So I am reading a book on relativity & differential geometry and in the text, they gave the Christoffel symbols in terms of the metric and its derivatives, but I wanted to derive it myself. However, when I derived it, I seem to be missing two terms. Can somebody spot where I messed up?
From the text, they said that the derivative of the basis vectors $\vec{e}_{\mu}$, denoted as $\vec{e}_{\mu, \nu} \equiv \partial_{\nu}\vec{e}_{\mu}$, can be written as a linear combination of these basis vectors and also a normal vector, i.e.
$$\vec{e}_{\mu,\nu}=\Gamma_{\mu\nu}^{\lambda}\vec{e}_{\lambda}+K_{\mu \nu} \vec{n}$$
I also know that the metric itself, $g_{\mu \nu}$ can be written as the dot product of these basic vectors as $$ g_{\mu \nu} = \vec{e}_{\mu} \cdot \vec{e}_{\nu}$$
So my logic was to take the derivative of the metric with this definition:
$$\begin{split}\partial_{\alpha} g_{\mu \nu} & =\partial_{\alpha} (\vec{e}_{\mu} \cdot \vec{e}_{\nu}) \\
& =\partial_{\alpha}\vec{e}_{\mu} \cdot \vec{e}_{\nu} + \vec{e}_{\mu} \cdot \partial_{\alpha} \vec{e}_{\nu} \\
& =\vec{e}_{\mu,\alpha} \cdot \vec{e}_{\nu} + \vec{e}_{\mu} \cdot \vec{e}_{\nu, \alpha} \\
& =(\Gamma_{\mu \alpha}^{\lambda} \vec{e}_{\lambda} +K_{\mu \alpha} \vec{n} ) \cdot \vec{e}_{\nu} +\vec{e}_{\mu} \cdot (\Gamma_{\nu \alpha}^{\lambda} \vec{e}_{\lambda} + K_{\nu \alpha} \vec{n}) \\
& =\Gamma_{\mu \alpha}^{\lambda} (\vec{e}_{\lambda} \cdot \vec{e}_{\nu}) + \Gamma_{\nu \alpha}^{\lambda} (\vec{e}_{\mu} \cdot \vec{e}_{\lambda}) \\
& =\Gamma_{\mu \alpha}^{\lambda} g_{\lambda \nu} + \Gamma_{\nu \alpha}^{\lambda} g_{\mu \lambda} \\
& =\Gamma_{\mu \alpha}^{\lambda} g_{\lambda \nu} + \Gamma_{\nu \alpha}^{\lambda} g_{\lambda \mu} \end{split}$$
In this, the only thing I used was that $\vec{n} \cdot \vec{e}_{\lambda} =0$ by definition and that the metric is symmetric, i.e. $g_{\mu \lambda} = g_{\lambda \mu}$.
So now that I have that equation for the derivative of the metric, I might as well play around with it and solve for the Christoffel symbols. The only thing I did was multiply the whole equation by $g^{\alpha \lambda}$ in an attempt to contract and eliminate some of the metric terms to isolate $\Gamma$:
$$ \begin{split} g^{\alpha \lambda} \partial_{\alpha} g_{\mu \nu} & = \Gamma_{\mu \alpha}^{\lambda} g^{\alpha \lambda} g_{\lambda \nu} + \Gamma_{\nu \alpha}^{\lambda} g^{\alpha \lambda} g_{\lambda \nu} \\
& =\Gamma_{\mu \alpha}^{\lambda} \delta_{\nu}^{\alpha} + \Gamma_{\nu \alpha}^{\lambda} \delta_{\mu}^{\alpha} \end{split}$$
Since this is just multiplying the metric by its inverse, it results in the identity matrix, or the Kronecker delta. Since this is $0$ when the indices are not equal to each other and $1$ when they are, we can write this as:
$$ g^{\alpha \lambda} \partial_{\alpha} g_{\mu \nu} = \Gamma_{\mu \nu}^{\lambda} + \Gamma_{\nu \mu}^{\lambda}$$
And lastly the Christoffel symbols are symmetric in their lower two indices so we finally get:
$$g^{\alpha \lambda} \partial_{\alpha} g_{\mu \nu} = 2 \Gamma_{\mu \nu}^{\lambda}$$ or $$\Gamma_{\mu \nu}^{\lambda} = \frac{1}{2} g^{\alpha \lambda} (\partial_{\alpha} g_{\mu \nu})$$
The problem is that the actual (correct) answer for $\Gamma$ involves three derivatives of the metric instead of my one. Where have I gone wrong here?
Best Answer
One defining property of Christoffel symbols of the second kind is
$d\mathbf{e}_i=\Gamma^k_{ij}\mathbf{e}_k dq^j$.
Accepting this as a definition for the object $\Gamma^k_{ij}$ one can show, looking at the second derivative of the line element, that $\Gamma$ is symmetrical in its lower indices $\Gamma^k_{ij}=\Gamma^k_{ji}$.
Now to the derivation of an expression for $\Gamma$: looking at the total derivative of the metric one can get to:
$dg_{ij}=d( \mathbf{e}_i \cdot \mathbf{e}_j )=(\Gamma^k_{jl}g_{ik}+\Gamma^k_{il}g_{jk})dq^l$.
But by definition the total derivativ of $g_{ij}$ is given by $dg_{ij}=\frac{\partial g_{ij}}{\partial q^l}dq^l$. By compare the coefficients we get to:
$\frac{\partial g_{ij}}{\partial q^l}=\Gamma^k_{jl}g_{ik}+\Gamma^k_{il}g_{jk}$.
EDIT: That is what was derived in the question by a different way. But now to isolate a single Christoffel symbol one needs to add this expression up with different indicies. The mistake in the derivation in the question was pointed out in the comments; it was a mistake concering the summation index $\lambda$.
Using that one can show using the symmetries of $\Gamma$ and $g$ that the following holds:
$\frac{\partial g_{ij}}{\partial q^l}+\frac{\partial g_{lj}}{\partial q^i}-\frac{\partial g_{il}}{\partial q^j}=2\Gamma^k_{li}g_{jk} $.
Now dividing by 2 and inverting with $g$ gets you to an expression for $\Gamma$:
$\Gamma^k_{li}=\frac{1}{2}g^{jm}(\frac{\partial g_{ij}}{\partial q^l}+\frac{\partial g_{lj}}{\partial q^i}-\frac{\partial g_{il}}{\partial q^j})$
This is one possible derivation where granted the step of summing up those 3 partial derivatives is not very intuitive.
I know one can get to an expression for the Christoffel symbols of the second kind by looking at the Lagrange equation of motion for a free particle on a curved surface. This basically get you the geodetic equation where $\Gamma$ shows up as well. Then the devining property would be the geodetic equation and one would need to do the above calculation to show that $d\mathbf{e}_i=\Gamma^k_{ij}\mathbf{e}_k dq^j$ actually holds for $\Gamma$.