The problem (from Misner, Thorne, and Wheeler's Gravitation, exercise 11.12) is as follows:
Calculate the components of the Riemann curvature tensor for the rotation group's manifold SO(3); use the basis of generators {e$_\alpha$}.
Here, by the basis of generators {e$_\alpha$}, they mean those given by
e$_1$$= \cos{\psi} \frac{\partial}{\partial \theta} - \sin{\psi} \left(\cot{\theta}\frac{\partial}{\partial \psi} - \frac{1}{\sin{\theta}} \frac{\partial}{\partial \phi} \right)$;
e$_2$$= \sin{\psi} \frac{\partial}{\partial \theta} + \cos{\psi} \left(\cot{\theta}\frac{\partial}{\partial \psi} - \frac{1}{\sin{\theta}} \frac{\partial}{\partial \phi} \right)$;
e$_3$$= \frac{\partial}{\partial \psi}$.
As they have already had me calculate in earlier problems, the commutation coefficients for this basis are
$c_{\alpha \beta}^{\gamma} = -\epsilon_{\alpha \beta \gamma}$,
and the connection coefficients for this basis are
$\Gamma^{\alpha}_{\beta \gamma} = \frac{1}{2}\epsilon_{\alpha \beta \gamma}$.
These values are independent of location in SO(3).
To find the components of Riemann, I tried using the equation
$R^{\alpha}_{\beta \gamma \delta} = \partial_\gamma \Gamma^{\alpha}_{\beta \delta} - \partial_\delta \Gamma^{\alpha}_{\beta \gamma} + \Gamma^{\alpha}_{\mu \gamma} \Gamma^{\mu}_{\beta \delta} - \Gamma^{\alpha}_{\mu \delta} \Gamma^{\mu}_{\beta \gamma} - \Gamma^{\alpha}_{\beta \mu} c_{\gamma \delta}^{\mu}$
for the components in a noncoordinate basis. Because the connection coefficients are the same everywhere, I assumed the $\partial \Gamma$ were zero--this is also what I found when I tried to calculate the terms more directly. Thus, the equation reduced to
$R^{\alpha}_{\beta \gamma \delta} = \Gamma^{\alpha}_{\mu \gamma} \Gamma^{\mu}_{\beta \delta} - \Gamma^{\alpha}_{\mu \delta} \Gamma^{\mu}_{\beta \gamma} - \Gamma^{\alpha}_{\beta \mu} c_{\gamma \delta}^{\mu}$.
Substituting the components of $\Gamma$ and $c$ from above, I got
$R^{\alpha}_{\beta \gamma \delta} = (\frac{1}{2} \epsilon_{\alpha \mu \gamma}) (\frac{1}{2} \epsilon_{\mu \beta \delta}) - (\frac{1}{2}\epsilon_{\alpha \mu \delta}) (\frac{1}{2}\epsilon_{\mu \beta \gamma}) - (\frac{1}{2}\epsilon_{\alpha \beta \mu}) (-\epsilon_{\gamma \delta \mu}) = \frac{1}{4} \epsilon_{\alpha \mu \gamma} \epsilon_{\mu \beta \delta} - \frac{1}{4}\epsilon_{\alpha \mu \delta} \epsilon_{\mu \beta \gamma} + \frac{1}{2}\epsilon_{\alpha \beta \mu} \epsilon_{\gamma \delta \mu}$.
The book uses the symbol $\delta^{\alpha \beta}_{\gamma \delta} \equiv (\delta^{\alpha}_{\gamma}\delta^{\beta}_{\delta} - \delta^{\alpha}_{\delta} \delta^{\beta}_{\gamma}) = \epsilon_{\mu \alpha \beta} \epsilon_{\mu \gamma \delta}$. Rewriting the above using this notation, I found
$R^{\alpha}_{\beta \gamma \delta} = -\frac{1}{4} \epsilon_{\mu \alpha \gamma} \epsilon_{\mu \beta \delta} + \frac{1}{4}\epsilon_{\mu \alpha \delta} \epsilon_{\mu \beta \gamma} + \frac{1}{2}\epsilon_{\alpha \beta \mu} \epsilon_{\gamma \delta \mu} = -\frac{1}{4} \delta^{\alpha \gamma}_{\beta \delta} + \frac{1}{4}\delta^{\alpha \delta}_{\beta \gamma} + \frac{1}{2}\delta^{\alpha \beta}_{\gamma \delta}$,
which I then simplified further via
$-\delta^{\alpha \gamma}_{\beta \delta} + \delta^{\alpha \delta}_{\beta \gamma} = -(\delta^{\alpha}_{\beta} \delta^{\gamma}_{\delta} - \delta^{\alpha}_{\delta} \delta^{\gamma}_{\beta}) + (\delta^{\alpha}_{\beta} \delta^{\delta}_{\gamma} - \delta^{\alpha}_{\gamma} \delta^{\delta}_{\beta}) = -\delta^{\alpha}_{\beta} \delta^{\delta}_{\gamma} + \delta^{\alpha}_{\delta} \delta^{\beta}_{\gamma} + \delta^{\alpha}_{\beta} \delta^{\delta}_{\gamma} - \delta^{\alpha}_{\gamma} \delta^{\beta}_{\delta} = \delta^{\alpha}_{\delta} \delta^{\beta}_{\gamma} - \delta^{\alpha}_{\gamma} \delta^{\beta}_{\delta} = \delta^{\alpha \beta}_{\delta \gamma}$.
Thus, I found
$R^{\alpha}_{\beta \gamma \delta} = \frac{1}{4} \delta^{\alpha \beta}_{\delta \gamma} + \frac{1}{2}\delta^{\alpha \beta}_{\gamma \delta} = \frac{1}{4} \delta^{\alpha \beta}_{\gamma \delta}$,
but the book says the answer should be
$R^{\alpha}_{\beta \gamma \delta} = \frac{1}{2} \delta^{\alpha \beta}_{\gamma \delta}$.
I cannot for the life of me figure out where the extra factor of 2 comes from. I am wondering if it was wrong to assume that the $\partial \Gamma$ terms were zero, but if so I can't figure out how to calculate them. When I use the equation
$\partial_\delta \Gamma^{\alpha}_{\beta \gamma} = <\omega^{\alpha}, e_\delta[e_\gamma[e_\beta]]> - \Gamma^{\mu}_{\beta \gamma} \Gamma^{\alpha}_{\mu \delta}$,
I invariably get $\partial \Gamma = 0$. Beyond that, I can't find anything wrong with my calculations, despite having gone over them numerous times. Any helps/hints/tips would be greatly appreciated!
Your result is right. You can double check it by brute force. Perform parallel transport of a tangent vector along a closed curve. Use eq. (1) in Box 11.7; page 281 of Misner et al. By choosing the basis vectors you can compute all the components $R^{\alpha}_{\beta\gamma\delta}$. You will get: 0, 1/4 or -1/4.