Tensor notation and rules I have a few questions about tensors:


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*I appreciate that $g^{\alpha\beta}=g^{\beta\alpha}$ but when contracting say $T^{\sigma}_{\mbox{    }\;\mu\nu\rho}$ to $T_{\;\;\mu\nu}$, first of all can it be written as $T_{\mu\nu}$ or does the gap need to be there? Secondly to contract one should multiply by $g^{\gamma\rho}g_{\sigma\gamma}$. Does the ordering of indices in anyway matter, e.g. should it be $g^{\rho\gamma}g_{\sigma\gamma}$ or $g_{\sigma\gamma}g^{\gamma\rho}$?

*When contracting to a scalar, e.g. $g^{\mu\nu}R_{\mu\nu}=R^{\mu}_{\mu}$, is the spacing ok or should it be $R^{\mu}_{\;\mu}$ and am I correct in thinking this is then equal to $1R$ as opposed to $4R$?
 A: 1) The contraction of $T^{\sigma}_{\mbox{    }\;\mu\nu\rho}$ is $T_{\mu\nu}$, (no gap).
There are 4 ways of contracting $g^{\delta\gamma}g_{\rho\sigma}$. Sum on the 1st indices,  sum on the 2nd indices, sum on 1st and 2nd indices, and sum on the 2nd and 1st indices. They are all different operations.
2) When contracting to a scalar the spacing does not matter: $g^{\mu\nu}R_{\mu\nu} = R^{\mu}_{\mu}$ = $R^{\mu}_{\;\mu}$ = $R_{\mu}^{\;\mu}$. Some authors & software consider the spacing to be significant, while others do not.
Mathematicians & theorists working with abstract tensors will often define $R^{\sigma}_{\mu}$ = $R^{\sigma}_{\;\mu}$ = $R_{\mu}^{\;\sigma}$, while computer scientists & engineers working with tensor representations often consider $R^{\sigma}_{\;\mu}$ to be the transpose of $R_{\mu}^{\;\sigma}$. 
If the author writes $R_\mu^\sigma$ or $\Gamma^{\delta}_{\gamma\rho\sigma}$ with the indices left justified, it is safe to assume that spacing is not meaningful. If the author writes $R_{\;\mu}^{\sigma}$ or $\Gamma^{\delta}_{\;\gamma\rho\sigma}$ it usually reasonable to assume that the spacing is intended to be meaningful.
If spacing is not meaningful, the order of the two tensors does not matter: $a^{\gamma\rho}b_{\sigma\gamma} = b_{\sigma\gamma}a^{\gamma\rho} = c^{\rho}_\sigma $.
If spacing is meaningful, the order of the two tensors does matter: $a^{\gamma\rho}b_{\sigma\gamma} = c^{\rho}_{\;\sigma} $, but $ b_{\sigma\gamma}a^{\gamma\rho} = c^{\;\rho}_{\sigma} $.
A: Thanks, I am stuck trying to contract $\delta R_{\mu\nu}$ to $\delta R^{\sigma}_{\mu\nu\rho}$ where 
$$\delta R^{\sigma}_{\mu\nu\rho}  =\delta\partial_{\nu}\Gamma^{\sigma}_{\mbox{ }\mu\rho}-\delta\partial_{\rho}\Gamma^{\sigma}_{\mbox{ }\mu\nu}$$ I can't show that it should become: $$\delta R_{\mu\nu}=[\partial_{\nu}(\delta\Gamma^{\sigma}_{\mu \sigma})-\partial_{\sigma}(\delta \Gamma^{\sigma}_{\mu\nu})]$$ MY ATTEMPT:
I thought to contract it by multiplying by  $g_{\sigma\alpha}g^{\alpha \rho}$ as$g_{\sigma\alpha}g^{\alpha\rho}\delta R^{\sigma}_{\mu\nu\rho}=\delta R_{\mu \nu}$ but then I get:$g_{\sigma \alpha}g^{\alpha \rho}(\delta\partial_{\nu}\Gamma^{\sigma}_{\mbox{ }\mu\rho}-\delta\partial_{\rho}\Gamma^{\sigma}_{\mbox{ }\mu\nu})=(\delta\partial_{\nu}\Gamma_{\mu\rho}-\delta \partial_{\rho}\Gamma^{\rho}_{\mbox{ }\mu\nu})$ 
