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?

  • 3
    $\begingroup$ This appears to be a pure math question; it belongs on Mathematics. $\endgroup$
    – ACuriousMind
    Commented Jun 16, 2016 at 17:33
  • 4
    $\begingroup$ You can't contract with $g^{\alpha\lambda}$ when $\lambda$ is already a dummy summation index. $\endgroup$
    – Evan Rule
    Commented Jun 16, 2016 at 17:42
  • $\begingroup$ It is one of the rules of Einstein summation notation: mathworld.wolfram.com/EinsteinSummation.html $\endgroup$
    – Evan Rule
    Commented Jun 16, 2016 at 18:22
  • 1
    $\begingroup$ $\Gamma^{\lambda}_{\mu\alpha}g^{\alpha\lambda}g_{\lambda\nu}$ has three $\lambda$s in it. Please explain how you think this summation should be executed. $\endgroup$
    – Evan Rule
    Commented Jun 16, 2016 at 18:23
  • $\begingroup$ Are you reading Zee? $\endgroup$
    – Ryan Unger
    Commented Jun 16, 2016 at 23:23

3 Answers 3


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 $\Gamma$.


In order to get the Christoffel symbols we should notice that when two vectors are parallelly transported along any curve then the inner product between them remains invariant under such operation. Therefore we should write your expression for three parameters $\alpha$, $\lambda$ and $\nu$ in a cyclic order to obtain the correct Christoffel symbols.


In my opinion this passage has a mistake:

$$ \begin{split} g^{\alpha \lambda} \partial_{\alpha} g_{\mu \nu} & = \Gamma_{\mu \alpha}^{\lambda} g^{\alpha \lambda} g_{\lambda \nu} + \Gamma_{\nu \alpha}^{\lambda} g^{\alpha \lambda} g_{\lambda \nu} \\ & =\Gamma_{\mu \alpha}^{\lambda} \delta_{\nu}^{\alpha} + \Gamma_{\nu \alpha}^{\lambda} \delta_{\mu}^{\alpha} \end{split}$$

You cannot contract a running index as you did, because you have chosen an index $\lambda$ but you contracted it with a pre-existent running $\lambda$ (this operation is fine for LHS). You can for example contract $g^{\nu l}$ with $\partial_a g_{\mu \nu}$ but it would be useful only for the first element of RHS since you would obtain $\Gamma^{\lambda}_{\mu \alpha} \delta_{\lambda}^{ \nu }$. For this reason for such demonstration many additions of "similar" tricks of this kind are required.


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