Lowering index on $\Gamma^\beta_{\mu\gamma}$ I was doing an exercise in Schutz (A First course in General Relativity). The exercise wanted the double covariant derivative calculated for a vector $V^\mu$ i.e. $\nabla_\alpha \nabla_\beta V^\mu$ i.e. $(V^\mu_{;\alpha})_{;\beta}$. This basically amounts to calculate the covariant derivative of a mixed (1,1) tensor $T^\alpha_\beta$. I was able to calculate the covariant derivative of this mixed tensor by converting it into $T^\alpha_\beta = A^\alpha B_\beta$ and it worked out.
But I originally tried to calculate it by lowering the index on a (2,0) tensor $T^{\alpha\beta}$ using a metric tensor and then calculating the covariant derivative, since the covariant derivative of metric tensor is zero. But it did not work out. Here is how I did it.
Given that,
$$
T^{\alpha \beta}_{;\gamma}= T^{\alpha \beta}_{,\gamma} + \Gamma^\alpha_{\mu\gamma}T^{\mu\beta}+\Gamma^\beta_{\mu\gamma}T^{\alpha\mu}
$$
I lowered the index,
$$
(T^{\alpha}_\lambda g^{\lambda\beta})_{;\gamma} =(T^{\alpha}_\lambda )_{;\gamma}g^{\lambda\beta}= (T^{\alpha \beta}_{,\gamma} + \Gamma^\alpha_{\mu\gamma}T^{\mu\beta}+\Gamma^\beta_{\mu\gamma}T^{\alpha\mu})g^{\lambda\beta}
$$
$$
(T^{\alpha}_\lambda)_{;\gamma} = g_{\lambda\beta}T^{\alpha \beta}_{,\gamma} + g_{\lambda\beta}\Gamma^\alpha_{\mu\gamma}T^{\mu\beta}+g_{\lambda\beta}\Gamma^\beta_{\mu\gamma}T^{\alpha\mu})
$$
The confusion is that $g_{\lambda\beta}\Gamma^\beta_{\mu\gamma}$ in the last term, after the contraction on $\beta$ becomes, $\Gamma_{\lambda\mu\gamma}$ which is not the correct term for the covariant derivative of $T^{\alpha}_\lambda$. 
So, I am missing something in the contraction of $\Gamma^\beta_{\mu\gamma}$ or may be that is not a valid operation(?). I have a hunch that the reason has something to do with $\Gamma^\beta_{\mu\gamma}$ not being a valid tensor but I cant place what it is mathematically/physically(?). Or may be there is an identity of $\Gamma^\beta_{\mu\gamma}$ that I am missing here(?).
 A: Leibniz' rule gives you
$$g_{\lambda\beta}T^{\alpha\beta}_{,\gamma} = (g_{\lambda\beta}T^{\alpha\beta})_{,\gamma} - g_{\lambda\beta,\gamma}T^{\alpha\beta}.$$
The problem is the term $g_{\lambda\beta,\gamma}$, which is not zero, in general. Actually, you have
$$g_{\lambda\beta,\gamma}=g_{\lambda \mu}\Gamma^\mu_{\beta\gamma}+g_{\beta\mu}\Gamma^\mu_{\lambda\gamma}$$
so that
\begin{align}
g_{\lambda\beta}T^{\alpha\beta}_{,\gamma}
&= (g_{\lambda\beta}T^{\alpha\beta})_{,\gamma} - g_{\lambda\beta,\gamma}T^{\alpha\beta} \\
&= (g_{\lambda\beta}T^{\alpha\beta})_{,\gamma} - g_{\lambda \mu}\Gamma^\mu_{\beta\gamma}T^{\alpha\beta} - g_{\beta\mu}\Gamma^\mu_{\lambda\gamma}T^{\alpha\beta}.
\end{align}
Substituting in the last equation of your post we get
\begin{align}
(T^{\alpha}_\lambda)_{;\gamma}
&= g_{\lambda\beta}T^{\alpha \beta}_{,\gamma} + g_{\lambda\beta}\Gamma^\alpha_{\mu\gamma}T^{\mu\beta}+g_{\lambda\beta}\Gamma^\beta_{\mu\gamma}T^{\alpha\mu} \\
&= (g_{\lambda\beta}T^{\alpha\beta})_{,\gamma} - g_{\lambda \mu}\Gamma^\mu_{\beta\gamma}T^{\alpha\beta} - g_{\beta\mu}\Gamma^\mu_{\lambda\gamma}T^{\alpha\beta} + g_{\lambda\beta}\Gamma^\alpha_{\mu\gamma}T^{\mu\beta}+g_{\lambda\beta}\Gamma^\beta_{\mu\gamma}T^{\alpha\mu}
\end{align}
The second term cancels the last one, so
$$
(T^{\alpha}_\lambda)_{;\gamma}
=
(g_{\lambda\beta}T^{\alpha\beta})_{,\gamma}  - g_{\beta\mu}\Gamma^\mu_{\lambda\gamma}T^{\alpha\beta} + g_{\lambda\beta}\Gamma^\alpha_{\mu\gamma}T^{\mu\beta}
$$
i.e.
$$
(T^{\alpha}_\lambda)_{;\gamma}
=
T^\alpha_{\lambda,\gamma}  - \Gamma^\mu_{\lambda\gamma}T^{\alpha}_{\mu} + \Gamma^\alpha_{\mu\gamma}T^{\mu}_{\lambda}
$$
as you wanted.
The problem is to prove the formula
$$g_{\lambda\beta,\gamma}=g_{\lambda \mu}\Gamma^\mu_{\beta\gamma}+g_{\beta\mu}\Gamma^\mu_{\lambda\gamma}.$$
This can be done directly, using the formula
$$g_{\lambda \mu}\Gamma^\mu_{\beta\gamma}
= \frac 1 2 (g_{\lambda\beta,\gamma}+g_{\lambda\gamma,\beta}-g_{\beta\gamma,\lambda}).$$
