Scalar curvature of a 2-sphere via the Ricci tensor Using the usual coordinates on a 2-sphere of radius $r$, I get the metric tensor $g_{\mu\nu}=\text{diag}(r^2, r^2\sin^2\theta)$ and so $g^{\mu\nu}=\text{diag}(1/r^2,1/r^2\sin^2\theta)$.
Hence the only nonzero Christoffel symbols are $\Gamma^\theta_{\phi\phi} = -\sin\theta\cos\theta$ and $\Gamma^\phi_{\theta\phi} = \Gamma^\phi_{\phi\theta} = \cot\theta$.
I'm fairly sure about these because I've been able to corroborate the same from various online sources.
Now using the formula for the Ricci tensor (e.g., Dirac's "General Theory of Relativity", Eq. (14.4)):
$$R_{\mu\nu} = \Gamma^\alpha_{\mu\alpha,\nu} - \Gamma^\alpha_{\mu\nu,\alpha} - \Gamma^\alpha_{\mu\nu}\Gamma^\beta_{\alpha\beta} + \Gamma^\alpha_{\mu\beta}\Gamma^\beta_{\nu\alpha}.\tag{14.4}$$
I get $R_{\mu\nu} = \text{diag}(-1, -\sin^2\theta)$, whence $R=g^{\mu\nu}R_{\mu\nu}=-2/r^2$.
Obviously the minus sign is wrong, but I've checked my work on $R_{\mu\nu}$ twice and I can't seem to find the error. Can anyone please spot it and advise?
 A: As per the comments, it appears that your equation (14.4) is the negative of what other authors define as the Ricci tensor: see e.g. this math SE post.
We can explicitly calculate e.g. $R_{\phi\phi}$ using equation (14.4), noting that most of the terms vanish:
$R_{\phi\phi}= 
-\partial_{\theta}{\Gamma^{\theta}}_{\phi\phi}
- {\Gamma^{\theta}}_{\phi\phi}{\Gamma^{\phi}}_{\theta\phi}
+ {\Gamma^{\phi}}_{\phi\theta}{\Gamma^{\theta}}_{\phi\phi}
+ {\Gamma^{\theta}}_{\phi\phi}{\Gamma^{\phi}}_{\phi\theta} \\
= -(\sin^2\theta- \cos^2\theta) -(-\sin\theta\cos\theta\cot\theta)+(-\sin\theta\cos\theta\cot\theta)+(-\sin\theta\cos\theta\cot\theta) \\
= -\sin^2\theta$,
and similarly for $R_{\theta\theta}$.
I also note that the claim in Dirac's book that the scalar curvature $R$ is defined in a way to be positive comes before the definition of eq. (14.4). I have not performed the calculation, but it is possible that the contraction of the Riemann tensor as defined in the book will yield a positive scalar curvature. Alternatively it is possible that Dirac simply made an (odd number of) sign errors.
