum_j \omega^j_i(X) e_j$. Consequently, physicists write the Cartan structure equations as $\Omega^a_{\ \ b} = d\omega^a_{\ \ b} + \omega^a_{\ \ c} \wedge \omega^c_{\ \ b}$ whereas mathematicians write them as $\Omega^{j}_i = d\omega^j_i - \sum_k \omega^k_i \wedge \omega^j_k$ with a different sign. I tried to derive the formalism myself, and accidentally picked a convention that is that is neither the mathematician's nor the physicists' convention. Will put this into a real answer soon.

1 Answer

+ 3 like - 0 dislike

The calculation is correct, but it appears that there are different conventions in use for the connection 1-form $\omega$. In the question, the definition for $ω$ used to the very right of the $\equiv$-sign in the question is not one of the standard conventions, that's why the signs differ.

Apparently, physicists tend use the notation $\nabla_X e_b = \omega^a_{\ \ b}(X) e_a$ (example), whereas mathematicians tend to use the notation $\nabla_X e_i = \sum_j \omega^j_i(X) e_j$ (example [pdf]).

Consequently, physicists write the Cartan structure equations for the curvature as $\Omega^a_{\ \ b} = d\omega^a_{\ \ b} + \omega^a_{\ \ c} \wedge \omega^c_{\ \ b}$ whereas mathematicians write them as $\Omega^{j}_i = d\omega^j_i - \sum_k \omega^k_i \wedge \omega^j_k$ with a different sign.

By the way, instead of calculating the commutator, the connection 1-form can also be calculated from the Cartan structure formula for the dual frame. Using the dual frame $θ^θ = dθ$ and $θ^ϕ = \sin θ \, dϕ$, the equations read
0 = dθ^θ + ω^θ_{\ ϕ} \wedge θ^ϕ = d(dθ) - \sin θ\, ω^ϕ_{\ θ} \wedge dϕ
= -\sin θ\, ω^ϕ_{\ θ} \wedge dϕ$$

0 = dθ^ϕ + ω^ϕ_{\ θ} \wedge θ^θ = d(\sin θ\, dϕ) - dθ \wedge ω^ϕ_{\ θ}
  = dθ \wedge (\cos θ\, dϕ - ω^ϕ_{\ θ})
The first equation implies that the form $ω^ϕ_{\ θ}$ is a multiple of the form $dϕ$. The second equation implies that the prefactor is $\cos θ$, so we have
    ω^ϕ_{\ θ} = \cos θ\, dϕ = \cot θ·θ^ϕ

answered Feb 15, 2015 by Greg Graviton (775 points) [ no revision ]

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