Vrftsjtryk

We use the following parameterisation for the unit sphere: $sigma(theta,phi)=(costhetacosphi,costhetasinphi,sintheta)$.
I have calculated the Christoffel symbols to be
$Gamma^1_{11}=Gamma^2_{11}=Gamma^1_{12}=0, Gamma^1_{22}=sinthetacostheta,Gamma^2_{22}=0$, which match the answers I am given in my notes. But when I calculate $Gamma^2_{12}$ I get $-sinthetacostheta$, which apparently is incorrect and should be $-tantheta$. My reasoning was that $Gamma^2_{12}=sigma_phi cdot sigma_{thetaphi}=(-costhetasinphi,costhetacosphi,0)cdot(sinthetasinphi,-sinthetacosphi,0)=-sinthetacostheta$. I am not sure what I am doing wrong - the same method worked for the other five symbols and I have no idea where a $tantheta$ term would come from. Any help would be appreciated.

$require{cancel}$
You can directly calculate the metric coefficients for this parameterization as ($x^1 = costheta, x^2 = phi$)

$$
(g_{munu}) = pmatrix{1 & 0 \ 0 & cos^2theta} ~~~mbox{and}~~
(g^{munu}) = pmatrix{1 & 0 \ 0 & 1/cos^2theta}
$$

From this is pretty straightforward to calculate $Gamma^{lambda}_{munu}$

$$
Gamma^{lambda}_{munu} = frac{1}{2}g^{lambdaalpha}left(frac{partial g_{mualpha}}{partial x^{nu}}+ frac{partial g_{alphanu}}{partial x^{mu}} - frac{partial g_{munu}}{partial x^{alpha}}right)
$$

Take $lambda = 2$, $mu = 1$ and $nu = 2$

begin{eqnarray}
Gamma^{2}_{12} &=& frac{1}{2}g^{2alpha}left(frac{partial g_{1alpha}}{partial x^{2}}+ frac{partial g_{alpha2}}{partial x^{1}} - frac{partial g_{12}}{partial x^{alpha}}right) = frac{1}{2}g^{22}left(cancelto{0}{frac{partial g_{12}}{partial x^{2}}} + frac{partial g_{22}}{partial x^{1}} - cancelto{0}{frac{partial g_{12}}{partial x^{2}}}right) \
&=& frac{1}{2}left(frac{1}{cos^2theta}right) frac{partial cos^2theta}{partial theta} = -tantheta
end{eqnarray}

You can calculate the other components the same way

$$
Gamma_{11}^1 = Gamma_{11}^2 = Gamma_{12}^1 = Gamma_{22}^2 = 0
$$

and

$$
Gamma_{22}^1 = sinthetacostheta
$$

There's another way, and I think this is what you were going for in your post, but is is completely equivalent to my other answer, the idea is to calculate

$$
Gamma^{lambda}_{munu} = frac{partial x^lambda}{partial sigma^alpha}frac{partial^2 sigma^alpha}{partial x^mu partial x^nu} tag{1}
$$

You can obtain the inverse mapping fairly easily

$$
theta = arcsin sigma^2 equiv x^1 ~~~mbox{and}~~~ phi = arctan left(frac{sigma^2}{sigma^1}right) equiv x^2 tag{2}
$$

With this we have

begin{eqnarray}
Gamma^{2}_{12} &=& frac{partial x^2}{partial sigma^alpha} frac{partial^2 sigma^alpha}{partial x^1 partial x^2} \
&=& frac{partial x^2}{partial sigma^1} frac{partial^2 sigma^1}{partial x^1 partial x^2} + frac{partial x^2}{partial sigma^2} frac{partial^2 sigma^2}{partial x^1 partial x^2} + frac{partial x^2}{partial sigma^3} frac{partial^3 sigma^1}{partial x^1 partial x^2} \
&=& -frac{sigma^2 sinthetasinphi}{(sigma^1)^2 + (sigma^2)^2}-frac{sigma^1sinthetacosphi}{(sigma^1)^2 + (sigma^2)^2} \
&=& - tantheta tag{3}
end{eqnarray}