# 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(?).

• I don't think it's common to lower the indices of $\Gamma$. But of course, it ought to be valid. Not being a tensor is no excuse. Lowering indices is just short-hand for multiplying with some $g$, after all. – Arthur Jul 13 '18 at 8:18
• Then I guess there must be some relation between $\Gamma^\beta_{\mu\gamma}$ and $\Gamma_{\beta\mu\gamma}$ to make the covariant derivative work. The actual answer is, $(T^{\alpha}_\lambda)_{;\gamma} = T^{\alpha}_{\lambda,\gamma} + \Gamma^\alpha_{\gamma\delta}T^{\delta}_{\lambda}-\Gamma^\delta_{\lambda\gamma}T^{\alpha}_{\delta}$ and what I got with contraction is, $(T^{\alpha}_\lambda)_{;\gamma} = T^{\alpha}_{\lambda,\gamma} + \Gamma^\alpha_{\mu\gamma}T^{\mu}_{\lambda}+\Gamma_{\lambda\mu\gamma}T^{\alpha\mu})$ – Shaz Jul 13 '18 at 8:39