Does $g_{\mu\mu}$ in an expression follow the Einstein summation convention? Assume that I have the expression for a Christoffel symbol:
$$ \Gamma^\mu_{\alpha \beta}=\frac{1}{2}g^{\mu \lambda}(\partial_\alpha g_{\beta \lambda}+\partial_\beta g_{\alpha \lambda} - \partial_\lambda g_{\alpha \beta}).\tag{1}$$
If the metric $g_{\mu\nu}$ is diagonal then the identity $$g^{\mu\lambda}g_{\lambda\nu}=\delta^\mu_\nu\tag{2}$$ simplifies to the expression $$g^{\mu\mu}g_{\mu\mu}=1.\tag{3}$$
Therefore is the following expression for the Christoffel symbol notationally correct?
$$\Gamma^\mu_{\alpha \beta}=\frac{1}{2g_{\mu \mu}}(\partial_\alpha g_{\beta \mu}+\partial_\beta g_{\alpha \mu} - \partial_\mu g_{\alpha \beta}).\tag{4}$$
Does it obey the Einstein summation convention correctly as the repeated $\mu$ index is not summed over?
If the expression is not correct how should it be written?
Addition
Ok I see the correct manipulation to get an all covariant form using Einstein notation:
\begin{eqnarray}\tag{5}
\Gamma_{\gamma\alpha\beta}&=&g_{\gamma\mu}\Gamma^\mu_{\alpha\beta}\\
&=&\frac{1}{2}g_{\gamma\mu}g^{\mu\lambda}(\partial_\alpha g_{\beta\lambda} + \partial_\beta g_{\alpha \lambda} - \partial_\lambda g_{\alpha\beta})\\
&=&\frac{1}{2}\delta^\lambda_\gamma(\partial_\alpha g_{\beta\lambda} + \partial_\beta g_{\alpha \lambda} - \partial_\lambda g_{\alpha\beta})\\
&=&\frac{1}{2}(\partial_\alpha g_{\beta\gamma} + \partial_\beta g_{\alpha \gamma} - \partial_\gamma g_{\alpha\beta})
\end{eqnarray}
That's correct isn't it?
 A: Your last expression is not valid, for two reasons: first, any given index can only occur twice per term in the Einstein convention, once as an upper index and once as a lower index. Remember that when an index is repeated, it means you sum over it with the metric:
$$T^a T_a = \sum_{a,b} g_{ab} T^a T^b$$
You have terms like $\frac{1}{2g_{\mu\mu}}\partial_\alpha g_{\beta\mu}$ which contain $\mu$ three times. That's meaningless in Einstein notation.
The other reason is that the metric doesn't necessarily commute with the partial derivative: $g\partial \neq \partial g$. So you can't convert a factor like $g^{\mu\lambda} \partial_{\alpha}g_{\beta \lambda}$ into $\partial_{\alpha}g^{\mu\lambda} g_{\beta\lambda}$, which I think is what you had in mind there. (You could put in an extra term to account for the commutator if you wanted to: $g\partial = \partial g + [g,\partial]$.)
By the way, it's not true that contracting the result of $g^{\mu\lambda}g_{\lambda\nu}$ gives you $1$. You get $g^{\mu\lambda}g_{\lambda\mu} = \delta^{\mu}_{\ \mu} = \sum_{\mu} 1$, which is the dimensionality of the space. In normal 3+1D space, this gives 4.
